System _ Technik _ Electronic LAB _ Semiconductor Components (2023).pdf

SEMICONDUCTOR COMPONENTS
4th
Edition

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Order no.: V 0103
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0.6.2

Introduction
The present manual
SEMICONDUCTOR COMPONENTS
serves for conducting basic experiments with hps training systems, giving an insight into the properties and charac-
teristics of the most important semiconductor components.
The different subjects are divided as follows:
· General / basic principles
· Section of experiments, including the task (experiment) and the procedure of the experiment
The chapter „General“ offers a short description on the subject of the respective experiment. A detailed theoretical
description has purposely been omitted here because of the large extent of the subject.
We refer you to the text books recommended by book-stores for studying the theory and as accompanying experi-
ment material.
All tables and diagrams necessary for solving the tasks set in the experiments are included. There is an extensive
solutions section in the appendix for checking your own answers to the tasks and questions set in the experiments.
hps SystemTechnik offer several training systems for conducting the experiments, designed to suit individual re-
quirements.

List of Contents
1. Rectifier Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Effect of the p-n junction in diodes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Representation of characteristics of diodes of different semiconductor materials . . . . . . . . . . . . . . . 3
1.3 Half-wave rectifier circuit M1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Bridge rectifier circuit B2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2. Zener Diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 On-state and off-state characteristics of Zener diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 DC voltage-limiting with Zener diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Series and series-opposed circuit of Zener diodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.4 AC voltage limitation and overvoltage protection with Zener diodes . . . . . . . . . . . . . . . . . . . . . . . 21
2.5 Voltage stabilization with Zener diodes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3. Diodes with Special Properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.1 LEDs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4. Bipolar Transistors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.1 Testing the layers and the rectifying behaviour of bipolar transistors . . . . . . . . . . . . . . . . . . . . . . 31
4.2 Current distribution in the transistor and control effect of the base current . . . . . . . . . . . . . . . . . . 35
4.3 Characteristics of the transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.4 Influence of the load resistance on the transistor properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5. Unipolar Transistors (Junction Field-Effect Transistors) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
5.1 Testing the layers and the rectifying behaviour of FETs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
5.2 On-state characteristic of the gate p-n junctions for FETs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
V 0103 Contents I

5.3 Control effect of the gate with n-channel FETs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.4 Output characteristic of the FET . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
6. MOSFET (IGFET). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.1 Control effect of the gate in the self-inhibiting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6.2 Output characteristics of the self-inhibiting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
7. Unijunction Transistors (UJT). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.1 Testing the line in the unijunction transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
7.2 Switching characteristics of the unijunction transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
7.3 Control characteristics of the unijunction transistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
8. Thyristors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
8.1 Diode thyristors (DIAC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
8.2 Triode thyristors (thyristor). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
8.3 Bidirectional (TRIAC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S 1 ... S 56
Appendix. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 1
Abbreviations and symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 1
List of plug-in components. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 3
Voltage supply requirements and measuring instruments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A 5
Foil
Front view of the ELECTRONIC BOARD (Type 1018) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F 1
II Contents V 0103

1. Rectifier Diodes
1.1 Effect of the P-N Junction in Diodes
1.1.1 General
Diodes are bipolar semiconductor components and
consist of an n-conducting and a p-conducting layer.
As free charge carriers, electrons are predominant in
the n-conducting layer and holes in the p-conducting
layer. The p-n junction between the two has an inter-
nal diffusion potential which prevents the union of the
free charge carriers. The diode is thus blocked.
By applying an external voltage the blocking effect
can be increased or eliminated. The semiconductor di-
ode transmits the current in one direction and blocks it
in the other direction.
1.1.2 Experiments
r Experiment
Investigate the effect of the p-n junction of a rectifier
diode on the current flowing through in dependence
on the applied voltage and its polarity.
Procedure
· Apply the DC voltages UF listed in Table 1.1.2.1 to
the diode as shown in Fig. 1.1.2.1 (polarity 1), mea-
sure the corresponding current IF and enter the val-
ues in Table 1.1.2.1. For this purpose, use the cur-
rent error measuring.
V 0103 Rectifier Diodes 1
U =
0 ... 30 V
0
R
680 W
V
~
-
mains
polarity
1
polarity
2
UF
IF
V1 V1
Fig. 1.1.2.1
anode cathode
graphical symbol:
P N
p-n transition
Fig. 1.1.1.1
V1
^
= hps Type 9114.3

· Then reverse the polarity of the diode (polarity 2) and repeat the experiment with the voltage values given in
Table 1.1.2.2. Doing this, the preceding voltage divider RV / P must be removed and the voltage must be set
directly on the power supply unit. For this purpose, use the voltage error measuring.
Accurate measurement of the off-state current IR is only possible with a highly sensitive multimeter
(100 nA full swing).
· Transpose the measured values from the two tables into the diagram of Fig. 1.1.2.2 to plot the diode characteristic.
Question: What do you call the voltage at which the diode becomes conductive?
Answer:
2 Rectifier Diodes V 0103
UR [V] 0 2.5 5 10 15 20 25 30
IR [nA]
Tab. 1.1.2.2
UF [V] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 0.75
IF [mA]
Tab. 1.1.2.1
25
20
15
10
5
40
80
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
30
I [mA]
F
U [V]
F
U [V]
R
I [nA]
R
20 10
60
20
0
Fig. 1.1.2.2

1.2 Characteristic Curves for Diodes of Different Semiconductor Materials
1.2.1 General
The diode characteristic can be displayed in full on the oscilloscope monitor if the diode is fed with an AC voltage
whose momentary value alternates periodically between zero and its peak values.
The diode potential is applied to the inverting X-amplifier, the diode current is tapped as a proportional voltage at a
series-connected resistor and applied to the Y amplifier.
1.2.2 Experiments
r Experiment
Record the characteristics of a silicon diode, of a ger-
manium diode and of a gallium-arsenide diode (LED)
with the oscilloscope.
Procedure
· Set up the circuit as shown in Fig. 1.2.2.1 and re-
cord the voltage course at the silicon diode, the ger-
manium diode and at the gallium-arsenide diode
(LED) with the oscilloscope (X/Y representation).
Note: Since both voltages are poled oppositely in
relation to the reference point, one of the two
deflection amplifiers (e. g. X-amplifier) must invert.
When using oscilloscopes without inverting facili-
ties, a mirror-inverted image is obtained.
Attention: The generator output or the oscilloscope
inputs must be potential-free to prevent
short-circuiting of the measuring voltages through
the common ground.
· Draw the course of the curve of the three diodes
into diagram 1.2.2.2.
R
330 W
u = 6 V
(sine)
f = 50 Hz
pp
V
X
Y
G
~
1
Fig. 1.2.2.1
V1
^
= hps Types:
9114.3 (silicon diode)
9114.2 (germanium diode)
9121.10.3 (gallium-arsenide diode)

Question 1: How do the characteristic data of the in-
dividual diodes basically differ?
Answer:
Threshold voltages:
Silicon diode: . . . . . . . . . . . . . . . . . .
Germanium diode: . . . . . . . . . . . . . . . . . .
Gallium-arsenide-diode (LED): . . . . . . . . . . . . . . . . . .
Question 2: Approximately how large are the internal
differential resistances of the individual diodes with a
voltage change of 1.5 ... 4.5 mA?
Answer:
The internal differential resistances Rdiff are:
Silicon diode:
R =
U
I
=
diff
F
F
D
D
Germanium diode:
R =
U
I
=
diff
F
F
D
D
Gallium-arsenide-diode (LED):
R =
U
I
=
diff
F
F
D
D
DUF = change of on-state potential in V
DIF = change of on-state current in A
DRdiff = differential internal resistance in W
Question 3: What is the advantage of dynamic char-
acteristic recording as opposed to the static method?
Answer:
3.0
1.5
0.5
I [mA]
F
U [V]
F
U [V]
R
0 1.0 1.5 2.0
-0.5
-1.0
-1.5
-2,0
4.5
Fig. 1.2.2.2

1.3 Half-Wave Rectifier Circuit M1
1.3.1 General
In circuits in which semiconductor diodes are in-
serted a current can only flow (on-state range) when
the ap-plied current has a certain polarity. If the polar-
ity of the voltage is reversed, the off-state range of the
diode becomes effective, preventing current from flow-
ing. If circuits such as these are fed with AC voltage,
current only flows at the half-wave at which the diode
is in the on-state. The other half-wave is suppressed.
The current in the circuit only flows in one direction.
1.3.2 Experiments
r Experiment
The rectifying effect of a semiconductor diode is to be
investigated in a half-wave rectifier circuit and its prop-
erties examined with the multimeter and the oscillo-
scope.
Procedure
· Set up the circuit according to Fig. 1.3.2.1 (without
smoothing capacitor). The input voltage Uin and the
DC voltage UDC are to be measured with a
multimeter and the ratio UDC to Uin to be calculated.
G
~
S
C
R
10 k W
L
Y2 Y1
V
Uin UDC
+
U = 10 V
f = 50 Hz
(sine)
rms
V
UDC
i
Uin
U in
t
U DC
t
Fig. 1.3.1.1 Half-wave rectifier circuit M1
Fig. 1.3.2.1
V ^
= hps Type 9114.3

· Then the input voltage Uin and the DC voltage UDC
are to be recorded with the oscilloscope and their
curve shape to be entered into the diagram of Fig.
1.3.2.2.
· Evaluate the peak-to-peak value and the frequency
of the ripple voltage Urip from the oscilloscope dia-
gram (Fig. 1.3.2.2).
Note: The ripple voltage is the AC voltage share of
the pulsing DC voltage UDC.
· Subsequently connect the smoothing capacitors CS
in parallel to the load resistor RL as per Table
1.3.2.1 and repeat the measurings.
Attention: Pay attention to the polarity in electro-
lytic capacitors!
· Enter all evaluated values into Table 1.3.2.1.
· Plot the curve of input voltage Uin and of DC volt-
age UDC which results using the smoothing ca-
pacitor 10 mF on the grid of Fig. 1.3.2.3.
Fig. 1.3.2.2
Settings:
Y1 = 10 V / div.
Y2 = 5 V / div.
X = 5 ms / div.
Fig. 1.3.2.3
Settings:
Y1 = 10 V / div.
Y2 = 5 V / div.
X = 5 ms / div.
– 0 (Y2)
^
= UDC
– 0 (Y1)
^
= Uin
Half-wave rectifier circuit M1
CS [mF] without 10 100 470
Uin [V]
UDC [V]
U
U
=
in
DC
urip pp [V]
frip [Hz]
Tab. 1.3.2.1
– 0 (Y1)
^
= Uin
– 0 (Y2)
^
= UDC

· Finally reverse the polarity of the diode in the circuit
(Fig. 1.3.2.1) and oscilloscope the voltages Uin and
UDC without smoothing capacitor.
· Enter the curve shapes of the two voltages into the
diagram (Fig. 1.3.2.4).
Question 1: What is the frequency of the ripple volt-
age Urip?
Answer:
Question 2: What happens if the polarity of the di-
ode in the circuit (Fig. 1.3.2.1) is reversed?
Answer:
Question 3: At which connection of the diode is the
plus pole of the resultant DC voltage UDC?
Answer:
Question 4: What is the off-state voltage effective on
the diode with smoothing capacitor CS?
Answer:
Question 5: What effect does the smoothing capaci-
tor have on the peak-to-peak value of the ripple volt-
age?
Answer:
Fig. 1.3.2.4
Settings:
Y1 = 10 V / div.
Y2 = 5 V / div.
X = 5 ms / div.
– 0 (Y2)
^
= UDC
– 0 (Y1)
^
= Uin

1.4 Bridge Rectifier Circuit B2
1.4.1 General
The half-wave rectifier circuit only makes use of one
half-wave of the AC voltage. This has the disadvan-
tage of a low DC voltage and a high ripple.
This disadvantage is avoided with the bridge rectifier
circuit B2: the opposite half-waves are reversed in po-
larity and added to the DC voltage.
1.4.2 Experiments
r Experiment
Measure the properties of a bridge rectifier with the
oscilloscope and the multimeter.
Procedure
· Set up the circuit according to Fig. 1.4.2.1 (without
smoothing capacitor). Measure the input voltage
Uin and the DC voltage UDC with a multimeter and
calculate the ratio UDC to Uin.
~ S
R
10 kW
L
V
Uin
UDC
1
a
V2
V3 V4
b
Y
+
U = 10 V
f = 50 Hz
(sine)
rms
G C
X
Fig. 1.4.2.1
UDC
Uin
Uin
t
UDC
t
Fig. 1.4.1.1 Bridge rectifier circuit B2
V1 - V4
^
= hps Type 9114.3

· Record the input voltage Uin and the DC voltage
UDC are to be recorded with the oscilloscope and
plot the curve on the grid of Fig. 1.4.2.2.
· Evaluate the peak-to-peak value and the frequency
of the ripple voltage Urip from the oscilloscope dia-
gram (Fig. 1.4.2.2).
Note: The ripple voltage is the part of the AC volt-
age of the pulsing DC voltage UDC.
· Subsequently connect the smoothing capacitors CS
in parallel to the load resistor RL as per Table
1.4.2.1 and repeat the measurings.
Attention: Pay attention to the polarity in electro-
lytic capacitors!
· Enter all evaluated values in Table 1.4.2.1.
· Plot the curve of the input voltage Uin and of the DC
voltage UDC, which results using the smoothing ca-
pacitor 10 mF on the grid of Fig. 1.4.2.3.
Fig. 1.4.2.3
Settings:
Y = 10 V / div.
X = 5 ms / div.
Fig. 1.4.2.2
Settings:
Y = 10 V / div.
X = 5 ms / div.
– 0 (Y)
^
= UDC
– 0 (Y)
^
= UDC
Bridge rectifier circuit B2
CS [mF] without 10 100 470
Uin [V]
UDC [V]
U
U
=
in
DC
urip pp [V]
frip [Hz]
Tab. 1.4.2.1
– 0 (Y)
^
= Uin
– 0 (Y)
^
= Uin

Question 1:
What is the ratio of the DC voltage UDC to the applied effective input voltage Uin (without smoothing capacitor)?
Answer:
Question 2: Which is the frequency of the ripple voltage Urip?
Answer:

2. Zener Diodes
2.1 On-State and Off-State Characteristics of Zener Diodes
2.1.1 General
Zener diodes – called after their discoverer Carl Zener – are silicon diodes whose on-state characteristic is the
same as that of rectifier diodes.
Zener diodes differ from rectifier diodes in the relatively low breakdown voltages in the off-state or backward range
(Z-voltage). When the breakdown voltage is exceeded, the current in reverse direction rises steeply (Z-effect).
Whereas this reverse current must be prevented under all circumstances in rectifier diodes, Zener diodes are oper-
ated in reverse direction.
R =
U - U
I + I
V
op Z
Z L
Uop = applied operating voltage
UZ = Zener voltage of the used type of diode
IZ = average admissible Z-current
IL = current across load resistor RL acting parallelly
to the Zener diode RL
2.1.2 Experiments
r Experiment
Plot the characteristic of a Zener diode and determine
the Z-voltage with the oscilloscope.
Procedure
· Apply a sinusoid AC voltage of Urms = 24 V;
f = 50 Hz to the circuit (Fig. 2.1.2.1).
· Switch the oscilloscope to X/Y representation.
Note: Since the two voltages are poled oppositely
in relation to the reference point, one of the two
deflection amplifiers (e. g. the X-amplifier) must in-
vert. Oscilloscopes without inverting facilities give
mirror-inverted images.
V 0103 Zener Diodes 11
graphical symbol
Fig. 2.1.1.1
~
mains
~
U = 24 V
f = 50 Hz
rms
R
680W
-X
Y
V1
V
Fig. 2.1.2.1
V1 ^
= hps Type 9114.8
The properties of Zener diodes make them suitable for
voltage-stabilising and voltage-limiting.

Attention:
The power supply unit or generator output or the
oscilloscope inputs must be potential-free to avoid
short-circuiting through the common ground.
· Enter the obtained oscilloscope image in diagram
2.1.2.2.
Question 1: What is the value of the Z-voltage UZ?
Answer:
UZ =
Question 2: What is the maximum current IZ?
Answer:
I =
U
R
Z max
R
=
Question 3: What is the value of the threshold voltage Uth?
Answer:
Uth =
12 Zener Diodes V 0103
Fig. 2.1.2.2
Settings:
Y = 10 V / div.
-X = 2 V / div.
I
0 (X)
– 0 (Y)

2.2 DC Voltage-Limiting with Zener Diodes
2.2.1 General
The steep current rise in the backward range of Zener diodes makes it possible to use the Zener diode for limiting
DC voltage.
To do this, a resistor at which the difference between the unstable input voltage and the limited output voltage
drops out is connected in series. The limited output voltage is equal to the Z-voltage and depends on the chosen
type of Zener diode.
2.2.2 Experiments
r Experiment 1
Investigate the dependence of the output voltage on
the input voltage in a limiter circuit assembled with
Zener diodes.
Procedure
· Set up the circuit according to Fig. 2.2.2.1, setting
the DC voltages Uin as per Table 2.2.2.1 one after
the other.
Measure the according output voltages with a
multimeter and enter the voltages in Table 2.2.2.1.
· Plot a graph on diagram (Fig. 2.2.2.2) showing the
dependence of output voltage Uout on input voltage
Uin.
Fig. 2.2.2.1
U =
0 ...15 V
DC V
~
-
+
-
mains Uout
Uin
1
V
R
330 W
V1 ^
= hps Type 9114.8

Uin [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Uout [V]
Tab. 2.2.2.1
1 2 3 4 5 6
0
0
7 8 9 10 11 12 13 14 15 16
1
2
3
4
5
6
7
8
9
10
11
U [V]
out
U [V]
in
Fig. 2.2.2.2

r Experiment 2
Investigate the dependence of the Z-current IZ on the
input voltage in a limiter circuit assembled with Zener
diodes.
Procedure
· Set up the circuit according to Fig. 2.2.2.3, setting
the DC voltages Uin as per Table 2.2.2.2 one after
the other.
Measure the according Z-currents IZ with a multi-
meter and enter the currents in Table 2.2.2.2.
· Plot a graph on diagram 2.2.2.4 showing the de-
pendence of the Z-current IZ on the input voltage
Uin.
Fig. 2.2.2.3
U =
0...15 V
DC
V
U out
~
-
+
-
mains
IZ
U in
1
R
330 W
V
V1 ^
= hps Type 9114.8

1 2 3 4 5 6
0
0
7 8 9 10 11 12 13 14 15 16
2
4
6
8
10
12
14
16
18
20
I [mA]
Z
U [V]
in
Fig. 2.2.2.4
Uin [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
IZ [mA]
Tab. 2.2.2.2

r Experiment 3
Measure the influence of load current IL on the
Z-current statically.
Procedure
· Set up the circuit according to Fig. 2.2.2.5 and set
the load currents IL as shown in Table 2.2.2.3 with
potentiometer P.
Note:
Resistor R1 must be increased to 1 kW, 2.2 kW,
4.7 kW and 10 kW with low load currents.
· Plot a graph on diagram (Fig. 2.2.2.6) showing the
dependence of the Z-current IZ on the load current
IL.
Fig. 2.2.2.5
U
= 15 V
DC
V
~
-
mains
IZ
R
680 W
L
1
P
1 k W
I
R
330 W
1
V
V1 ^
= hps Type 9114.8

Question 1: Under what circumstances does the output voltage remain constant in a limiter circuit with Zener
diode?
Answer:
Question 2: When does the Z-current IZ begin to flow?
Answer:
Question 3: Under what circumstances is the limiting effect maintained even under load?
Answer:
1 2 3 4 5 6
0
0
7 8 9 10 11 12 13 14 15 16
2
4
6
8
10
12
14
16
18
20
I [mA]
Z
I [mA]
L
Fig. 2.2.2.6
IL [mA] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
IZ [mA]
Tab. 2.2.2.3

2.3 Series and Series-Opposed Circuit of Zener Diodes
2.3.1 General
Zener diodes can be connected in series whereby their voltages are added. This series circuit is useful to combine
Zener diodes with positive and negative temperature coefficients in such a way that the total Z-voltage is as tem-
perature-independent as possible.
For the same reason, rectifier and Zener diodes can be connected in series (see Fig. 2.3.2.1).
2.3.2 Experiments
r Experiment
Connect different Zener and rectifier diodes in series
and series-opposed, measure their total voltages and
calculate their temperature coefficients.
Procedure
· Set up the circuit according to Fig. 2.3.2.1, connect-
ing the series connections consisting of Zener and
rectifier diodes as per Fig. 2.3.2.1 A ... C. Doing
this, measure voltage Utot with the multimeter and
enter the measured voltages into Table 2.3.2.1.
· Then calculate the temperature-dependent voltage
changes of the different circuits and enter the val-
ues into Table 2.3.2.1 using the manufacturer spec-
ifications of Table 2.3.2.2 as a basis.
U
= 20 V
DC
U tot
~
-
mains
X
Y
A B C
X
Y
V1 V1
V2
V2
V3
V
R
470 W
V2
Fig. 2.3.2.1
V1 ^
= hps Type 9114.14
V2 ^
= hps Type 9114.8
V3
^
= hps Type 9114.3

Question 1: How is the temperature-dependent voltage change DU/DJ obtained?
Answer:
Question 2: What is the value of the total temperature-dependent voltage change with the respective series circuit?
Answer:
Series circuit A: DU/DJ = (DU1/DJ) + (DU2/DJ) =
Series circuit B: DU/DJ = (DU1/DJ) + (DU2/DJ) =
Series circuit C: DU/DJ = (DU1/DJ) + (DU2/DJ) + (DU3/DJ) + (DU4/DJ) =
Question 3: In circuit C, would there be any advantage in using a single diode with UZ = 12 V for which the
manufacturer specifies a Z-voltage of between 11.4 V and 12.7 V and a Z-voltage temperature coefficient of
a × ×
= + 9 10 K
-4 -1
?
Answer:
UZ = 3.3 V
(hps Type 9114.14)
-3 10 K
-4 -1
× × -0.99 mV / K
UZ = 10 V
(hps Type 9114.8)
+8 10 K
-4 -1
× × +8 mV / K
US = 0.7 V
(hps Type 9114.3)
-3 10 K
-4 -1
× × -2.4 mV / K
Tab. 2.3.2.2
Circuit Utot [V] DU / DJ [mV / K]
A
B
C
Tab. 2.3.2.1

2.4 AC Voltage Limitation and Overvoltage Protection with Zener Diodes
2.4.1 General
As long as the voltage active on a Z diode is less than the Z-voltage, no Z-current is flowing. This does not cut in
until the value of the Z-voltage is equalled or exceed-ed.
These components therefore provide facilities for protecting other components (e. g. MOS components) against
high voltages.
To prevent a rectifying effect caused by the low threshold voltage of the on-state range, two Zener diodes are con-
nected series-opposed and thus a voltage-limiting effect which is independent of the polarity is obtained.
2.4.2 Experiments
r Experiment 1
Record the characteristic Uout = f (Uin) of two series-
opposed Zener diodes.
Procedure
· Set up the circuit according to Fig. 2.4.2.1 and ad-
just the DC voltages Uin one after the other as
shown in Table 2.4.2.1.
Measure the respective output voltages Uout with
the multimeter and enter the voltage values in Ta-
ble 2.4.2.1.
· Subsequently reverse the polarity of the input volt-
age Uin and repeat the measurings above. Enter
the respective output voltages Uout in Table 2.4.2.2.
· Show the dependence of the output voltage Uout on
the input voltage Uin by plotting a graph in the grid
of diagram 2.4.2.2.
U =
0...15 V
DC
V
-
+
-
mains
V1
2
~ U in Uout
V
R
330 W
Fig. 2.4.2.1
V1 ^
= hps Type 9114.8
V2 ^
= hps Type 9114.14

10
8
6
4
2
-4
-8
2 4 6 8 10 14 16
-6 -2
-4
-8
-10
U [V]
in
-2
-6
-10
-12
-14
U [V]
out
-16 12
-U [V]
out
-U [V]
in
Fig. 2.4.2.2
-Uin [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
-Uout [V]
Tab. 2.4.2.2
Uin [V] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Uout [V]
Tab. 2.4.2.1

r Experiment 2
Examine the voltage limiting effect of two series-
opposed Zener diodes with an oscilloscope.
Procedure
· Apply a sinusoid AC voltage Urms = 10 V;
f = 50 Hz to the circuit (Fig. 2.4.2.3) and set an in-
put voltage of Uin rms = 2 V with the potentiometer.
· Record the input voltage Uin and the output voltage
Uout with the oscilloscope and plot a graph in dia-
gram of Fig. 2.4.2.4.
· Then increase the input voltage to Uin rms = 10 V.
· Repeat the measurings above and enter the values
in the diagram of Fig. 2.4.2.5.
~
mains
~
V
V1
2
P
1 kW
Y2 Y1
Uin Uout
U = 10 V
f = 50 Hz
rms
V
R
680 W
470 W
Fig. 2.4.2.3
– 0 (Y2)
Fig. 2.4.2.4
Settings: Remarks:
X = 5 ms / div. Y1 = input voltage Uin
Y1 = 2 V / div. Y2 = output voltage Uout
Y2 = 2 V / div.
– 0 (Y1)
– 0 (Y2)
Fig. 2.4.2.5
Settings: Remarks:
X = 5 ms / div. Y1 = input voltage Uin
Y1 = 20 V / div. Y2 = output voltage Uout
Y2 = 5 V / div.
– 0 (Y1)
V1 ^
= hps Type 9114.8
V2 ^
= hps Type 9114.14

Question 1: Give a possible application for a Zener diode.
Answer:
Question 2: What is the advantage of the series-opposed connection of two Zener diodes?
Answer:

2.5 Voltage Stabilization with Zener Diodes
2.5.1 General
Zener diodes not only stabilize long-term voltage fluctuations, but also short-term fluctuations as they are
caused by the residual ripple of a rectified, presmoothed AC voltage.
2.5.2 Experiments
r Experiment
Examine the stabilizing effect of a Zener diode on a
DC voltage with pronounced ripple.
To do this, rectify an AC voltage with a rectifier diode
and slightly presmooth it with a charging capacitor.
Procedure
· Set up the circuit according to Fig. 2.5.2.1, oscillo-
scope the two voltages Uin and Uout and plot the
curve on the grid of diagram 2.5.2.2. The inputs on
the oscilloscope should be set to DC.
Attention: Pay attention to the polarity of electro-
lytic capacitors!
· Subsequently set the inputs of the oscilloscope to
AC (Y2 = 0.1 V/div.). The DC components of both
voltages are then suppressed and only the ripple
parts are displayed. Plot the graph of voltages Uin
and Uout on diagram 2.5.2.3.
~
mains
~ V2
C = 10 mF
2 Y1
Uin Uout
+
U = 24 V
f = 50 Hz
rms
1
V
Y
S
V
R
330 W
Fig. 2.5.2.1
V1 ^
= hps Type 9114.3
V2 ^
= hps Type 9114.8

Question 1: How high is the ripple voltage DUin on
the smoothing capacitor CS (measure in Fig. 2.5.2.2)?
Anwer:
DUin =
DUin = peak-to-peak value of the input voltage Uin
Question 2: How high is the ripple voltage DUout on
the Zener diode (measure in Fig. 2.5.2.3)?
Answer:
DUout =
DUout = peak-to-peak value of the output voltage
Uout
Question 3: What is the value of the smoothing fac-
tor G (absolute stabilizing factor)?
Answer:
G =
U
U
=
in
out
D
D
G = smoothing factor
Question 4: What is the value of the relative stabiliz-
ing factor S?
Answer:
S =
U U
U U
= G
U
U
=
in out
out in
out
in
D
D
×
×
×
S = relative stabilizing factor
Measure the voltages Uin and Uout with the multimeter.
– 0 (Y2)
Fig. 2.5.2.3
Settings: Remarks:
X = 5 ms / div Y1 = voltage Uin
Y1 = 5 V / div. without DC component
Y2 = 0.1 V / div. Y2 = voltage Uout
(inputs on AC) without DC component
– 0 (Y1)
Fig. 2.5.2.2
Settings: Remarks:
X = 5 ms / div. Y1 = voltage Uin
Y1 = 10 V / div. Y2 = voltage Uout
Y2 = 5 V / div.
(inputs on DC)
– 0 (Y1)
– 0 (Y2)

3. Diodes with Special Properties
General
In practice there is a number of special diodes available in addition to the standard diodes examined here:
Tunnel diode, Backward diode, Schottky diode, PIN diode, Impatt diode, Trapatt diode
Their special properties can only be investigated with expensive measuring set-ups and with special measuring
equipment which is uncommon in the training sec-tor. The following experiments therefore treat only the
light-emitting diodes (LEDs) and the variable capacitance diodes (varicaps).
3.1 LEDs
3.1.1 General
In the case of semiconductor diodes made of
intermetallic connections such as gallium-arsenide or
gallium phosphide, a part of the fed electrical energy
is converted not into heat as in other semiconductor
components, but into light beams with a much shorter
wavelength. The colour of the radiated light can be
determined by choosing appropriate materials and by
doping. It may be infrared, red, yellow, orange, green
or even blue.
3.1.2 Experiments
r Experiment 1
Record the characteristic of an LED with the oscillos-
cope.
Procedure
· Apply a sinusoid AC voltage Urms = 10 V; f = 50 Hz
to the circuit (Fig. 3.1.2.1) and oscilloscope the de-
pendence of the current on the voltage with the os-
cilloscope in X/Y representation.
· Plot the oscilloscope image on the diagram
(Fig. 3.1.2.2).
V 0103 Diodes with Special Properties 27
~
mains
~
-X
Y
LED
U = 10 V
f = 50 Hz
rms
V
V
R
330 W
Fig. 3.1.2.1
graphical symbol
Fig. 3.1.1.1
V ^
= hps Type 9121.10.3

r Experiment 2
The influence of the diode voltage UF, the diode cur-
rent IF and the polarity on the light emission of an LED
is to be examined.
Procedure
· Set up the circuit as shown in Fig. 3.1.2.3 and ad-
just the DC voltage Uin in steps according to Table
3.1.2.1.
Measure the diode voltage UF and the diode cur-
rent IF with the multimeter and find out the light
emission of the LED (none, low, middle, bright).
Enter the values to be evaluated in Table 3.1.2.1.
· Subsequently reverse the polarity of the diode and
observe the light emission.
28 Diodes with Special Properties V 0103
Fig. 3.1.2.3
~
-
mains
LED
R
330 W
V
+
-
U
0 ... 10 V
in
V IF
UF
( )
Fig. 3.1.2.2
Settings:
Y = 2 V / div.
-X = 2 V / div.
I
0 (X)
– 0 (Y)
Note:
Since the two voltages are poled oppositely in relation
to their reference points, one of the two deflection am-
plifiers of the oscilloscope (e. g. the X-amplifier) must
invert. With oscilloscopes without inverting facility a
mirror-inverted image is displayed.
Attention:
The power supply unit and the oscilloscope may not
have a common ground.
V ^
= hps Type 9121.10.3

Question 1: What is the minimum current required by the LED for weak light emission?
Answer:
Question 2: How does the light emission increase in the range between 15 mA and 20 mA?
Answer:
Question 3: How does the light emission behave when operating voltage polarity is reversed?
Answer:
Question 4: An LED is to be operated with an operating voltage of 5 V. What dropping resistance is necessary
for a current of 15 mA?
Answer:
V 0103 Diodes with Special Properties 29
Uin [V] UF [V] IF [mA] light
emission
1
2
3
4
5
6
7
8
9
10
Tab. 3.1.2.1

Notes:
30 Diodes with Special Properties V 0103

4. Bipolar Transistors
4.1 Testing the Layers and the Rectifying Behaviour of Bipolar Transistors
4.1.1 General
Transistors are three-pole semiconductor components
in which either a thin p-conducting layer is embedded
between two n-conducting layers (n-p-n transistor) or
a thin n-conducting layer between two p-conducting
layers (p-n-p transistor).
The p-n junctions between the middle layer (base)
and the two outer layers (emitter and collector) have a
rectifier effect which can be investigated as with any
rectifier diode.
4.1.2 Experiments
r Experiment
Examine the effect of the p-n junctions of an n-p-n
transistor on the current flowing through it, in relation
to the applied voltage and its polarity.
Repeat the experiment with a p-n-p transistor, and
demonstrate the basic differences between this and
the n-p-n transistor.
Procedure
· Set up the circuit as shown in Fig. 4.1.2.1 (diagram 1).
Using potentiometer P in conjunction with the
multimeter, set the voltages UF consecutively ac-
cording to Table 4.1.2.1. Measure each corre-
sponding current IF and enter the values in Table
4.1.2.1. On the diagram (Fig. 4.1.2.2), plot a graph
showing the dependence of the current IF on the
voltage UF.
V 0103 Bipolar Transistors 31
Fig. 4.1.1.1
U =
0 ... 30 V
DC
R
4.7 k W
P
1 k W
V
~
-
+
-
mains
A
B
A
B
A
B
diagram 1
diagram 3
diagram 2
diagram 4
base / emitter-
line
collector / base-
line
IF
UF (U )
R
(I )
R
V
V
( )
1
1
B
A V1
B
A
V1
B
P
P
N
circuit symbol
N
N
P
B = base
C = collector
E = emitter
B
C C
E E
circuit symbol
Fig. 4.1.2.1
V1 ^
= hps Type 9118.2 (NPN), V2 ^
= hps Type 9118.7 (PNP)
Note:
Always use the correct
measuring method:
voltage error or current
error circuit

· Set up the circuit as shown in Fig. 4.1.2.1 (diagram 2). Set the voltages UF consecutively according to Table
4.1.2.3. Measure each corresponding current IF and enter the values in Table 4.1.2.3. On the diagram
(Fig. 4.1.2.3), plot a graph showing the dependence of the current IF on the voltage UF.
· For the next set of measurements, remove potentiometer P from the circuit (Fig. 4.1.2.1) and set the voltage
directly on the power supply unit. Resistor RV should remain connected for safety reasons.
· Set up the circuit as shown in Fig. 4.1.2.1 (diagram 3). Set the voltages UR consecutively according to Table
4.1.2.2. Measure each corresponding current IF and enter the values in Table 4.1.2.2. On the diagram
(Fig. 4.1.2.2), plot a graph showing the dependence of the current IR on the voltage UR.
· Set up the circuit as shown in Fig. 4.1.2.1 (diagram 4). Set the voltages UR consecutively according to Table
4.1.2.4. Measure each corresponding current IR (multimeter with 0.1 -mA measuring range required) and enter
the values in Table 4.1.2.4. On the diagram (Fig. 4.1.2.3), plot a graph showing the dependence of the current IR
on the voltage UR.
32 Bipolar Transistors V 0103
UF [V] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 0.75* 0.76*
IF [mA]
2
4
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
4
12
I [mA]
F
U [V]
F
U [V]
R
I [mA]
R
10 2
6
1
3
8
14
1.0
2.0
0.5
1.5
0
Fig. 4.1.2.2
UR [V] 0 2 4 6 8 8.1 8.2 8.3
IR [mA]
Tab. 4.1.2.2 Diagram 3 (base/emitter line)
* Because of transistor tolerances, the voltage values UF in the vicinity of the threshold voltage may have to be specified differently.
Tab. 4.1.2.1 Diagram 1 (base/emitter line)

2.0
1.5
1.0
0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
30
I [mA]
F
U [V]
F
U [V]
R
I [nA]
R
20 5
10
15
25
35
80
60
40
20
0
UR [V] 0 5 10 15 20 25 30
IR [nA]
Tab. 4.1.2.4 Diagram 4 (collector/base line)
UF [V] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 0.75 0,8*
IF [mA]
Tab. 4.1.2.3 Diagram 2 (collector/base line)
Fig. 4.1.2.3
* Because of transistor tolerances, the voltage values UF in the vicinity of the threshold voltage may have to be specified differently.

· Then repeat the measurements with a p-n-p transistor in such a way as to demonstrate at which base/emitter
line polarity and at which collector/base line polarity the junctions of p-n-p and n-p-n transistors are conducting or
blocked. Enter the results in Table 4.1.2.5.
Question 1: What are the basic properties common to both p-n junctions of a transistor?
Answer:
Question 2: What properties differentiate the p-n junction between base and emitter from the p-n junction
between base and collector?
Answer:
Question 3: What must be taken into consideration when switching a circuit over from n-p-n transistors to
p-n-p transistors?
Answer:
Polarity N-P-N Type P-N-P Type
Base/Emitter Line
(conducting or blocked)
base + / emitter -
(polarity 1)
base - / emitter +
(polarity 3)
Collector/Base Line
(conducting or blocked)
collector - / base +
(polarity 2)
collector + / base -
(polarity 4)
Tab. 4.1.2.5

4.2 Current Distribution in the Transistor and Control Effect of the
Base Current
4.2.1 General
Charge carriers, which are accelerated from the emitter through the conducting p-n junction into the extremely thin
base zone, penetrate the opposite, blocked p-n junction between collector and base and can drain to the collector.
The base current IB is smaller by the amount of this collector current IC than the emitter current IE.
I = I - I
B E C
On the other hand, the value of the collector current is influenced by the base current. The ratio between the
two currents is known as the current gain factor B.
B =
I
I
C
B
The small signal current gain b is: b =
I
I
C
B
D
D
The wiring of a transistor’s connections with negative or positive operating voltage depends on the transistor layer-
ing. In n-p-n transistors, base and collector are positive in relation to the emitter, in p-n-p transistors negative.
4.2.2 Experiments
r Experiment 1
Examine the influence of the collector current on the
base current statically. Carry out the experiment with
an n-p-n transistor.
Procedure
· Apply a DC voltage of UDC = 20 V to the circuit
shown in Fig. 4.2.2.1. Measure the base current IB
with interrupted collector line (potentiometer re-
moved) and enter its value in Table 4.2.2.1.
· Replace the potentiometer and set the collector cur-
rent values listed in Table 4.2.2.1 Enter the corre-
sponding base current values in Table 4.2.2.1.
U = 20 V
DC
R
4.7 k W
R
1 k W
1
~
-
+
-
mains
RV
R
10W
3
I C
2
P
1 kW
IB
V
( )
Fig. 4.2.2.1
V ^
= hps Type 9118.2

· On the diagram (Fig. 4.2.2.2), plot a graph showing the dependence of the base current IB on the collector
current IC (at a constant base/emitter voltage).
IB = f (IC), UBE constant
RV [W] ¥ 1000 680 470 470 330 220
IC [mA] 0 20 25 30 40 50 60
IB [mA]
Tab. 4.2.2.1
0 50 60
2.5
I [mA]
B
I [mA]
C
40
30
20
10
0.5
1.5
2
0
1
3.5
4
3
Fig. 4.2.2.2

r Experiment 2
Examine the influence of the base current on the collector current statically. Carry out the experiment with an
n-p-n transistor.
Procedure
· Set up the circuit as shown in Fig. 4.2.2.3. Using the potentiometer, vary the base current according to the
values given in Table 4.2.2.2. Measure the corresponding collector currents IC and enter the values in Table
4.2.2.2.
· On the diagram (Fig. 4.2.2.4), plot a graph showing the dependence of the collector current on the base current.
IC = f (IB )
Fig. 4.2.2.3
U = 20 V
DC
R
4.7 kW
P
1 k W
1
~
-
+
-
mains
R2
R
10 W
3
I C
I B
V
100 W
( )
V ^
= hps Type 9118.2

Question 1: What does the characteristic (Fig.
4.2.2.4) demonstrate?
Answer:
Question 2: What is the current gain factor B
when IC = 55 mA (see Fig. 4.2.2.4)?
Answer:
B =
I
I
=
C
B
Question 3: What is the small signal current gain b
(see Fig. 4.2.2.4)?
Answer:
Small signal current gain when DIC = 40 mA - 20 mA:
b =
I
I
C
B
D
D
=
Small signal current gain when DIC = 80 mA - 70 mA:
b =
I
I
=
C
B
D
D
0.1 0.2 0.3
0
0
0.4 0.5 0.6
10
20
30
40
50
60
70
80
90
100
I [mA]
C
I [mA]
B
110
Fig. 4.2.2.4
IB [mA] 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
IC [mA]
Tab. 4.2.2.2

4.3 Characteristics of the Transistor
4.3.1 General: The transistor properties can be represented by 4 characteristics:
Input Characteristic
The input characteristic shows the base current IB dependent on the base/emitter voltage UBE (with short-circuit
output).
Output Characteristic
The output characteristic shows the dependence of the collector current IC on the collector/emitter voltage UCE at
different constant base currents.
Control Characteristic
The control characteristic provides information on the dependence of the collector current IC on the base current IB.
Feedback Characteristic
The feedback characteristic shows the dependence of the base/emitter voltages UBE, corresponding to the different
constant base currents, on the collector/emitter voltage UCE.
4.3.2 Experiments
r Experiment
Measure the electrical values of a transistor and draw
the four characteristic fields of the transistor
(four-quad-rant representation).
Procedure
· Set up the circuit according to Fig. 4.3.2.1. Using
the potentiometer, set the base currents IB consec-
utively according to Table 4.3.2.1. Measure the cor-
responding collector currents IC and enter their val-
ues in Table 4.3.2.1. Plot a graph showing the de-
pendence of the collector current on the base cur-
rent IC = f (IB) in the second quadrant of the dia-
gram (Fig. 4.3.2.5).
Note:
Considerable deviations in the measured values oc-
cur in these and the following measurements due to
self-heating of the transistor. Steps cannot be taken
to stabilize temperatures, since these would falsify
the basic characteristic curves. It is therefore advis-
able to reduce the currents to zero for approx. 30
seconds after every measurement and to read the
values quickly after resetting. Temperature-related
dispersions of measured values can be compen-
sated by interpolation when plotting the curves.
Fig. 4.3.2.1
U = 10 V
DC
R
4.7 k W
P
1 k W
1
~
-
+
-
mains
R
10 W
2
IC
I B
V
( )
V ^
= hps Type 9118.2

· Modify the circuit according to Fig. 4.3.2.2. Using the potentiometer, set the base/emitter voltages UBE consecu-
tively according to Table 4.3.2.2. Measure the corresponding base currents IB and enter their values in Table
4.3.2.2. Plot a graph showing the dependence of the base current on the base/emitter voltage IB = f(UBE) in the third
quadrant of the diagram (Fig. 4.3.2.5).
· Then modify the circuit according to Fig. 4.3.2.3. Measure the collector current IC with varying base currents IB
and collector/emitter voltages UCE, according to Table 4.3.2.3. Enter the values in Table 4.3.2.3. Plot a graph
showing the dependence of the collector current on the collector/emitter voltage IC = f (UCE) with varying base
currents in the first quadrant of the diagram (Table 4.3.2.5).
U = 10 V
DC
R
4.7 k W
P
1 k W
1
~
-
+
-
mains
R
10 W
2
U BE
I B
V
( )
Fig. 4.3.2.2
U = 15 V
DC
R
4.7 kW
P
1 k W
1
~
-
+
-
mains
R
10 W
2
U CE
I C
U = 0 ... 30 V
DC
~
-
+
mains
-
I B
V
( )
( )
R
10 kW
3
Fig. 4.3.2.3
V ^
= hps Type 9118.2
V ^
= hps Type 9118.2

· Then modify the circuit according to Fig. 4.3.2.4. Measure the base/emitter voltages UBE with varying base currents IB
and collector/emitter voltages UCE, according to Table 4.3.2.4.
Enter the values in Table 4.3.2.4. Plot a graph showing the dependence of the base/ emitter voltage on the collec-
tor/emitter voltage UBE = f (UCE) with varying base currents in the fourth quadrant of the diagram (Table 4.3.2.5).
· Finally, examine the influence of resistor R2. To do this, set a collector current of IC = 3 mA in the circuit shown
in Fig. 4.3.2.1, bypass the resistor briefly (approx. 2-3 seconds) and at the same time observe the two amme-
ters.
Question 1: What function does resistor R2 have in the emitter line?
Answer:
Question 2: With what value can collector current IC be controlled?
Answer:
U = 15 V
DC
R
4.7 k W
P
1 k W
1
-
+
-
mains
R
10 W
2
U = 0 ... 20 V
DC
-
+
mains
-
~
~
I B
U CE
V
( )
( )
U BE
R
10 k W
3
Fig. 4.3.2.4
V ^
= hps Type 9118.2

UCE = 10 V
UBE [V] 0 0.5 0.6 0.65 0.7 approx. 0.71 approx. 0.71 approx. 0.71 approx. 0.71 approx. 0.71
IB [mA]
Tab. 4.3.2.2
UCE = 10 V
IB [mA] 0 20 40 60 80 100 120 140 160
IC [mA]
Tab. 4.3.2.1
UCE [V] 2 4 6 8 10 12 14
UBE [V]
at IB = 20 mA
UBE [V]
at IB = 40 mA
UBE [V]
at IB = 60 mA
Tab. 4.3.2.4
UCE [V] 0 0.2 0.5 2 4 6 8 10 12 14
IC [mA]
at IB = 20 mA
IC [mA]
at IB = 40 mA
IC [mA]
at IB = 60 mA
IC [mA]
at IB = 80 mA
Tab. 4.3.2.3

20
2
4
6
8
30
I
[mA]
C
40
0.5
0.6
10
12
14
CE
200
160
120
80
40
B
BE
0.7
0
Fig. 4.3.2.5

4.4 Influence of the Load Resistance on the Transistor Properties
4.4.1 General
The change in collector current IC of a transistor caused by the base current IB is converted into a voltage change
UCE on a series-connected load resistor RL. The base current change IB is caused by a change in the base/emitter
voltage UBE.
The ratio of the two voltage changes gives the voltage gain of the transistor:
v =
U
U
U
CE
BE
D
D
Since the collector/emitter voltage change UCE depends on the load resistor RL. this resistor also influences the
voltage gain.
4.4.2 Experiments
r Experiment
Use experiments to examine the influence of the load
resistance on the voltage gain and the frequency
behaviour of a transistor amplifier.
Procedure
· Set up the circuit according to Fig. 4.4.2.1.
f = 1 kHz
RL = 100 W
Operate the transistor as an AC voltage amplifier.
Capacitors C1 and C2 keep DC voltage compo-
nents away from the input and output.
Note:
Before carrying out each actual measurement, ad-
just the collector rest current with the potentiometer
so that the amplitude shown on the oscilloscope
monitor is as large and sinusoidal as possible (op-
erating point setting).
· Apply AC voltages Uin (f = 1 kHz) to the input
(point A) with varying load resistance RL according
to Table 4.4.2.1, and measure the corresponding
output voltage Uout with the oscilloscope.
U = 10 V
DC
R
4.7 kW
P
1 k W
1
~
-
+
-
mains
C
1 mF
1
RL
V
to point A
u =
0 ... 1 V
pp
f = 10 Hz
... 50 kHz
A
DU = U
^
BE in
G
D U = U
^
CE out
2
C
0.22 mF
( )
Y1 Y2
R
10 W
2
Fig. 4.4.2.1
V ^
= hps Type 9118.2

· Then calculate the gain factor using the following formula:
vU =
U
U
out
in
· Enter the values in Table 4.4.2.1.
· Plot the curve of input voltage Uin and output voltage Uout (when RL = 4.7 kW) on the grid (Table 4.4.2.2).
· Plot a graph on the diagram (Fig.4.4.2.3) showing the dependence of the gain factor on the load resistance.
– 0 (Y1)
– 0 (Y2)
Fig. 4.4.2.2
Settings: Remarks:
X = 0.1 ms / div. Y1 = input voltage Uin
Y1 = 20 mV / div. Y2 = output voltage Uout
Y2 = 2 V / div.

0.1 1 10 30
R [ kW]
L
1
10
100
vU
200
300
0.2 2 20
2
20
Fig. 4.4.2.3
RL [kW] 0.1 0.47 1 4.7 10 22
U in pp [mV]* 400 200 100 40 20 20
U out pp [V]
vU
Tab. 4.4.2.1 * These voltage values must be reduced depending on the transistor type

· In the next experiment, measure the influence of the load resistance on the upper cut-off frequency. To do this,
set the output voltage when f = 1 kHz to a fixed value, which is equated to 100 %. Then increase the frequency
until the output voltage has dropped back to 70.7 %.
The frequency at which this occurs is the upper cut-off frequency. Determine the upper cut-off frequency for
each load resistance shown in Table 4.4.2.2 and enter its value in Table 4.4.2.2. Care should be taken to ensure
that the set input voltage Uin does not overdrive the transistor.
· Plot a graph to show the dependence of the upper cut-off limit on the load resistance on the diagram
(Fig. 4.4.2.4).
0.1 1 10 30
R [ kW]
L
0.1
0.2
1
2
5
10
0.5
f [MHz]
0.2 2 20
Fig. 4.4.2.4
RL [kW] 0.1 0.47 1 4.7 10 22
fup [kHz]*
Tab. 4.4.2.2 * A function generator with a frequency range between 100 kHz and 5 Mhz is required for this experiment.

Question 1: What influence does the load resistance have on the gain factor?
Answer:
Question 2: What influence does the load resistance have on the upper cut-off frequency?
Answer:
Question 3: How large is the phase shift between the input and output voltages?
Answer:

5. Unipolar Transistors (Junction Field-Effect Transistors)
5.1 Testing the Layers and the Rectifying Behaviour of FETs
5.1.1 General
With field-effect transistors (FETs), the charge-carrier
currents do not negotiate p-n junctions between dif-
ferent conductive layers but flow in a uniform channel.
This is why they are called unipolar transistors. The
charge-carrier currents are controlled by electrodes
(gates), which are isolated from the channel either by
p-n junctions (junction FET) or by crystal layers
(MOSFET).
5.1.2 Experiments
r Experiment
Investigate the properties of the p-n junctions between
the gate electrodes and the main electrodes (source
and drain) of an n-channel FET. Using the multimeter,
measure the dependence of the current on the applied
voltage. Then repeat the experiment with a p-channel
FET.
Procedure
· Set up the circuit according to Fig. 5.1.2.1 (diagram 1)
and use the multimeter to determine whether the
p-n junction is conducting or blocked.
Repeat the measurement (diagrams 2, 3 and 4).
Enter the results (conducting/blocked) in Table
5.1.2.1.
V 0103 Unipolar Transistors 49
circuit symbol circuit symbol
N
P P
P
N N
D D
G G
S S
n-channel-FET p-channel-FET
Fig. 5.1.2.1
U = 2 V
DC
R
1 k W
P
1 k W
~
-
+
-
mains
A
B
diagram 1
diagram 3
diagram 2
diagram 4
gate / source
line
drain / gate
line
n-channel type
A
B
I
( )
F (I )
R
UF (U )
R
B
A
A
B
B
A
Fig. 5.1.1.1
hps Type 9118.8

· Then replace the n-channel FET with a p-channel FET (Fig. 5.1.2.2). Determine the states of the p-n junctions by
measuring the currents (diagrams 1 to 4) and enter the results in Table 5.1.2.1.
Note:
With the Field Effect Transistor 2 N 3820 (Type 9118.20) tolerances > 100% may occur.
Question 1: When are the p-n junctions of the n-channel FET blocked?
Answer:
Question 2: When are the p-n junctions of the p-channel FET blocked?
Answer:
50 Unipolar Transistors V 0103
Diagram 1 2 3 4
N-channel type
P-channel type
Tab. 5.1.2.1
U = 2 V
DC
R
1 k W
P
1 k W
~
-
+
-
mains
A
B A
B
diagram 1
diagram 3
diagram 2
diagram 4
gate / source-
line
drain / gate-
line
p-channel type
A
B
I (I )
( )
F R
U (U )
F R
B
A
B
A
Fig. 5.1.2.2
hps Type 9118.20

5.2 On-State Characteristic of the Gate P-N Junctions for FETs
5.2.1 General
There is a rectifying effect between the gate and the channel of a junction FET. Although this is of no practical
significance, its on-state characteristic must be known in order to understand certain features of the control
behaviour of FETs.
5.2.2 Experiments
r Experiment 1
Measure and examine the on-state characteristic of
the p-n junctions between the gate and the channel
connections of a junction field-effect transistor. This
experiment is only to be carried out on an n-channel
FET. Its results are also valid for p-channel types, ex-
cept that the polarity is reversed.
Procedure
· Set up the circuit according to Fig. 5.2.2.1 (with
gate/source line). Set the voltages UF consecutively
according to Table 5.2.2.1. Measure the corre-
sponding currents IF with the multimeter and enter
their values in Table 5.2.2.1.
· Repeat the measurements with the drain/gate line
and enter the current values IF in Table 5.2.2.2.
· Plot a graph showing the on-state characteristics of
the p-n junctions IF = f (UF) on the diagram (Fig.
5.2.2.2).
U =
0 ... 30 V
DC
R
1 k W
P
1 k W
~
-
+
-
mains
A
B
gate / source
line
drain / gate
line
UF
IF
A
B
A
B
( )
Fig. 5.2.2.1
hps Type 9118.8

Question: What is the significance of the deviations between the two on-state characteristics?
Answer:
0.2 0.4 0.6
0
0
0.8 1
2
4
6
8
10
12
14
16
18
20
I [mA]
F
U [V]
F
Gate/drain line
UF [V] 0 0.2 0.4 0.6 0.7 0.75 0.8 0.85 0.9 1.0
IF [mA]
Tab. 5.2.2.2
Fig. 5.2.2.2
Gate/source line
UF [V] 0 0.2 0.4 0.6 0.7 0.75 0.8 0.85 0.9 1.0
IF [mA]
Tab. 5.2.2.1

5.3 Control Effect of the Gate with N-Channel FETs
5.3.1 General
The current flowing through the channel of the field-effect transistor (source/drain) can be controlled with the gate
potential. Unlike bipolar transistors, no power is required for this as long as the p-n junction between the gate and
the channel remains blocked.
The input characteristic or control characteristic of a FET identifies the relationship between the gate/source volt-
age UGS and the drain current ID.
This can be used to determine the grade of steepness S, which is a direct gauge for the voltage gain:
S =
I
U
D
GS
D
D
S = steepness in mA/V
DID = drain current change in mA
DUGS = gate/source voltage change in mA
5.3.2 Experiments
r Experiment
Experiment to examine the influence of the gate/
source voltage on the gate current and the drain cur-
rent. Construct the control characteristics:
IG = f (UGS); ID = f (UGS)
Procedure
· Set up the circuit according to Fig. 5.3.2.1 and ad-
just the gate/source voltage UGS in steps according
to Table 5.3.2.1.
Measure each corresponding gate current IG and
drain current ID with the multimeter and enter the
values in Table 5.3.2.1 or 5.3.2.2 as applicable.
U = +15 V
DC
~
-
+
-
mains
U = -15 V
DC
P
1 k W
R
1 k W
~
-
+
-
mains
2
R
1 kW
1
V
ID
UGS
IG
( )
Fig. 5.3.2.1
V ^
= hps Type 9118.8

· Plot a graph showing the dependence of the gate current IG on the gate/source voltage UGS on the diagram
(Fig. 5.3.2.2).
IG = f (UGS)
UGS [V] -4 -3 -2 -1 0 +0.5 +0.6 +0.7 +0.75
IG [mA]
Tab. 5.3.2.1 UDS = 15 V
0.4
0.2
1 2
0.6
I [mA]
G
0.8
U [V]
GS
-4 -3 -2 -1 0
Fig. 5.3.2.2

· Plot a graph showing the dependence of the drain current ID on the gate/source voltage UGS on the diagram
(Fig. 5.3.2.3): ID = f (UGS)
Question 1: How great is the steepness S of a FET
when the gate/source voltage change UGS = 1.5 V
and the corresponding drain current change DID =
4.5 mA?
Answer:
Question 2: When is the field-effect transistor con-
trolled without power?
Answer:
8
4
1 2
10
I [mA]
D
12
U [V]
GS
-4 -3 -2 -1
20
14
2
6
16
18
22
0
Fig. 5.3.2.3
UGS [V] -4 -3 -2 -1 0 +0.5 +0.6 +0.7 +0.75
ID [mA]
Tab. 5.3.2.2 UDS = 15 V

5.4 Output Characteristics of the FET
5.4.1 General
Semiconductor manufacturers specify for FETs not only the control characteristic but also the more significant out-
put characteristics which show the dependence of the drain current on the drain/source voltage at different con-
stant gate/source voltages.
The output characteristics are recorded without load resistance (static values). The load resistance used in practice
is drawn as a straight line in the characteristic field and provides information on the voltage gain.
5.4.2 Experiments
r Experiment 1
Statically measure the dependence of the drain cur-
rent on the drain/source voltage at different gate/
source voltages.
ID = f (UDS)
Procedure
· Set up the circuit according to Fig. 5.4.2.1. Set the
gate/source voltage UGS and the drain/source volt-
age UDS as shown in Table 5.4.2.1 and measure
the corresponding drain current ID with the
multimeter.
Reverse the polarity of the power supply unit volt-
age (15 V) to measure the drain/source voltage
when UGS = 0.5 V.
Enter the drain current values ID in Table 5.4.2.1.
U = 0 ... 30 V
DC
~
-
+
-
U = 15 V
DC
P
1 kW
R
1 kW
~
-
+
-
ID
V
mains
mains
UGS
UDS
( )
~
-
+
-
mains
R
1 kW
V2
V1
U =
15 V
DC
Fig. 5.4.2.1
V1 ^
= hps Type 9118.8
V2 ^
= hps Type 9114.3

· On the diagram (Fig. 5.4.2.2), plot a graph showing the dependence of the drain current ID on the drain/source
voltage UDS at different gate/source voltages UGS.
UDS [V] 0 0.5 1 1.5 2 3 4 6 8 10 12 14 16 18 20
ID [mA]
at UGS = -2 V
ID [mA]
at UGS = -1 V
ID [mA]
at UGS = 0 V
ID [mA]
at UGS = 0.5 V
Tab 5.4.2.1

58 Unipolare Transistoren V 0103
2
4
6
0
0
8
10
12
14
16
2
4
10
I
[mA]
D
U
[V]
DS
18
20
6
8
12
14
16
Fig. 5.4.2.2

r Experiment 2
Examine the influence of the load resistance on the
field-effect transistor.
Procedure
· Set up the circuit according to Fig. 5.4.2.3. Set the
load resistance RL and different input voltages Uin
as shown in Table 5.4.2.2, and measure the corres-
ponding output voltage Uout with the multimeter.
Enter all results in Table 5.4.2.2.
· Then use the following formula to calculate the gain v
with the different load resistances RL:
v =
D
D
U
U
out
in
DUout = Uout 1 - Uout 2
DUin = Uin 1 - Uin 2
U = 15 V
DC
~
-
+
-
mains
U = 15 V
DC
P
1 kW
R
1 kW
~
-
+
-
mains
RL
GS
in
V
1
U
= U
ID
DS
out
U
= U
( )
Fig. 5.4.2.3
V ^
= hps Type 9118.8

· On the diagram (Fig. 5.4.2.4), plot a graph showing the dependence of the voltage gain v on the load
resistance RL.
· On the diagram (Fig. 5.4.2.2), draw in the load resistance line for RL = 2 kW.
1 10
R [kW]
L
v
4
8
12
16
20
24
28
32
2 5 20 40
RL [kW] 1 2 x 1 4.7 10 22
-Uin 1 [V] 0.5 1.0 1.5 2.0 2.5
-Uin 2 [V] 1.0 1.5 2.0 2.5 3.0
Uout 1 [V]
Uout 2 [V]
DUin [V]
DUout [V]
v = DUout / DUin
Fig. 5.4.2.4
Tab. 5.4.2.2

Question 1: How does the gain factor v behave as the load resistance RL increases?
Answer:
Question 2: How large is the gain factor v when Uin = -1 V -(-2 V)? Refer to Fig. 5.4.2.2.
Answer:
Question 3: What do the negative results for Uout signify?
Answer:

Notes:

6. MOSFET (IG FET)
6.1 Control Effect of the Gate in the Self-Inhibiting MOSFET
6.1.1 General
The gate of a MOSFET (MOS field-effect transistor) is insulated from the channel by a thin layer of crystal.
It therefore has no rectifying properties, unlike the junction FET.
Electrical charge can be applied to the gate from outside to control the channel flow between source and drain.
In practice, a distinction is made between self-inhibiting types (enhancement types) and self-conducting types
(depletion types).
The following experiments should be carried out with an enhancement type.
V 0103 MOSFET (IG FET) 63
enhancement type
p-channel n-channel
depletion type
p-channel n-channel
Fig. 6.1.1.1

6.1.2 Experiments
r Experiment
Measure and investigate the control effect of the gate
on the drain current in a self-inhibiting MOSFET and
construct the control characteristic.
The properties and dependences of the p-channel
type are also fundamentally valid for n-channel types
if the operating and gate/source voltage polarities are
reversed.
Procedure
· Apply a drain/source voltage -UDS = 10 V to the cir-
cuit (Fig. 6.1.2.1). Adjust the gate/source voltage
-UGS in steps as shown in Table 6.2.2.1 and mea-
sure each corresponding drain current -ID with the
multimeter.
Enter the drain values -ID in Table 6.1.2.1.
· On the diagram (Fig. 6.1.2.2), plot a graph showing
the dependence of the drain current -ID on the
gate/source voltage -UGS. This gives the control
characteristic of the MOSFET.
ID = f (UGS)
64 MOSFET (IG FET) V 0103
Fig. 6.1.2.1
U =
10 V
DC
~
-
+
-
mains
R
-ID
P
1 kW -UGS
V
-U = 10 V
DS
( ) 2.2 kW
Attention: -UGS max = -3.2 V !
V ^
= hps Type 9118.11

-UDS = 10 V
-UGS [V] 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0
-ID [mA]
Tab. 6.1.2.1
20
2.2
-U [V]
GS
-I [mA]
D
0
1.8
0 1.4
40
60
80
2.6 3.0 3.4 3.8
100
Fig. 6.1.2.2

Question 1: When is the channel between source and drain built up?
Answer:
Question 2: What is the course of the characteristic ID = f (UGS) above a gate/source voltage of UGS = -2.7 V?
Answer:
Question 3: What is the steepness of the MOSFET when the gate/source voltage is DUGS = 0.2 V
(see Fig. 6.2.2.2: between -3.0 and -3.2 V)?
Answer:
Steepness:
S =
I
U
D
GS
D
D
=

6.2 Output Characteristics of the Self-Inhibiting MOSFET
6.2.1 General
For the MOSFET also, the output characteristics are the most informative of all the characteristics.
They show the dependence of the drain current on the drain/source voltage for different gate/source voltages.
The characteristics are recorded without load resistance (static characteristics), and after drawing in the straight
line for the load resistance they allow the user to read off the values actually arising during practical application.
6.2.2 Experiments
r Experiment 1
In an experiment, measure the dependence of the
drain current on the drain/source voltage at different
constant gate/source voltages. Then construct the
output characteristics.
Procedure
· Set up the circuit as shown in Fig. 6.2.2.1. Set the
gate/source voltages UGS and drain/source volt-
ages UDS as listed in Table 6.2.2.1 and measure
the corresponding drain current ID with the multi-
meter.
Enter the values of the drain current ID in Table
6.2.2.1.
Fig. 6.2.2.1
U = 15 V
DC
~
-
+
-
mains
R
330 W
P
10 k W
-UDS
2
P
1 k W
1
V
-ID
-UGS
( )
V ^
= hps Type 9118.11

-UDS [V] 0 0.25 0.5 1 1.5 2 2.5 3 3.5 4
-ID [mA]
at UGS = -3.0 V
-ID [mA]
at UGS = -3.1 V
-ID [mA]
at UGS = -3.2 V
-ID [mA]
at UGS = -3.3 V
-ID [mA]
at UGS = -3.4 V
Tab. 6.2.2.1
· On the diagram (Fig. 6.2.2.2), plot a graph showing the dependence of the drain current ID on the drain/source voltage
UDS for each of the gate/source voltages.

Fig. 6.2.2.2
0.5
1.0
1.5
0
0
2.0
2.5
3.0
3.5
4.0
5
25
-I
[mA]
D
-U
[V]
DS
4.5
5.0
30
35
40
10
15
20

r Experiment 2
In a further experiment, examine the influence of the
load resistance on the MOSFET.
Procedure
· Set up the circuit as shown in Fig. 6.2.2.3. With the
load resistance values RL listed in Table 6.2.2.2
and varying input voltages Uin, measure the
corresponding output voltage Uout with the multi-
meter. Enter all results in Table 6.2.2.2.
· Then use the following formula to calculate the gain
v with the different load resistance values RL:
v =
U
U
out
in
D
D
DUout = Uout 1 - Uout 2
DUin = Uin 1 - Uin 2
Fig. 6.2.2.3
U = 10 V
DC
~
-
+
-
mains
R
P
1 kW
L
V
-ID
-UDS
=
^ -Uout
UGS
=
^ Uin
( )
V ^
= hps Type 9118.11
RL [kW] 0.33 0.68 1
-Uin 1 [V] 2.7 2.7 2.7
-Uin 2 [V] 3.1 3.1 3.1
-Uout 1 [V]
-Uout 2 [V]
DUin [V]
DUA [V]
v = DUout / DUin
Tab. 6.2.2.2

· On the diagram (Fig. 6.2.2.4), plot a graph showing the dependence of the voltage gain v on the load
resistance RL.
v = f (RL)
Fig. 6.2.2.4
0
R [kW]
L
v
10
15
20
30
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
0

Question 1: How does the voltage gain v change with increasing load resistance RL?
Answer:
Question 2: How large is the voltage gain when the load resistance RL = 1 kW?
Answer:
Input voltage change:
DU = U - U =
in in 1 in 2
Output voltage change:
DU = U - U =
out out 1 out 2
Gain factor:
v =
U
U
=
out
in
D
D
Question 3: What do the negative values for Uout indicate?
Answer:

7. Unijunction Transistor (UJT)
7.1 Testing the Interbase Line in the Unijunction Transistor
7.1.1 General
This semiconductor component has two contacts on
an n-conducting crystal as base terminals and also a
p-conducting emitter layer. Accordingly, it only has
one p-n junction and is also referred to as a double-
base transistor or double-base diode. It is a compo-
nent with a negative differential resistance and, be-
cause of its switch properties, it is more like a thyristor
than a transistor.
The ohmic base terminals of the UJT do not form p-n
junctions and prevent a rectifying effect between
themselves and the rest of the crystal.
7.1.2 Experiments
r Experiment
Determine what influence the size of the current active
between the base terminals (interbase line) has on the
current and the resistance.
Procedure
· Set up the circuit as shown in Fig. 7.1.2.1. With the
interbase voltages UBB shown in Table 7.1.2.1,
measure the corresponding interbase current IBB
and enter the values in Table 7.1.2.1.
· Then calculate the corresponding resistance values
RBB and also enter in Table 7.1.2.1.
V 0103 Unijunction Transistor (UJT) 73
p-type n-type
Fig. 7.1.1.1
U =
0...30 V
DC
R
220 W
~
-
+
-
mains
E
B2
B1
V
I BB
1
UBB
( )
Fig. 7.1.2.1
V ^
= hps Type 9118.9

· On the diagram (Fig. 7.1.2.2), plot a graph showing the dependence of the interbase current IBB on the interbase
voltage UBB.
Question: What does the slight curve of the cur-rent/voltage characteristic indicate?
Answer:
74 Unijunction Transistor (UJT) V 0103
UBB [V] 4 8 12 16 20 24
IBB [mA]
RBB [kW]
Tab. 7.1.2.1
2
1
4 8 12 16
3
I [mA]
BB
4
20
U [V]
BB
0
0 24
Fig. 7.1.2.2

7.2 Switching Characteristics of the Unijunction Transistor
7.2.1 General
The properties of the unijunction transistor are specified by the characteristic field.
It shows the dependence of the emittercurrent IE on the emitter/base1voltage UEB 1 for different constant interbase
voltages UBB.
7.2.2 Experiments
r Experiment
Determine the characteristic field of the unijunction
transistor and plot a representative graph.
Procedure
· Set up the circuit as shown in Fig. 7.2.2.1. With dif-
ferent interbase voltages UBB and emitter/base 1
voltages UEB 1 as listed in Tables 7.2.2.1 to 7.2.2.5,
measure the corresponding emitter current IE with
the multimeter. Enter the emitter current values IE
in Tables 7.2.2.1 to 7.2.2.2.
the dependence of the emitter current IE on the
emitter/base 1 voltage UEB 1 for each different
interbase voltage UBB.
This gives the characteristic field (switching charac-
teristic field) of the unijunction transistor.
U =
0...30 V
DC
~
-
+
-
mains
E
R
220 W
1
R
1 kW
UEB 1
U = 15 V
DC
~
-
+
-
mains
R
470 W
2
B2
B1
V
UBB
I E
( )
Fig. 7.2.2.1
V ^
= hps Type 9118.9

UBB = 5 V
UEB 1 [V] 0 1.0 2.0 4.0 4.5 1.1 1.2 1.3
IE [mA]
Tab. 7.2.2.2
UBB = 10 V
UEB 1 [V] 0 2.0 4.0 6.0 8.5 1.2 1.3
IE [mA]
Tab. 7.2.2.3
UBB = 15 V
UEB 1 [V] 0 4.0 8.0 10.0 11.0 1.2 1.3
IE [mA]
Tab. 7.2.2.4
UBB = 0 V
UEB 1 [V] 0 0.2 0.4 0.6 0.7 0.8 0.9 1.0 1.1 1.2
IE [mA]
Tab. 7.2.2.1
UBB = 18 V
UEB 1 [V] 0 4.0 8.0 12.0 13.0 1.2 1.3
IE [mA]
Tab. 7.2.2.5

Question 1: What is the value of the on/off ratio or
the distance ratio h respectively of the UJT when
UBB = 10 V and Ubd = 8.5 V?
Answer: On/off ratio of the UJT:
h =
U - U
U
=
bd th
BB
Uth = threshold voltage of the emitter diode (0.7 V)
Ubd = breakdownvoltage
(when this voltage is exceeded, the unijunction
transistor becomes low-resistance)
UBB = interbase voltage
Question 2: How high is the resistance of the emit-
ter/base 1 line of a unijunction transistor before and
after reaching breakdown voltage (see Fig. 7.2.2.2)?
Answer:
5
0
0
10
5
10
25
I [mA]
E
U [V]
EB 1
30
15
20
15
Fig. 7.2.2.2

7.3 Control Characteristics of the Unijunction Transistor
7.3.1 General
The base 2 current IB 2 of a switched-on UJT depends on the interbase voltage UBB and also on the emitter
current IE.
This dependence can be shown by the control characteristic. However, the control effect of the emitter on the base
2 has no practical significance.
7.3.2 Experiments
r Experiment
Examine the dependence of the base 2 current on the
interbase voltage with different constant emitter cur-
rents, then construct the control characteristics.
Procedure
· Set up the circuit as shown in Fig. 7.3.2.1. Using
different interbase voltages UBB and emitter cur-
rents IE as listed in Table 7.3.2.1, measure the cor-
responding base 2 current IB 2 with the multimeter.
Enter the base 2 current values in Table 7.3.2.1.
Note: Before beginning each series of measure-
ments, increase the emitter/base 1 voltage UEB 1
until the unijunction transistor switches over to the
low-resistance state.
the dependence of the base 2 current IB 2 on the
interbase voltage UBB with different constant emit-
ter currents IE.
· On the diagram (Fig. 7.3.2.2), draw in the load re-
sistance line for an operating voltage of 12 V and a
resistance RL = 680 W.
U =
0...20 V
DC
~
-
+
-
mains
E
R
220 W
1
P
1 kW
U EB 1
U = 15 V
DC
~
-
+
-
mains
R
470 W
2
B2
B1
V
UBB
I E
I B2
( ) ( )
Fig. 7.3.2.1
V ^
= hps Type 9118.9

UBB [V] 0 2 4 6 8 10 12 14 16 18 20
IB 2 [mA]
at IE = 0 mA
IB 2 [mA]
at IE = 2 mA
IB 2 [mA]
at IE = 5 mA
IB 2 [mA]
at IE = 10 mA
4 8 12
0
0
16 20
2
4
6
8
10
12
14
16
18
20
I [mA]
B 2
U [V]
BB
22
24
6 10 14
2 18 22 24
Fig. 7.3.2.2

Question 1: Why is it pointless to exploit the control effect of the emitter current on the base 2 current for
practical purposes?
Answer:
Question 2: What value can be read off the diagram (Fig. 7.3.2.2) for the interbase voltage UBB and the base 2
current IB 2 when IE = 5 mA and RL = 680 W?
Answer:
UBB =
IB 2 =

8. Thyristors
8.1 Diode Thyristor (DIAC)
8.1.1 General
Diode thyristors, also known as DIACs (diode alternat-
ing current switches), change their switching state ac-
cording to the applied voltage U. At a certain voltage
(breakdown voltage) they flip from a state of high re-
sistance to a state of low resistance. They remain
low-resistant as long as the current does not drop be-
low the holding current IH. The change of state occurs
with both polarities of the applied voltage. DIACs are
mainly used to switch on TRIACs (triode thyristors)
which are manufactured for high currents and
voltages.
8.1.2 Experiments
r Experiment 1
Make a static record of the characteristic of a DIAC.
Procedure
· Set up the circuit as shown in Fig. 8.1.2.1. Adjust
the applied voltage in steps according to Table
8.1.2.1 and measure the corresponding current with
the multimeter. Enter the current values in Table
8.1.2.1. In order to set voltages under 15 V, remove
the 15 V voltage supply.
· Then repeat the measurements with the polarity of
the DIAC reversed and enter the current values in
Table 8.1.2.2.
V 0103 Thyristors 81
Fig. 8.1.2.1
U = 0 ... 30 V
DC
~
-
+
-
mains
U = 15 V
DC
R
2.2 k W
R
4.7 k W
~
-
+
-
mains
1
2
V
U
I
( )
Fig. 8.1.1.1
DIAC
circuit symbol
V ^
= hps Type 9114.4

· On the diagram (Fig. 8.1.2.2), plot a graph showing the dependence of the current on the voltage.
This gives the characteristic of the DIAC.
82 Thyristors V 0103
1.0
0.8
0.6
0.4
0.2
10 20 30
-I [mA]
I [mA]
-U [V]
1.0
0.8
0.6
0.4
0.2
10
20
30
U [V]
A
B
40 40
B
A
0
-U [V] 0 10.0 20.0 30.0 31.9 32.0 31.0 30.5 30.0 29.5
-I [mA]
Tab. 8.1.2.2
Fig. 8.1.2.2
+U [V] 0 10.0 20.0 30.0 31.9 32.0 31.0 30.5 30.0 29.5
+I [mA]
Tab. 8.1.2.1

r Experiment 2
Make a dynamic record of the characteristic of a DIAC using an oscilloscope.
Procedure
· Set up the circuit as shown in Fig. 8.1.2.3 (switch the oscilloscope to X/Y operation). Set the voltage with the
potentiometer so as to eliminate the multiple breakdowns which occur at higher currents.
Note:
Since both voltages are oppositely poled in relation to connection point A, one of the two deflection amplifiers
(e. g. the X-amplifier) must invert.
Please note:
There should be no conducting connection (earth) be-
tween the power supply unit and the
oscilloscope, otherwise the measuring voltage will be
short-circuited.
· Draw the characteristic displayed on the monitor in
the grid (Fig. 8.1.2.4) and compare it with the static
characteristic.
~
mains
~
U = 24 V
f = 50 Hz
rms
R
1 kW
R
2.2 k W
1
2
V
A
P
1 kW
Y
X
Fig. 8.1.2.3
Fig. 8.1.2.4
Settings: Remark:
Y = 5 V / div. X-amplifier inverted
X = 10 V / div.
I
0 (X)
– 0 (Y)
V ^
= hps Type 9114.4

Question 1: What is the voltage value of the DIAC breakdown voltage (see Fig. 8.1.2.2)?
Answer:
Question 2: How high are the resistance values of the DIAC in the high-resistance state (point A) and in the
low-resistance state (point B)? See Fig. 8.1.2.2.
Answer:
R =
Question 3: Which measures cause the DIAC to return to the high-resistance state?
Answer:

8.2 Triode Thyristor (Thyristor)
8.2.1 General
Triode thyristors (abbreviated to thyristors) have one
layer leading out as a control electrode (gate). This
enables the cathode/anode line of the component to
be „fired“, i. e. switched to the low-resistance state.
The thyristor can also be switched to the
low-resistance state by its cathode/anode voltage.
However, this method should be avoided if possible
since it may destroy the thyristor.
Once fired, the thyristor remains in the low-resistance
state even when the voltage on the control electrode
is switched off again. The cathode/anode line only
goes back to the high-resistance state, „quenching“
the thyristor, when the anode current drops below a
minimum value (holding current IH).
When the thyristor is driven in an AC circuit, firing and quenching occur as the alternating current passes through
zero. Special measures are needed to quench the thyristor in an AC circuit and these will be dealt with in the fol-
lowing experiment.
8.2.2 Experiments
r Experiment
Examine the influence of the gate/cathode voltage of
a thyristor on the gate current and the anode current.
In addition, examine procedures for quenching the
fired thyristor.
Procedure
Set up the circuit as shown in Fig. 8.2.2.1 (UGC = 0 V).
Increase the gate/cathode voltage UGC in steps ac-
cording to Table 8.2.2.1 and reduce again once firing
has taken place (lamp lights up). Measure each gate
current value IG with the multimeter. Enter the gate
current values in Table 8.2.2.1.
U = 30 V
DC
~
-
+
-
U = 15 V
DC P
1 k W
R
1 kW
~
-
+
-
IG
UGC
R
470 W
2
N
1
H
V
mains
mains
( )
S
Fig. 8.2.1.1
thyristor
circuit symbol
Fig. 8.2.2.1
V ^
= hps Type 9117.1

· On the diagram (Fig. 8.2.2.2), plot a graph showing the dependence of the gate current IG on the gate/cathode
voltage UGC.
· For the quenching experiment, open switch S (Fig. 8.2.2.1) – when UGC = 0 ... 0.5 V – whilst observing lamp H.
20
2
4
0.3
0.4
0.5
0.6
0.7
0.8
0.2
I [mA]
G
U [V]
GC
0.3 0.4 0.5 0.6 0.7 0.2
U [V]
GC
-I [mA]
G
6
8
10
not fired fired
15
10
5
0 0.1 0.1 0
Fig. 8.2.2.2
UGC [V]
before firing the thyristor after firing the thyristor
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
IG [mA]
Tab. 8.2.2.1

· Next, set up the circuit as shown in Fig. 8.2.2.3 (UGC = 0 V). Measure the anode currents IA corresponding to the
gate/cathode voltages UGC shown in Table 8.2.2.2. Enter the values in Table 8.2.2.2.
· Then set up the circuit as shown in Fig. 8.2.2.4 (UGC = 0 V). Measure the cathode currents IC corresponding to
the gate/cathode voltages UGC shown in Table 8.2.2.2. Enter the cathode current values in Table 8.2.2.2.
· On the diagram (Fig. 8.2.2.5), plot a graph showing the dependence of the anode current IA on the gate/cathode
voltage UGC.
· For the quenching experiment, close switch S (Fig. 8.2.2.3) – when UGC = 0 ... 0.5 V – whilst observing lamp H.
Fig. 8.2.2.4
U = 30 V
DC
~
-
+
-
mains
U = 15 V
DC
P
1 kW
R
1 k W
~
-
+
-
mains
R
470 W
2
1
IC
S
H
V
R
1 kW
3
C
0.22 mF
UGC
( )
Fig. 8.2.2.3
U = 30 V
DC
~
-
+
-
mains
U = 15 V
DC P
1 k W
R
1 kW
~
-
+
-
mains
R
470 W
2
1
S
H
V
R
22 W
3
UGC
IA
( )
N
V ^
= hps Type 9117.1
V ^
= hps Type 9117.1

· On the diagram (Fig. 8.2.2.5), plot a graph showing the dependence of the cathode current IC on the gate/cath-
ode voltage UGC.
· For the quenching experiment, close switch S (Fig. 8.2.2.4) – when UGC = 0 ... 0.5 V – whilst observing lamp H.
· Finally, try to fire the thyristor when the operating voltage polarity is reversed.
20
0.3
0.4
0.5
0.6
0.7
0.8
0.2
U [V]
GC
0.3 0.4 0.5 0.6 0.7 0.2
U [V]
GC
100
80
I I [mA]
C
10
40
30
70
60
50
90
0 0.1 0.1 0
A /
Fig. 8.2.2.5
UGC [V]
before firing the thyristor after firing the thyristor
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
IA [mA]
IC [mA]
Tab. 8.2.2.2

Question 1: Is the fired thyristor quenched when the gate/cathode voltage is switched off?
Answer:
Question 2: When UGC = 0 ... 0.5 V, what happens to the fired thyristor when the switch in Fig. 8.2.2.1 is
opened?
Answer:
Question 3: When UGC = 0 ... 0.5 V, how would the fired thyristor behave if a sinusoidal AC voltage
Urms = 24 V/ 50 Hz were applied to the load circuit (R2 + H + V)?
Answer:
closed?
Answer:
closed?
Answer:
Question 6: What properties does the thyristor have when it is operated with reversed-polarity voltages?
Answer:

8.3 Bidirectional Thyristor (TRIAC)
8.3.1 General
In bidirectional thyristors, also known as TRIACs (tri-
ode alternating current switches), two one-directional
thyristors (one n-controlled and one p-controlled) are
doped in such a way that their gates can be combined
and fed outwards. This means that the switching prop-
erties are the same in backward and forward directions.
Note:
The terms „cathode“ and „anode“ for the electrodes of
the TRIAC are incorrect, since they alternately trans-
mit and receive electrons depending on the polarity of
the voltage. The electrodes of a TRIAC are therefore
referred to as A1 and A2 (anode 1 and anode 2).
The TRIAC is fired by the control electrode (gate) and flips from a high-resistance to a low-resistance state.
The TRIAC remains fired until the anode current drops below a minimum value (holding current IH).
Then the TRIAC flips back to a high-resistance state and is quenched.
8.3.2 Experiments
r Experiment
Experiment to measure the influence of the gate volt-
age of a TRIAC on the gate current and the load cur-
rent. Then examine the influence of the voltage polar-
ity on these values and also determine procedures for
quenching a fired TRIAC.
Procedure
· Set up the circuit as shown in Fig. 8.3.2.1
(UG = 0 V). Increase the gate voltage UG in steps
according to Table 8.3.2.1 and reduce again once
firing has taken place (lamp lights up). Measure
each gate current IG and load current IL value with
the multimeter. Enter the gate current IG and load
current IL values in Table 8.3.2.1.
Fig. 8.3.2.1
U = 30 V
DC
~
-
+
-
mains
U = 15 V
DC P
1 kW
R
1 kW
~
-
+
-
mains
1
H
2 3
V
UG
IG
IL
( )
R
100 W
R
330 W
N
Fig. 8.3.1.1
TRIAC
circuit symbol
2
A
1
A
G
V ^
= hps Type 9117.2

· On the diagram (Fig. 8.3.2.2), plot a graph showing the dependence of the gate current IG
on the gate voltage UG.
4
2
2
4
0.4
0.5
0.6
0.7
0.8
U [V]
G
0.4 0.5 0.6 0.7
U [V]
G
6
8
0.2 0.3
6
I [mA]
G
0.2
0.3
8
0.1 0
0.1
0
-I [mA]
G
Fig. 8.3.2.2
UG [V]
before firing the TRIAC after firing the TRIAC
0 0.2 0.4 0.6 0.7 0.75 0.8 0.75 0.7 0.6 0.4 0.2 0
IG [mA]
IL [mA]
Tab. 8.3.2.1 positive operating voltage

· On the diagram (Fig. 8.3.2.3), plot a graph showing the dependence of the load current IL
· Then test to find out which of the following measures quenches the TRIAC:
- Switching off the gate voltage
- Interrupting the reference electrode (A1)
- Interrupting the load current
- Short-circuiting the line between anode 1 and anode 2.
80
20
0.4
0.5
0.6
0.7
U [V]
G
0.4 0.5 0.6 0.7 0.8
U [V]
G
0.2 0.3
100
I [mA]
L
0.2
0.3
40
60
0.1 0
0.1
0
Fig. 8.3.2.3

· Next, set up the circuit as shown in Fig. 8.3.2.4 (with reversed operating voltage polarity). Repeat the measure-
ments carried out in the first part of the experiment.
Enter the gate current IG and load current IL values in Table 8.3.2.2.
Fig. 8.3.2.4
U = 15 V
DC
~
-
-
mains
U = 30 V
DC P
1 k W
R
1 kW
~
-
-
+
mains
1
H
2 3
V
-UG
-IG
-IL
( )
R
100 W
R
330 W
+
N
V ^
= hps Type 9117.2

On the diagram (Fig. 8.3.2.5), plot a graph showing the dependence of the gate current IG
4
2
2
4
0.4
0.5
0.6
0.7
0.8
U [V]
G
0.4 0.5 0.6 0.7
U [V]
G
6
8
0.2 0.3
6
I [mA]
G
0.2
0.3
8
0.1 0
0.1
0
-I [mA]
G
Fig. 8.3.2.5
UG [V]
before firing the TRIAC after firing the TRIAC
0 0.2 0.4 0.6 0.7 0.75 0.8 0.75 0.7 0.6 0.4 0.2 0
IG [mA]
IL [mA]
Tab. 8.3.2.2 negative operating voltage

· On the diagram (Fig. 8.3.2.6), plot a graph showing the dependence of the load current IL
Question: How can the TRIAC be quenched?
Answer:
0.4
0.5
0.6
0.7
-U [V]
G
0.4 0.5 0.6 0.7
-U [V]
G
0.2 0.3 0.2
0.3
0.8
20
40
60
80
100
0.1
0 0.1 0
-I [mA]
L
Fig. 8.3.2.6

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