Transistors
EL 102 / 5:00 – 8:00
Electronics Laboratory
Computer Engineering IV
Bipolar Transistors
Two PN junctions joined together
Two types available – NPN and PNP
The regions (from top to bottom) ar...
Operation
 Begin by reverse biasing the CB
junction
 Here we are showing an NPN transistor
as an example
 Now we apply ...
Currents
Conventional View
Origin of the names
the Emitter 'emits' the electrons which
pass through the device
the Collector 'collects' them again ...
Original Manufacture
Base Thickness
 The thickness of the unmodified Base region
has to be just right.
 Too thin, and the Base would essentia...
Amplification Properties
The C-B voltage junction operates near
breakdown.
This ensures that a small E-B voltage causes
...
Common Base NPN
Common Emitter NPN
Common Collector NPN
How does IC vary with VCE for various IB?
Note that both dc sources are variable
Set VBB to establish...
Collector Characteristic Curve
 If VCC = 0, then IC = 0 and VCE = 0
 As VCC ↑ both VCE and IC ↑
 When VCE ≈ 0.7 V, base...
NPN Characteristic Curves
PNP Characteristic Curves
Load Line
For a constant load, stepping IB gives different currents (IC) predicted by
where the load line crosses the char...
Saturation and Cut-off
Note that the load line intersects the 75 mA curve below the
plateau region. This is saturation and...
Example
We adjust the base current to 200 µA and note
that this transistor has a β = 100
Then IC = βIB = 100(200 X 10-6
...
Example
Now adjust IB to 300 µA
Now we get IC = 30 mA
And VCE = 10 V – (30 mA)(220 Ω) = 3.4 V
Finally, adjust IB = 400...
Plot the load line
VCE IC
5.6 V 20 mA
3.4 V 30 mA
1.2 V 40 mA
Gain as a function of IC
As temperature increases, the gain increases
for all current values.
Operating Limits
There will be a limit on the dissipated power
PD(max) = VCEIC
VCE and IC were the parameters plotted o...
Example
 Assume PD(max) = 0.5 W
VCE(max) = 20 V
IC(max) = 50 mA
PD(max) VCE IC
0.5 W 5 V 100 mA
10 50
15 33
20 25
Operating Range
Operating
Range
Voltage Amplifiers
Common Base PNP
Now we have added an ac source
The biasing of the junctions are:
BE is forward biased b...
Equivalent ac Circuit
gainvoltage== V
in
out
A
V
V
rE = internal ac emitter
resistance
IE = Vin/rE (Ohm’s Law)
Vout = ICRC...
Current Gains
 Common Base
 α = IC/IE < 1
 Common Emitter
 β = IC/IB
βα
1
1
1
I
I
1
I
I
III
LawCurrentsKirchhoff'From
...
Example
If β = 50, then α = 50/51 = 0.98
Recall α < 1
Rearranging,
β = α + αβ
β(1-α) = α
β = α/(1-α)
Transistors as Switches
The operating points
We can control the base current using VBB (we
don’t actually use a physical switch). The circuit
then...
Details
In Cut-off
All currents are zero and VCE = VCC
In Saturation
IB big enough to produce IC(sat) ≈ βIB
Using Kir...
Example
a) What is VCE when Vin = 0 V?
Ans. VCE = VCC = 10 V
b) What minimum value of IB is
required to saturate the trans...
Example
LED
If a square wave is input for VBB,
then the LED will be on when the
input is high, and off when the
input is l...
Transistors with ac Input
Assume that β is such that
IC varies between 20 and 40
mA. The transistor is
constantly changing...
Pt. A corresponds to the positive peak. Pt. B
corresponds to the negative peak. This graph shows
ideal operation.
Distortion
The location of the point Q (size of the dc
source on input) may cause an operating
point to lie outside of th...
Base Biasing
 It is usually not necessary to provide two
sources for biasing the transistor.
The red arrows follow the ba...
Base Biasing
 Use Kirchhoff’s Voltage Law on the black
arrowed loop of the circuit
VCC = ICRC + VCE
So, VCE = VCC – ICRC
...
Example
 Let RC = 560 Ω @ 25 °C β = 100
RB = 100 kΩ @ 75 °C β = 150
VCC = +12 V
mA11.3A)(100)(113II BC === µβ
V5.67
)A)(5...
Transistor Amplifiers
Amplification
The process of increasing the strength of a
signal.
The result of controlling a rel...
Class A
Entire input waveform is faithfully
reproduced.
Transistor spends its entire time in the
active mode
Never reac...
Class A
Class B
No DC bias voltage
The transistor spends half its time in active
mode and the other half in cutoff
Push-pull Pair
Transistor Q1 "pushes" (drives the output voltage in a positive direction with
respect to ground), while tr...
Class AB
Between Class A (100% operation) and
Class B (50% operation).
Class C
IC flows for less than half then cycle. Usually get
more gain in Class B and C, but more distortion
Common Emitter Transistor Amplifier
Notice that VBB forward biases the emitter-base junction and dc
current flows through ...
Details
 At positive peak of input, VBB is adding to the
input
 Resistance in the transistor is reduced
 Current in the...
More details
As the input goes to the negative peak
Transistor resistance increases
Less current flows
Less voltage is...
PNP Common Emitter Amplifier
NPN Common Base Transistor
Amplifier
Signal that adds to VBB causes transistor current to increase
Signal that subtracts f...
PNP Common Base Amplifier
NPN Common Collector Transistor
Amplifier
Also called an Emitter Follower circuit – output on emitter is almost a replica ...
PNP Common Collector Transistor
Amplifier
Gain Factors
E
C
I
I
=α Usually given for common base amplifier
B
C
I
I
=β Usually given for common emitter amplifier
B
E
...
Gamma
 Recall from Kirchhoff’s Current Law
 IB + IC = IE
γβ1
I
I
I
I
1I
B
E
B
C
B
=+
=+÷
α
γ
α
αα
γ
α
α
α
α
β
-1
1
-1
-1...
Bringing it Together
Type Common
Base
Common
Emitter
Common
Collector
Relation
between
input/output
phase
0° 180° 0°
Volta...
Upcoming SlideShare
Loading in...5
×

Transistor 1 lecture

521

Published on

Published in: Technology, Business
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
521
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
46
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide

Transcript of "Transistor 1 lecture"

  1. 1. Transistors EL 102 / 5:00 – 8:00 Electronics Laboratory Computer Engineering IV
  2. 2. Bipolar Transistors Two PN junctions joined together Two types available – NPN and PNP The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E) Base Collector Emitter
  3. 3. Operation  Begin by reverse biasing the CB junction  Here we are showing an NPN transistor as an example  Now we apply a small forward bias on the emitter-base junction  Electrons are pushed into the base, which then quickly flow to the collector  The result is a large emitter-collector electron current (conventional current is C-E) which is maintained by a small E-B voltage  Some of the electrons pushed into the base by the forward bias E-B voltage end up depleting holes in that junction  This would eventually destroy the junction if we didn’t replenish the holes  The electrons that might do this are drawn off as a base current
  4. 4. Currents
  5. 5. Conventional View
  6. 6. Origin of the names the Emitter 'emits' the electrons which pass through the device the Collector 'collects' them again once they've passed through the Base ...and the Base?...
  7. 7. Original Manufacture
  8. 8. Base Thickness  The thickness of the unmodified Base region has to be just right.  Too thin, and the Base would essentially vanish. The Emitter and Collector would then form a continuous piece of semiconductor, so current would flow between them whatever the base potential.  Too thick, and electrons entering the Base from the Emitter wouldn't notice the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there'd be no Emitter-Collector current.
  9. 9. Amplification Properties The C-B voltage junction operates near breakdown. This ensures that a small E-B voltage causes avalanche Large current through the device
  10. 10. Common Base NPN
  11. 11. Common Emitter NPN
  12. 12. Common Collector NPN How does IC vary with VCE for various IB? Note that both dc sources are variable Set VBB to establish a certain IB
  13. 13. Collector Characteristic Curve  If VCC = 0, then IC = 0 and VCE = 0  As VCC ↑ both VCE and IC ↑  When VCE ≈ 0.7 V, base-collector becomes reverse-biased and IC reaches full value (IC = βIB)  IC ~ constant as VCE ↑. There is a slight increase of IC due to the widening of the depletion zone (BC) giving fewer holes for recombinations with e¯ in base.  Since IC = βIB, different base currents produce different IC
  14. 14. NPN Characteristic Curves
  15. 15. PNP Characteristic Curves
  16. 16. Load Line For a constant load, stepping IB gives different currents (IC) predicted by where the load line crosses the characteristic curve. IC = βIBworks so long as the load line intersects on the plateau region of the curve. Slope of the load line is 1/RL
  17. 17. Saturation and Cut-off Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = βIB doesn’t work in this region. Cut-off
  18. 18. Example We adjust the base current to 200 µA and note that this transistor has a β = 100 Then IC = βIB = 100(200 X 10-6 A) = 20 mA Notice that we can use Kirchhoff’s voltage law around the right side of the circuit VCE = VCC – ICRC = 10 V – (20 mA)(220 Ω) = 10 V – 4.4 V = 5.6 V
  19. 19. Example Now adjust IB to 300 µA Now we get IC = 30 mA And VCE = 10 V – (30 mA)(220 Ω) = 3.4 V Finally, adjust IB = 400 µA IB = 40 mA and VCE = 1.2 V
  20. 20. Plot the load line VCE IC 5.6 V 20 mA 3.4 V 30 mA 1.2 V 40 mA
  21. 21. Gain as a function of IC As temperature increases, the gain increases for all current values.
  22. 22. Operating Limits There will be a limit on the dissipated power PD(max) = VCEIC VCE and IC were the parameters plotted on the characteristic curve.  If there is a voltage limit (VCE(max)), then you can compute the IC that results  If there is a current limit (IC(max)), then you can compute the VCEthat results
  23. 23. Example  Assume PD(max) = 0.5 W VCE(max) = 20 V IC(max) = 50 mA PD(max) VCE IC 0.5 W 5 V 100 mA 10 50 15 33 20 25
  24. 24. Operating Range Operating Range
  25. 25. Voltage Amplifiers Common Base PNP Now we have added an ac source The biasing of the junctions are: BE is forward biased by VBB - thus a small resistance BC is reverse biased by VCC – and a large resistance Since IB is small, IC ≈ IE
  26. 26. Equivalent ac Circuit gainvoltage== V in out A V V rE = internal ac emitter resistance IE = Vin/rE (Ohm’s Law) Vout = ICRC ≈ IERC E C EE CE V r R rI RI A == Recall the name – transfer resistor
  27. 27. Current Gains  Common Base  α = IC/IE < 1  Common Emitter  β = IC/IB βα 1 1 1 I I 1 I I III LawCurrentsKirchhoff'From C B C E BCE += += += β β α βαβ αβαβ β β α + = += += + = 1 )1( 11
  28. 28. Example If β = 50, then α = 50/51 = 0.98 Recall α < 1 Rearranging, β = α + αβ β(1-α) = α β = α/(1-α)
  29. 29. Transistors as Switches
  30. 30. The operating points We can control the base current using VBB (we don’t actually use a physical switch). The circuit then acts as a high speed switch.
  31. 31. Details In Cut-off All currents are zero and VCE = VCC In Saturation IB big enough to produce IC(sat) ≈ βIB Using Kirchhoff’s Voltage Law through the ground loop VCC = VCE(sat) + IC(sat)RC but VCE(sat) is very small (few tenths), so IC(sat) ≈VCC/RC
  32. 32. Example a) What is VCE when Vin = 0 V? Ans. VCE = VCC = 10 V b) What minimum value of IB is required to saturate the transistor if β = 200? Take VCE(sat) = 0 V IC(sat) ≈ VCC/RC = 10 V/1000 Ω = 10 mA Then, IB = IC(sat)/β = 10 mA/200 = 0.05mA
  33. 33. Example LED If a square wave is input for VBB, then the LED will be on when the input is high, and off when the input is low.
  34. 34. Transistors with ac Input Assume that β is such that IC varies between 20 and 40 mA. The transistor is constantly changing curves along the load line.
  35. 35. Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows ideal operation.
  36. 36. Distortion The location of the point Q (size of the dc source on input) may cause an operating point to lie outside of the active range. Driven to saturation Driven into Cutoff
  37. 37. Base Biasing  It is usually not necessary to provide two sources for biasing the transistor. The red arrows follow the base-emitter part of the circuit, which contains the resistor RB. The voltage drop across RB is VCC – VBE (Kirchhoff’s Voltage Law). The base current is then… C BECC R VV − =BI and IC = βIB
  38. 38. Base Biasing  Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuit VCC = ICRC + VCE So, VCE = VCC – ICRC VCE = VCC – βIBRC  Disadvantge  β occurs in the equation for both VCE and IC  But β varies – thus so do VCE and IC  This shifts the Q-point (β-dpendent)
  39. 39. Example  Let RC = 560 Ω @ 25 °C β = 100 RB = 100 kΩ @ 75 °C β = 150 VCC = +12 V mA11.3A)(100)(113II BC === µβ V5.67 )A)(560(100)(113-V12 RIVV CBCCCE = = −= Ωµ β @ 75 °C IB is the same IC = 16.95 mA VCE = 2.51 V IC increases by 50% VCE decreases by 56% A113 100,000 V0.7-V12 I C25@ B µ Ω == − = ° B BECC R VV
  40. 40. Transistor Amplifiers Amplification The process of increasing the strength of a signal. The result of controlling a relatively large quantity of current (output) with a small quantity of current (input). Amplifier Device use to increase the current, voltage, or power of the input signal without appreciably altering the essential quality.
  41. 41. Class A Entire input waveform is faithfully reproduced. Transistor spends its entire time in the active mode Never reaches either cutoff or saturation. Drive the transistor exactly halfway between cutoff and saturation. Transistor is always on – always dissipating power – can be quite inefficient
  42. 42. Class A
  43. 43. Class B No DC bias voltage The transistor spends half its time in active mode and the other half in cutoff
  44. 44. Push-pull Pair Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground). Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform.
  45. 45. Class AB Between Class A (100% operation) and Class B (50% operation).
  46. 46. Class C IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion
  47. 47. Common Emitter Transistor Amplifier Notice that VBB forward biases the emitter-base junction and dc current flows through the circuit at all times The class of the amplifier is determined by VBB with respect to the input signal. Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease
  48. 48. Details  At positive peak of input, VBB is adding to the input  Resistance in the transistor is reduced  Current in the circuit increases  Larger current means more voltage drop across RC (VRC = IRC)  Larger voltage drop across RC leaves less voltage to be dropped across the transistor  We take the output VCE – as input increases, VCE decreases.
  49. 49. More details As the input goes to the negative peak Transistor resistance increases Less current flows Less voltage is dropped across RC More voltage can be dropped across C-E The result is a phase reversal Feature of the common emitter amplifier The closer VBB is to VCC, the larger the transistor current.
  50. 50. PNP Common Emitter Amplifier
  51. 51. NPN Common Base Transistor Amplifier Signal that adds to VBB causes transistor current to increase Signal that subtracts from VBB causes transistor current to decrease • At positive peak of input, VBB is adding to the input • Resistance in the transistor is reduced • Current in the circuit increases • Larger current means more voltage drop across RC (VRC = IRC) • Collector current increases • No phase reversal
  52. 52. PNP Common Base Amplifier
  53. 53. NPN Common Collector Transistor Amplifier Also called an Emitter Follower circuit – output on emitter is almost a replica of the input Input is across the C-B junction – this is reversed biased and the impedance is high Output is across the B-E junction – this is forward biased and the impedance is low. Current gain is high but voltage gain is low.
  54. 54. PNP Common Collector Transistor Amplifier
  55. 55. Gain Factors E C I I =α Usually given for common base amplifier B C I I =β Usually given for common emitter amplifier B E I I =γ Usually given for common collector amplifier
  56. 56. Gamma  Recall from Kirchhoff’s Current Law  IB + IC = IE γβ1 I I I I 1I B E B C B =+ =+÷ α γ α αα γ α α α α β -1 1 -1 -1 LCD -1 1 -1 sinceAnd == + =+ = Ex. For β = 100 α = β/(1+β) = 0.99 γ = 1 + β = 101
  57. 57. Bringing it Together Type Common Base Common Emitter Common Collector Relation between input/output phase 0° 180° 0° Voltage Gain High Medium Low Current Gain Low (α) Medium (β) High (γ) Power Gain Low High Medium Input Z Low Medium High Output Z High Medium Low
  1. A particular slide catching your eye?

    Clipping is a handy way to collect important slides you want to go back to later.

×