A Single stage transistor amplifier has one transistor, bias circuit and other auxiliary components. The following circuit diagram shows how a single stage transistor amplifier looks like. When a weak input signal is given to the base of the transistor as shown in the figure, a small amount of base current flows.

1. Electronic Devices and Circuits
UNIT – I
SVEC19

2. The High-Pass RC Circuit
o
V (t) = 0; thengain A= 𝑜
𝑉 (𝑡)
𝑉𝑖(𝑡)
= 0.
decreases, then
When f increases, XC
outputand gain increases.
When f= ∞; XC = 0 (Capacitor is short
circuited);
Thecircuitwhich transmits only high- frequencysignals and attenuatesor
stops low frequencysignals.
1
𝑋𝐶 =
2𝜋𝑓𝐶
When f= 0; XC = ∞ (Capacitor is open
circuited);
M.Balaji, Departmentof ECE, SVEC 2

3. Sinusoidal input
Laplace transformed High- Pass RC circuit frequency response
M.Balaji, Departmentof ECE, SVEC 3

4. 𝑉𝑂 𝑠
𝐴 =
𝑉𝑖 𝑠
=
𝑅
𝑅 + 1
𝐶𝑠
=
1
1 + 1
𝑅𝐶𝑠
Putting s= j𝜔,
𝐴 =
1
1
=
1 − 𝑗 𝜔𝑅𝐶
1
1
1 − 𝑗 2𝜋𝑓𝑅𝐶
∴ 𝐴 =
1
1 +
1
2𝜋𝑓𝑅𝐶
2
𝑎𝑛𝑑 𝜃 = −𝑡𝑎𝑛−1
1
2𝜋𝑓𝑅𝐶
M.Balaji, Departmentof ECE, SVEC 4

5. 𝐴𝑡 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑡𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑓1, 𝐴 =
1
2
1 +
1
2𝜋𝑓1𝑅𝐶
2
=
1 1
2
Squaring on both sides and equating the denominators,
1
2𝜋𝑓1𝑅𝐶
= 1
1
∴ 𝑓1 =
2𝜋𝑅𝐶
M.Balaji, Departmentof ECE, SVEC 5

6. Step Voltage input
t =
0-
t =
0+
t < 0 t > 0
t = 0
M.Balaji, Departmentof ECE, SVEC 6
A step signal is one which maintains the value
zero for all times t < 0; and maintains the value
V forall times t > 0.
The transition between the two voltage
levels takes placeat t = 0.
i
V = 0, immediately before t = 0 (to be
referred to as time t= 0-) and Vi = V,
immediately after t= 0 (to be referred to as
time t= 0+).

7. Step input
Step input
M.Balaji, Departmentof ECE, SVEC 7
Step response fordifferent timeconstants
𝑉𝑓 − 𝑉𝑖 𝑒−𝑡/𝑅𝐶
𝑉0 𝑡 = 𝑉𝑓 −
When Vf = 0
V 𝑒−𝑡/𝑅𝐶

8. Pulse input
⚫The pulse is equivalenttoa positivestep followed bya delayed negativestep.
Pulse waveform Pulsewaveform in terms of step
M.Balaji, Departmentof ECE, SVEC 8

9. RC >> tp
M.Balaji, Departmentof ECE, SVEC 9

10. RC comparable to tp
M.Balaji, Departmentof ECE, SVEC 10

11. RC << tp
M.Balaji, Departmentof ECE, SVEC 11

12. (a) RC >> tp
M.Balaji, Departmentof ECE, SVEC 12
(b) RC comparable to tp (c) RC << tp

13. Square wave input
⚫A square wave is a periodic waveform which maintains itself at one constant
level V’ with respect to ground for a time T1, and then changes abruptly to
another level V’’, and remains constant at that level for a time T2, and repeats
itself at regular intervalsof timeT1+T2.
⚫ A squarewave may be treated as a series of positiveand negative pulses.
M.Balaji, Departmentof ECE, SVEC 13

14. M.Balaji, Departmentof ECE, SVEC
14
(a) Square wave input (b) When RC is arbitrarily large

15. (c) RC > T
M.Balaji, Departmentof ECE, SVEC 15

16. (d) Outputwhen RC comparable toT
M.Balaji, Departmentof ECE, SVEC 16

17. (e) Outputwhen RC << T
M.Balaji, Departmentof ECE, SVEC 17

18. Under steady state conditions, the capacitor charges and discharges to the
same voltage levels in each cycle. So, the shape of the output waveform is
fixed.
𝑓𝑜𝑟 0 < 𝑡 < 𝑇1, theoutput isgiven by 𝑣01 = 𝑉1𝑒−𝑡/𝑅𝐶
Att=T1 ; 𝑣01 = 𝑉1𝑒−𝑇1/𝑅𝐶
𝑓𝑜𝑟 𝑇1 < 𝑡 < 𝑇1 + 𝑇2, theoutput isgiven by 𝑣02 = 𝑉2𝑒−(𝑡−𝑇1)/𝑅𝐶
At t=T1 +T2, 𝑣02 = 𝑉2𝑒−𝑇2/𝑅𝐶
Also 𝑉′ − 𝑉2= 𝑉 𝑎𝑛𝑑 𝑉1 − 𝑉′ = 𝑉
1 2
M.Balaji, Departmentof ECE, SVEC 18

19. Expression for the percentage tilt
The expression for the percentage tiltcan be derived when the timeconstant RC
of thecircuit isvery largecompared to the period of the inputwaveform, i.e RC
>> T.
Fora symmetrical squarewavewith zero averagevalue
𝑉1 = −𝑉2,𝑖. 𝑒 𝑉1 = 𝑉2
2
𝑉′
𝑉′ = −𝑉′, 𝑖. 𝑒 𝑉′ =
1 2 1
𝑇
𝑇1 = −𝑇2 =
2
M.Balaji, Departmentof ECE, SVEC 19

20. The output waveform for RC >> T is
M.Balaji, Departmentof ECE, SVEC 20

21. 𝑉′ = 𝑉 𝑒−𝑇/2𝑅𝐶
1 1
𝑎𝑛𝑑 𝑉′ = 𝑉 𝑒−𝑇/2𝑅𝐶
2 2
2
𝑉1 − 𝑉′ = 𝑉
𝑉1− 𝑉2𝑒−𝑇/2𝑅𝐶 = 𝑉1 + 𝑉1𝑒−𝑇/2𝑅𝐶 = 𝑉
𝑉
𝑉1 =
1 + 𝑒−𝑇/2𝑅𝐶
𝑉 = 𝑉1(1 + 𝑒−𝑇/2𝑅𝐶)
% 𝑇𝑖𝑙𝑡, 𝑃 =
𝑉1 − 𝑉′
𝑉
2
1
× 100
=
𝑉1− 𝑉1𝑒−𝑇/2𝑅𝐶
1
𝑉 (1 + 𝑒−𝑇/2𝑅𝐶 )
× 200%
1 − 𝑒−𝑇/2𝑅𝐶
= × 200%
1 + 𝑒−𝑇/2𝑅𝐶
M.Balaji, Departmentof ECE, SVEC 21

22. When the time constant is very large, i.e,
𝑇
𝑅𝐶
≪ 1
1 − 1 +
−𝑇 −𝑇 1
+ ( )2 + ⋯
1 + 1 +
−𝑇
2𝑅𝐶 2𝑅𝐶 2!
−𝑇 1
+ ( )2 + ⋯
𝑃 = 2𝑅𝐶 2𝑅𝐶 2! × 200%
𝑇
= 2𝑅𝐶 × 200%
=
2
𝑇
2𝑅𝐶
× 100%
𝑓
𝜋𝑓1
× 100%
1
M.Balaji, Departmentof ECE, SVEC 22
Where 𝑓1 = 2𝜋𝑅𝐶
𝑖𝑠 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑢𝑡 − 𝑜𝑓𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ − 𝑝𝑎𝑠𝑠 𝑐𝑖𝑟𝑐𝑢𝑖𝑡

23. Ramp input
𝑡 = 𝛼𝑡,
When a high- pass RC circuit isexcited with a ramp input, i.e 𝑣𝑖
𝑤ℎ𝑒𝑟𝑒 𝛼 is theslopeof the ramp
then, 𝑉𝑖 𝑠 =
𝛼
𝑠2
𝑉0 𝑠 = 𝑉𝑖 𝑠
From the Laplace transformed high-pass circuit, we get
𝑅
1
𝑅 + 𝐶𝑠
𝛼 𝑅𝐶𝑠
=
𝑠2
=
1 + 𝑅𝐶𝑠
𝛼
1
𝑠(𝑠 + 𝑅𝐶)
1
𝛼𝑅𝐶
𝑠
−
1
1
𝑠 + 𝑅𝐶
M.Balaji, Departmentof ECE, SVEC 23

24. Taking inverse Laplace transform on both sides, weget
𝑉0 𝑡 = 𝛼𝑅𝐶 1 − 𝑒−𝑡/𝑅𝐶
Fortime twhich areverysmall in comparison with RC, we have
𝑉0 𝑡 = 𝛼𝑅𝐶 1 − 1 + +
−𝑡 −𝑡
𝑅𝐶 𝑅𝐶
−𝑡
2 1 3 1
+ + ⋯
2! 𝑅𝐶 3!
= 𝛼𝑅𝐶
𝑡 𝑡2
− + ⋯
𝑅𝐶 2(𝑅𝐶)2
= 𝛼𝑡 −
𝛼𝑡2
2𝑅𝐶
= 𝛼𝑡 1 −
𝑡
M.Balaji, Departmentof ECE, SVEC 24
2𝑅𝐶

25. The above figures shows the response of the high-pass circuit for a ramp
input when (a) RC >> T and (b) RC << T, where T is the duration of the
ramp.
For small values of T,the output signal falls away slightly from the input.
M.Balaji, Departmentof ECE, SVEC 25

26. High-Pass RC circuit as a differentiator
M.Balaji, Departmentof ECE, SVEC 26
⚫Sometimes, a square wave may need to be converted into
sharp positive and negative spikes (pulses of short
duration).
⚫By eliminating the positive spikes, we can generate a train
of negative spikes and vice-versa.
⚫The pulses so generated may be used to trigger a
multivibrator.
⚫If in a circuit, the output is a differential of the input
signal, then thecircuit is called adifferentiator.

27. 𝑟 = 𝑅𝐶 ≪ 𝑇 then thecircuitworksas adifferentiator
1
1/f ; here frequency must besmall.
At low frequencies, 𝑋𝑐 = 2𝜋𝑓𝐶
= 𝑙𝑎𝑟𝑔𝑒 𝑤ℎ𝑒𝑛 𝑐𝑜𝑚𝑝𝑎𝑟𝑒𝑑 𝑡𝑜 𝑅
Thevoltagedropacross R is verysmall when compared tothedropacross C.
1
𝑉𝑖 =
𝐶
∫ 𝑖(𝑡)𝑑𝑡 + 𝑖 𝑡 𝑅
𝑖
𝐶
1
𝑉 = ∫ 𝑖(𝑡)𝑑𝑡
𝑖
𝑉 =
1
0
∫ 𝑉 𝑑𝑡
𝑅𝐶 𝑅
(𝑖 𝑡 = 𝑉0
)
Differentiating on both sides weget
𝑑𝑉𝑖
=
1
𝑑𝑡 𝑅𝐶
𝑉0
𝑉0 = 𝑅
𝑑𝑉
𝑖
𝐶 𝑑𝑡
Thus , theoutput is proportional to thederivativeof the input.
V0 issmall
M.Balaji, Departmentof ECE, SVEC 27

28. Numerical Problem 1
A 1KHz symmetrical square wave of ±10𝑉 is applied to an RC circuit having 1ms
time constant. Calculate and plot the output for the RC configuration as a High-
Pass RC circuit.
Solution: Given 𝑓 = 1𝐾𝐻𝑧, 𝑇 = 1𝑚𝑠
∴ 𝑇𝑂𝑁 = 0.5𝑚𝑠 𝑎𝑛𝑑 𝑇𝑂𝐹𝐹 = 0.5𝑚𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐𝑎𝑙 𝑠𝑞𝑢𝑎𝑟𝑒 𝑤𝑎𝑣𝑒
𝑟 = 𝑅𝐶 = 1𝑚𝑠
Peak-to –peak amplitudeVPP = 10- (-10)=20V
Since RC is comparable toT, thecapacitorcharges and dischargesexponentially.
M.Balaji, Departmentof ECE, SVEC 28

29. High-Pass RC circuit
M.Balaji, Departmentof ECE, SVEC 29

30. When the 1 KHz square wave shown by dotted line is applied to the RC high
pass circuit.
Under steady state conditions the output waveform will be as shown by the
thick line.
M.Balaji, Departmentof ECE, SVEC 30
Since the inputsignal is a symmetrical squarewave, we have
𝑉1 = −𝑉2 𝑎𝑛𝑑 𝑉′ = −𝑉′
1 2
1
𝑉′ = 𝑉1𝑒−𝑇/2𝑅𝐶 = 𝑉1𝑒−0.5/1 = 0.6065 𝑉1
2
𝑉′ = 𝑉2𝑒−𝑇/2𝑅𝐶 = 𝑉2𝑒−0.5/1 = 0.6065 𝑉2

31. 1
𝑉′ − (−𝑉2) = 20
0.6065 𝑉1+ 𝑉1= 20
20
∴ 𝑉1=
1.6065
= 12.449𝑉
𝑉2= −𝑉1 = −12.449𝑉
M.Balaji, Departmentof ECE, SVEC 31
1
𝑉′ = 20 − 𝑉1 = 20 − 12.449 = 7.551𝑉
𝑉′ = −𝑉′ = −7.551 𝑉
2 1

32. Numerical Problem 2
If a square wave of 5 KHz is applied to an RC high-pass circuit and the resultant
waveform measured on a CRO was tilted from 15V to 10V, find out the lower 3-
dB frequency of the high-pass circuit.
Sol: f= 5 KHz
𝑉1= 15V
𝑉′ = 10𝑉
1
𝑓1 = ? 𝑙𝑜𝑤𝑒𝑟 3 𝑑𝐵 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
M.Balaji, Departmentof ECE, SVEC 32

33. Peak-to-peak value of input 𝑉 = 𝑉1+ 𝑉2
= 15 + 15 = 30𝑉
𝑃 =
𝑉1− 𝑉′
2
1
× 100
=
15 − 10
2
× 100 = 33.33%
𝑓
Also% tilt 𝑃 = 𝜋𝑓1
× 100
5 × 103
3.14𝑓1
33.33 = × 100
𝑓1 =
33.33 × 5 × 103
3.14 × 100
= 530.56𝐻𝑧
M.Balaji, Departmentof ECE, SVEC 33