Analog Circuits

Tutorial

Unit 1: BJT and MOSFET Amplifiers

1. A transistor has IB = 200μA and IC = 4mA. Find (i) β of the transistor and emitter current.
(ii) If IB changes by +25μA and IC changes by +0.6mA, find the new value of β. (4M)

[SRGEC – EEE – June 2022]

Answer:

(i) β = IC / IB = 4mA / 200μA = 20

(ii) Emitter Current = IE = IC + IB = 4 mA + 200μA = 4.2 mA

(iii) β = 4.6 mA / 225 μA => β = 20.4

2. A transistor has IB = 100μA and IC = 2mA. Determine (i) β of the transistor. (ii) α of the
transistor. (iii) Emitter current. ( iv) γ of the transistor. (5M)

[SRGEC – EEE – June 2022]

Answer:

(i) β = IC / IB = 2mA / 100μA = 20

(ii) α of the transistor = α = β / β + 1 = 20 / 21 = 0.95

(iii) Emitter Current = IE = IC + IB = 2 mA + 100μA = 2.1 mA

(iii) γ of the transistor = 1 + β = 21

3. Determine IDQ, VGSQ, and VDS for the given Enhancement mode MOSFET circuit. (10M)

[SRGEC – EEE – June 2022]

Answer:

From the Drain current equation find the value of Kn/2 * (W/L)

3 x 10-3 = (Kn/2) (W/L) [(10-5)2]

=> (Kn/2) (W/L) = 0.12 x 10 -3 A/V

(i) Finding VGSQ :

Applying KVL in the input Loop: VG = VGS + VS => VGS = VG - VS

As a rule of thumb, VS = VDD / 10 = 4 V

Finding VG by applying voltage divider rule,

VG = VDD * [R2 / (R1 + R2)] = 40 * 18 x106 / 40 x 106 = 18 V

Therefore, VGSQ = 18- 4 = 14 V

(ii) Finding IDQ :

Now, applying the value of VGSQ = 14 V in Drain current equation:

IDQ = 0.12 x 10 -3 (14 – 5)2 = 9.72 mA

(iii) Finding VDS :

Applying KVL in the output Loop:

VDD – ID (RD + RS) – VDS = 0

 40 – 9.72 x 10 -3 (3.82 x 103) – VDS = 0

VDS = 40 – 37.13 = 2.87 V

4. Explain the characteristic parameters of single-stage BJT amplifiers. (5 M)

[SRGEC – ECE – Aug. 2021]

Answer:

5. A CS amplifier using an NMOS transistor for which gm = 2 mA/V is found to have an

overall voltage gain Go of -16 V/V. What value should a resistance R S inserted in the
source lead to reduce the voltage gain by a factor of 4? (5M)

: [SRGEC – ECE – Aug. 2021]

6. In the circuit shown, Vsig is a small sine wave signal with zero average. The transistor β is
100. (i) Find the value of RE to establish a dc emitter current of about 0.5 mA. (ii) Find RC to
establish a dc collector voltage of about +5 V. (iii) For R L = 10 kΩ and the transistor ro = 200
kΩ, draw the small-signal equivalent circuit of the amplifier and determine its overall voltage
gain. (10 M)

[SRGEC – ECE – Aug. 2021]

Answer:

Finding RC:

RC = (VCC - VC ) / IC = (15 – 5) / 0.495 x 10-3

RC = 20.20 KΩ
7. Design a CE amplifier to operate between a 10 KΩ source and a 2 kΩ load with a gain Vo/ Vsig
of -8 V/V. The power supply available is 9 V. Use an emitter current of approximately 2 mA and
a current of about one-tenth of that in the voltage divider that feeds the Base, with the dc voltage
at the base about one-third of the supply. The transistor available has β = 100 and V A = 100 V.
Neglect capacitive effects. (10 M)

[SRGEC – ECE – June. 2022]

Answer:
Given Vcc = 9 V ; VB = 3 V ; Vo / Vsig = -8 ; IE = 2 mA = IC (approx.)

IB = IC / β = 2 mA / 100 = 0.02 mA

Current through voltage divider: 0.2 mA ; VBE = 0.7 V

β = 100 ; VA = 100 V ; Output resistance ro = VA / IC = 100 / 2mA = 50 KΩ

Transconductance gm = IC / VT = 2 mA / 25 mV =

(i) Finding resistance R2 = VB / 0.2 mA = 3/0.2 x 10-3 = 15 KΩ

(ii) Finding resistance R1 :

Total voltage across Voltage divider = 6 V

Voltage across R1 = 9 V – 3 V = 6V

Resistance R1 = 6 V / 0.2 mA = 30 KΩ

(iii) Finding resistance RE = VE / IE

Voltage across RE = VE = Vcc / 10 (A rule of thumb) = 0.9 V

RE = VE / IE = 0.9/ 2 mA = 0.45 KΩ = 450 Ω.

(iv) Finding resistance Rc = (Vcc – VCE – VE ) / IC = (9 – 4.5 – 0.9) / 2 mA = 1.8 KΩ.

8. A CS amplifier has Cgs = 2 pF, Cgd = 0.1 pF, CL = 1 pF, gm = 5 mA/V, and Rsig = RL’ = 20 KΩ.
Find the mid-band gain AM, the input capacitance Cin using the Miller equivalence, and hence an
estimate of the 3-dB frequency fH. (4M)

[SRGEC – ECE – June. 2022]

Answer:
(i) Mid-band Gain Am = Vo / Vgs = - gm RL’ = - (5 x 10-3 * 20 x 103 ) = - 100

(ii) Input Capacitance using Miller equivalence:

Cin = Cgs + Cgd (1 + gm RL’) = 2 pF + 0.1 pF (1 + 100) = 12.1 pF

(iii) fH = 1 / 2πCin Rsig = 1 / (2 * 3.14 * 12.1x10-12 * 20x103 ) = 109 / 1519.6 = 658 KHz

9. A BJT is operated at Ic =1mA, VT=25mV with β=50. For fβ =10MHz, find fT & Cµ if Cπ= 3pF.

(6M)

[SRGEC – ECE – OCT-202]

10. A CS amplifier is to be designed to provide a 0.5 V peak output signal across a 50 kΩ load that
can be used as a drain resistor. If a gain of at least 5 V/V is needed, what gm is required? Using a
dc supply of 3 V, what values of ID and VOV would you choose? What W/L ratio is required if µn
Cox = 100 µA/V2 ? If Vt = 0.8 V, find VGS.

[SRGEC – ECE – FEB. 2022]

Soln:

Vopeak = 0.5 V ; RD = 50 KΩ; If Av = 5 ; Gm?

(i) Finding transconductance gm:

Voltage gain, Av = - gm * RD => gm = Av / RD = 5 / 50 K = 0.1 m A/ V

(ii) Finding ID and VOV:

Vcc = 3 V ; ID = ? VOV = VGS – VT = ?

(iii) µn Cox = 100 µA/V2

W/L ratio = ?

ID = µn * Cox (W/L) * VOV2

(W/L) = ID / [( µn Cox) * VOV2 ] =

(iv) Finding VGS:

Vt = 0.8 V

VOV = VGS – Vt => VGS = VOV +Vt = + 0.8

11. In the circuit shown, Vsig is a small sine-wave signal. Find Rin and the gain Vo/Vsig. Assume
β = 100. If the amplitude of the signal vbe is to be limited to 5 mV, what is the largest signal at the
input? What is the corresponding signal at the output? (10M)

[SRGEC – ECE – FEB. 2022]

Answer:

(i) Finding Rin : Rin = rπ || RB

rπ = VT / IB

IB = IC / β = IE / β = 0.1 x 10 -3 / 100 = 1 µA [Note : β = 100]

rπ = VT / IB = 25 mV / 1 µA = 25 KΩ

Rin = rπ || RB = 25 KΩ || 1 MΩ = 25 x 10 3 x 1000 x 10 3 / 1025 x 103 = 24.29 KΩ

(ii) Finding overall voltage Gain, Vo / Vsig :

Vo / Vsig = - gm * (Rin / Rin + Rsig) * (ro || Rc || RL)

We know that ro = |VA| / IC ; VA - Early Voltage is not given, hence assuming, channel
length modulation effect =0, ro = ∞, Therefore,

Vo / Vsig = - gm * (Rin / Rin + Rsig) * (Rc || RL)

gm = IC / VT = 0.1 / 25 = 4 mA / V

IE = 0.1 mA ; Ic = 0.1 mA

Rc = 20 KΩ ; RL = 20 KΩ; (Rc || RL) = 10 KΩ

Vo / Vsig = - 4 x10 -3 * (24.29 KΩ / 24.29 KΩ + 20 K) * 10 KΩ = - 21.937

(iii) Largest signal at input: when vbe = 5 mV

Vbe = Vin * (Rin / Rin + Rsig) => Vin = Vbe * (Rin + Rsig) / Rin

Vin = 5 x 10-3 * (24.29 KΩ + 20 K) / 24.29 KΩ = 9.12 mV

(iv) When vbe = 5 mV, Vo ?

Vo = - gm * Vbe (Rc || RL) = 4 mA/V * 5 mV * 10 KΩ

Vo = 200 mV