Please help with this BJT problem. 2) For the BJT amplifier shown,a) Draw the small signal circuit of the amplifier.b) Calculate the small signal input resistance (Rin) looking into the base of Q1.c) Calculate the small signal gain of the amplifier Av=vo/vi.d) What type of small signal amplifier is this CC, CE, CEWR, CB ?Beta = 100 for all transistors. VA=109 volts for Q1, VA= 5 volts for Q2 and Q3.182352.6

Answered: 2) For the BJT amplifier shown, a) Draw…

Introductory Circuit Analysis (13th Edition)

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ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Please help with this BJT problem.

2) For the BJT amplifier shown,
a) Draw the small signal circuit of the amplifier.
b) Calculate the small signal input resistance (Rin) looking into the base of Q1.
c) Calculate the small signal gain of the amplifier Av=vo/vi.
d) What type of small signal amplifier is this CC, CE, CEWR, CB ?
Beta = 100 for all transistors. VA=109 volts for Q1, VA= 5 volts for Q2 and Q3.
1823
52.6

Expert Solution

Step 1

"According to company policy we need to answer the first four subparts of a question, if you want more to get solved you should post them to a different question."

Given a BJT amplifier circuit.

We need to draw a.c equivalent circuit.

We need to calculate the small signal input resistance R_in looking into base of transistor Q1.

We also need to calculate the voltage gain of the amplifier.

Given:

V_A=109 for Q1

V_A=5 for Q1

Step 2

To determine the a.c resistance and output resistance by dc analysis:

D.C analysis:

Capacitors acts as open circuit and a.c voltages are deactivated.

Applying KVL:

$10 - I_{1} \times 1823 - V_{B E} = 0 I_{1} = \frac{10 - V_{B E}}{1823} = 5.1 mA . N o w, I_{1} = I_{C_{3}} + I_{B_{3}} + I_{B_{2}} = I_{C_{3}} + 2 I_{B_{3}} (as transistors Q2 and Q3 are identical so I_{B_{3}} = I_{B_{2}}) = I_{C_{3}} [1 + \frac{2}{β}] I_{C_{3}} = \frac{I_{1}}{[1 + \frac{2}{β}]} = 4.9 mA I_{C_{3}} = I_{C_{2}} = 4.9 mA I_{E_{1}} = I_{C_{1}} = 4.9 mA I_{B_{1}} = \frac{I_{E_{1}}}{β + 1} = \frac{4.9}{101} = 48.5 μA I_{C_{1}} = β I_{B_{1}} = 4.85 mA a.c resistance across base of transistor Q_{1}, r_{π_{1}} = \frac{V_{T}}{I_{B_{1}}} = \frac{26 m V}{48.5 μ A}, {where V}_{T} is the thermal volatge which is equal to 26 mV at T = 300 K = 0.536 k Ω Output resistance due to early voltage of transistor Q_{1}, r_{o_{1}} = \frac{V_{A}}{I_{C_{1}}} = \frac{109}{4.85 \times 10^{- 3}} = 20.33 k Ω Output resistance due to early voltage of transistor Q_{2}, r_{o_{2}} = \frac{V_{A}}{I_{C_{2}}} = \frac{5}{4.9 \times 10^{- 3}} = 1.02 k Ω$

Thus, Output ac resistance across base of transistor Q1 is $r_{π_{1}}$ 0.536 kilo-ohm

Output resistance due to early voltage of transistor Q1 is $r_{o_{1}}$ 20.33 kilo-ohm.

Output resistance due to early voltage of transistor Q2 is $r_{o_{2}}$ 1.02 kilo-ohm.

Step 3

a) To draw the a.c equivalent of the circuit:

The a.c equivalent circuit is:

Clearly from the a.c equivalent circuit the no input voltage across the base in transistor Q₂ & Q₃ so $i_{b_{2}} = 0 & i_{b_{3}} = 0$ .

So now the equivalent circuit is:

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