Question 19.17: Consider the common-emitter amplifier circuit of Fig. 19.59....

1. Define. In an initial look at this problem, it appears to be clearly stated. However, when we are asked to determine the input impedance and the voltage gain, do they refer to the transistor or the circuit? As far as the current gain and the output impedance are concerned, they are the same for both cases.
We ask for clarification and are told that we should calculate the input impedance, the output impedance, and the voltage gain for the circuit and not the transistor. It is interesting to note that the problem can be restated so that it becomes a simple design problem: Given the h parameters, design a simple amplifier that has a gain of −60.

2. Present. Given a simple transistor circuit, an input voltage of 3.2 mV, and the h parameters of the transistor, calculate the output voltage.

3. Alternative. There are a couple of ways we can approach the problem, the most straightforward being to use the equivalent circuit
shown in Fig. 19.57. Once you have the equivalent circuit you can use circuit analysis to determine the answer. Once you have a solution, you can check it by plugging in the answer into the circuit equations to see if they are correct. Another approach is to simplify the right-hand side of the equivalent circuit and work backward to see if you obtain approximately the same answer . We will use that approach here.

4. Attempt. We note that R_s = 0.8  kΩ \text{ and } R_L = 1.2  kΩ. We treat the transistor of Fig. 19.59 as a two-port network and apply Eqs. (19.70) to (19.79).

(19.70):    A_i=\pmb{\frac{ I _c}{ I _b}}=\frac{h_{f e}}{1+h_{o e} R_L}

(19.71):    \pmb{I _b}=\frac{h_{o e} \pmb{V _c}}{\frac{h_{f e}}{1+h_{o e} R_L}-h_{f e}}

(19.72):    \pmb{\frac{ V _b}{ V _c}}=\frac{h_{o e} h_{i e}}{\frac{h_{f e}}{1+h_{o e} R_L}-h_{f e}}+h_{r e} =\frac{h_{i e}+h_{i e} h_{o e} R_L-h_{r e} h_{f e} R_L}{-h_{f e} R_L}

(19.73):    A_\nu=\pmb{\frac{ V _c}{ V _b}}=\frac{-h_{f e} R_L}{h_{i e}+\left(h_{i e} h_{o e}-h_{r e} h_{f e}\right) R_L}

(19.74):    \pmb{\frac{ V _b}{ I _b}}=h_{i e}-h_{r e} R_L \pmb{\frac{ I _c}{ I _b}}

(19.75):     Z_{\text {in }}=\pmb{\frac{ V _b}{ I _b}}=h_{i e}-\frac{h_{r e} h_{f e} R_L}{1+h_{o e} R_L}

(19.76):      h_{r e}(1)=- \pmb{I _b}\left(R_s+h_{i e}\right) \quad \Rightarrow \quad \pmb{I _b}=-\frac{h_{r e}}{R_s+h_{i e}}

(19.77):      \pmb{I _c}= \pmb{h _{o e}}(1)+h_{f e} \pmb{I _b}

(19.78):      \pmb{I _c}=\frac{\left(R_s+h_{i e}\right) h_{o e}-h_{r e} h_{f e}}{R_s+h_{i e}}

(19.79):      Z_{ out }=\frac{R_s+h_{i e}}{\left(R_s+h_{i e}\right) h_{o e}-h_{r e} h_{f e}}

h_{i e} h_{o e}-h_{r e} h_{f e}=10^3 \times 20 \times 10^{-6}-2.5 \times 10^{-4} \times 50 = 7.5 × 10^{−3} \\ \space \\ A_\nu=\frac{-h_{f e} R_L}{h_{i e}+\left(h_{i e} h_{o e}-h_{r e} h_{f e}\right) R_L}=\frac{-50 \times 1200}{1000+7.5 \times 10^{-3} \times 1200} = -59.46

A_v is the voltage gain of the amplifier = V_o/V_b. To calculate the gain of the circuit we need to find V_o/V_s. We can do this by using the mesh equation for the circuit on the left and Eqs. (19.71) and (19.73).

– V _s+R_s I _b+ V _b=0

or

V_s=800 \frac{20 \times 10^{-6}}{\frac{50}{1+20 \times 10^{-6} \times 1.2 \times 10^3}-50}-\frac{1}{59.46} V _o \\ \space \\ = −0.03047 V_o.

Thus, the circuit gain is equal to −32.82. Now we can calculate the output voltage.

V_o=\text { gain } \times V_s=-\pmb{105.09 \underline{/0^{\circ}}  mV} \text {. } \\ \space \\ A_i=\frac{h_{f e}}{1+h_{o e} R_L}=\frac{50}{1+20 \times 10^{-6} \times 1200}=48.83 \\ \space \\ Z_{\text {in }}=h_{i e}-\frac{h_{r e} h_{f e} R_L}{1+h_{o e} R_L}\\ \space \\ =1000-\frac{2.5 \times 10^{-4} \times 50 \times 1200}{1+20 \times 10^{-6} \times 1200} \\ \space \\ = 985.4  Ω

You can modify Z_{in} to include the 800-ohm resistor so that

Circuit input impedance = 800 + 985.4 = 1785.4 Ω.

\left(R_s+h_{i e}\right) h_{o e}-h_{r e} h_{f e} \\ \space \\  =(800+1000) \times 20 \times 10^{-6}-2.5 \times 10^{-4} \times 50=23.5 \times 10^{-3} \\ \space \\ Z_{\text {out }}=\frac{R_s+h_{i e}}{\left(R_s+h_{i e}\right) h_{o e}-h_{r e} h_{f e}}=\frac{800+1000}{23.5 \times 10^{-3}}=76.6  k \Omega

5. Evaluate. In the equivalent circuit, h_{oe} represents a resistor of 50,000 Ω. This is in parallel with a load resistor equal to 1.2 kΩ. The size of the load resistor is so small relative to the h_{oe} \text{ resistor that } h_{oe} can be neglected. This then leads to

I _c=h_{f e} I _b=50 I _b, \quad V _c=-1200 I _c,

and the following loop equation from the left-hand side of the circuit:

-0.0032+(800+1000) I _b+(0.00025)(-1200)(50) I _b=0 \\ \space \\ I _b=0.0032 /(1785)=1.7927  \mu A .\\ \space \\ I _c=50 \times 1.7927=89.64  \mu A \text { and } V _c=-1200 \times 89.64 \times 10^{-6} \\ \space \\ = −107.57  mV

This is a good approximation to −105.09 mV

Voltage gain = −107.57 / 3.2 = −33.62

Again, this is a good approximation to 32.82.

Circuit input impedance = 0.032 / 1.7927 × 10^{−6} = \pmb{1785  Ω}

which clearly compares well with the 1785.4 Ω we obtained before.
For these calculations, we assumed that Z_{out} = ∞  Ω. Our calculations produced 72.6 kΩ. We can test our assumption by calculating the equivalent resistance of this and the load resistance.

72,600 × 1200 / (72,600 + 1200) = 1,180.5 = 1.1805 kΩ

Again, we have a good approximation.

6. Satisfactory? We have satisfactorily solved the problem and checked the results. We can now present our results as a solution to the problem.