Question 19.2: Determine the small-signal voltage gain, input resistance an......

The first step in this problem is to determine the small-signal equivalent circuit of the amplifier.

Voltage gain

In order to determine the behaviour of the circuit, we need to establish the values of g_{m} and h_{ie} . To do this, we must know the DC operating conditions as both are affected by the quiescent current. Fortunately, we have already investigated the DC conditions of the circuit in Example 19.1, from which we know that I_{C} is 1.02 mA. Therefore, as I_{E} \approx I_{C} , it follows that I_{E} \approx 1.02  mA and thus

and

From Equation 19.15 we have

voltage gain = \frac{v_{o}}{v_{i}} = -g_{m} \left( \frac{1}{h_{oe}} //R_{C} \right) = -g_{m} \frac{R_{C}}{h_{oe}R_{C}  +  1}     (19.15)

voltage gain = \frac{v_{o}}{v_{i}} = -g_{m} \frac{R_{C}}{h_{oe}R_{C}  +  1}

and substituting for the component values gives

voltage gain = -40.8 \times 10^{-3} \frac{4700}{10 \times 10^{-6} \times 4700  +  1} \approx -183

If we consider that 1/h_{oe} is large compared with R_{C} and assume that the voltage gain is equal to -g_{m}R_{C} , this gives a value of -192. Given the inaccuracies in our calculations, this is probably a reasonable approximation. Therefore

voltage gain = \frac{v_{o}}{v_{i}} \approx -g_{m}R_{C}

Input resistance

From the equivalent circuit it is clear that the small-signal input resistance is simply R_{B}//h_{ie} . As R_{B} ⪢ h_{ie} , it is reasonable to say

Output resistance

The small-signal output resistance is the resistance seen ‘looking into’ the output terminal of the circuit. As the idealised current generator has an infinite internal resistance, the output resistance is simply the parallel combination of R_{C}   and 1/h_{oe} . Thus

and again it is reasonable to use the approximation that r_{o} \approx R_{C} .