II$.$ Using a computer simulation, verify the results of Example $11\mathrm{.}13$ for the simple op$-$amp circuit.

EXAMPLE $11\mathrm{.}13$

Objective: Calculate the input resistance and the small$-$signal voltage gain of a Darlington pair.

Consider the circuit shown in Figure $11\mathrm{.}46,$ with parameters $I_{C7}=I_Q=0\mathrm{.}2mA,I_{C8}=1mA,R_4=10k,$ and $R_3=0\mathrm{.}2k.$ Assume $=100$ for all transistors, and the Early voltage for $Q_{11}$ is $100$ V $.$

Solution: The input resistance, given by Equation $(11\mathrm{.}126),$ is

$R_i=\frac{2(1+)V_T}{I_Q}=\frac{2(101)(100)(0\mathrm{.}026)}{0\mathrm{.}2}=>2\mathrm{.}63M$

The small$-$signal voltage gain is a function of $R_{L7},$ which in turn is a function of $R_{c11}$ and $R_{b8}.$ We can find that

$r_{11}=\frac{V_T}{I_Q}=(100)\frac{0\mathrm{.}026}{0\mathrm{.}2}=13k$

such that

$R_E^{\text{'}}=13||0\mathrm{.}2=0\mathrm{.}197k$

Also

$g_{m11}=\frac{I_Q}{V_T}=\frac{0\mathrm{.}2}{0\mathrm{.}026}=7\mathrm{.}69m\frac{A}{V}$

and

$r_{o11}=\frac{V_A}{I_Q}=\frac{100}{0\mathrm{.}2}=500k$

Therefore,

$R_{c11}=r_{o11}(1+g_{m11}{R}_E^{\text{'}})=500[1+(7\mathrm{.}69)(0\mathrm{.}197)]=>1\mathrm{.}26M$

We can determine that

$r_8=\frac{V_T}{I_{C8}}=(100)\frac{0\mathrm{.}026}{1}=2\mathrm{.}6k$

Then

$R_{b8}=r_8+(1+)R_4=2\mathrm{.}6+(101)(10)=>1\mathrm{.}01M$

Consequently, resistance $R_{L7}$ is

$R_{L7}=R_{c11}||R_{b8}=1\mathrm{.}26||1\mathrm{.}01=0\mathrm{.}561M$