Exercise 8.5 A simple opamp schematic

a)

An answer:

$$\begin{array}{rcll}& i_{\mathit{OUT}}& =\frac{I_{\mathit{TAIL}}}{2}\cdot (\frac{\mathit{\beta fe}}{\mathit{\beta fe}+2}-1)& \\ & & =-\frac{I_{\mathit{TAIL}}}{2+\mathit{\beta fe}}\cong 0& \end{array}$$

b)

An answer:
Assuming that this differential input voltage is small, you may use the transconductance of the differential pair:

$$\begin{array}{rcll}& i_{\mathit{OUT}}& \approx g_m{v}_1& \\ & & =\frac{q}{\mathit{kT}}\frac{I_{\mathit{TAIL}}}{2}{v}_1& \\ & & =200\mathit{\mu A}& \end{array}$$

Actually, this 10mV is quite small for this differential pair at room temperature, but to get the exact value the large signal transfer function can be used. This large signal transfer function can relatively easily be derived (see book), leading to:

$$\begin{array}{rcll}& i_{\mathit{OUT}}& =I_{\mathit{TAIL}}\cdot \tanh <!--\; FUNCTION\; APPLICATION\; -->\left(\frac{q{v}_1}{2\mathit{kT}}\right)& \\ & & =197\mathit{\mu A}& \end{array}$$

c)

An answer:
$$\begin{array}{rcll}& v_{\mathit{OUT}}& v_{\mathit{out}}+V_{\mathit{OUT},0}& \\ & & \approx g_m{v}_1\cdot (R_1∕∕R_2)+V_{\mathit{OUT},0}& \\ & & =8.5V& \end{array}$$

d)

An answer:
A straight forward reasoning results in $i_{\mathit{OUT}}\cong I_{\mathit{TAIL}}$ from which it follows that the output voltage rises by $5V$ from its value for $v_1=0$. Then $v_{\mathit{OUT}}\cong 12.5V$.

e)

An answer:
See the previous answer; $i_{\mathit{OUT}}\cong -I_{\mathit{TAIL}}$ and consequently $v_{\mathit{OUT}}\cong 2.5V$.

f)

An answer:

With a zero input voltage difference, the current in both branches is $0.5\cdot I_{\mathit{TAIL}}$.

In one branch the current is $0.95\cdot I_{\mathit{TAIL}}$. Here the extra voltage drops due to the higher current is:

$$\begin{array}{rcll}& & \mathrm{\Delta }{V}_{\mathit{RE}}=0.45\cdot R_E{I}_{\mathit{TAIL}}& \\ & & \mathrm{\Delta }{V}_{\mathit{BE}}=\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->\left(1.9\right)& \end{array}$$

In the other branch the current is $0.05\cdot I_{\mathit{TAIL}}$. Here the extra voltage drops due to the higher current is:

$$\begin{array}{rcll}& & \mathrm{\Delta }{V}_{\mathit{RE}}=-0.45\cdot R_E{I}_{\mathit{TAIL}}& \\ & & \mathrm{\Delta }{V}_{\mathit{BE}}=\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->\left(0.1\right)& \end{array}$$

The difference between the two (the differential input voltage) then yields the max (and -min) value of the input signal:

$$\begin{array}{rcll}& \mathrm{\Delta }{V}_{\mathit{IN}}& =0.9\cdot R_E{I}_{\mathit{TAIL}}+\frac{\mathit{kT}}{q}\ln <!--\; FUNCTION\; APPLICATION\; -->\left(19\right)& \\ & & =524\mathit{mV}& \end{array}$$