A differential pair

a)

An answer:

\begin{array}{rcl} v_{in, cm} & = \frac{v_{1}}{2} \end{array}

b)

An answer:

\begin{array}{rcl} i_{C, 1} = i_{C, 2} & = \frac{I_{TAIL}}{2} \cdot \frac{βfe}{βfe + 1} \end{array}

c)

An answer:

\begin{array}{rcl} i_{C 2} & = I_{TAIL} \cdot \frac{βfe}{βfe + 1} \cdot \frac{1}{1 + e^{\frac{q v_{1}}{kT}}} \\ \to 1 + e^{\frac{q v_{1}}{kT}} = 20 \to \\ v_{1} & = \frac{kT}{q} \ln (19) ≅ 74 mV \end{array}

Alternatively, calculate the change in $v_{BE}$ for both transistors to get a change in the collector current by:

This yields $v_{1} ≅ \frac{kT}{q} \ln (10) + \frac{kT}{q} \ln (1.9) = 74 mV$

d)

An answer:
Same answer as for the previous question EXCEPT for the sign: $v_{1} ≅ - 74 mV$ .

e)

An answer:
$4.7 V$

f)

An answer:
$0.3 V$

g)

An answer:
The maximum current that can be delivered by the differential pair is $1 mA$ . For the capacitor we get:

\begin{array}{rcl} SR & \equiv {| \frac{\partial V_{C_{M}}}{∂t} |}_{\max} \\ = \frac{I_{C_{M}, \max}}{C_{M}} \end{array}

From which the numerical value follows: $SR = 1 0^{7} \frac{V}{s}$

h)

An answer:

\begin{array}{rcl} a_{v} & = \frac{v_{out}}{v_{1}} \\ v_{out} = R_{C} \frac{g_{m}}{2} \cdot v_{1} \\ g_{m} = \frac{q}{kT} \frac{I_{TAIL}}{2} \\ a_{v} & = R_{C} \frac{q}{kT} \frac{I_{TAIL}}{4} = 47 \end{array}

The last two steps were not explicitly asked for, so these are optional.

i)

An answer:

\begin{array}{rcl} \frac{v_{1} - v_{2}}{v_{1}} & = \frac{v_{1} - β v_{out}}{v_{1}} \\ v_{out} = a_{v} \cdot (v_{1} - v_{2}) \\ = a_{v} \cdot v_{1} - a_{v} β \cdot v_{out} \\ = \frac{a_{v}}{1 + a_{v} β} \cdot v_{1} \\ \frac{v_{1} - v_{2}}{v_{1}} & = \frac{1}{1 + a_{v} β} \end{array}

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