An actively loaded differential pair

PHYSICS · Multi-part · Fermi University Physics · Electronics, Instrumentation & Communication · Semiconductor devices and analogue circuits · Semiconductor devices and analogue circuits

Inside the Instrument, Ch Multistage amplifiers, the Miller effect and output stages

Problem

An actively loaded differential pair, built as in Chapter 2, has both transistors biased at

I_C = 40\,\mu\mathrm{A}

with Early voltage

V_A = 50\,\mathrm{V}

. Its output node therefore carries the parallel combination of the two transistors' small-signal output resistances

r_o = V_A/I_C

. ((a)) Find the pair's combined output resistance

R_o

at that node. ((b)) The pair's differential transconductance is

g_{m1}=I_C/V_T = 40\,\mu\mathrm{A}/25.9\,\mathrm{mV}

. This node drives directly (with no follower) into a common-emitter gain transistor biased at

I_{C2}=80\,\mu\mathrm{A}

with

\beta_F = 120

, whose input resistance is

r_{\pi 2}=\beta_FV_T/I_{C2}

. Find the loaded first-stage gain

A_1 = g_{m1}(R_o \| r_{\pi2})

. ((c)) A follower (same

\beta_F=120

) is inserted between the pair and the gain transistor, its emitter loaded only by

r_{\pi2}

. Find the follower's input resistance

R_{i2}=(\beta_F+1)r_{\pi2}

and the new loaded gain

A_1' = g_{m1}(R_o\|R_{i2})

. ((d)) With the follower in place, a stray capacitance of

3\,\mathrm{pF}

loads the pair's output node. Using the resistance found by open-circuit reasoning at that node (

R_o\|R_{i2}

), estimate the pole this capacitance contributes.

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