A two-stage amplifier has a first stage (as in Chapter 2) of transconductance driving…

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

A two-stage amplifier has a first stage (as in Chapter 2) of transconductance gm,pre=1.8 mSg_{m,\rm pre}=1.8\,\mathrm{mS} driving node 1, which presents R1=2 MΩR_1=2\,\mathrm{M}\Omega in parallel with C1=1.5 pFC_1=1.5\,\mathrm{pF} to ground. Node 1 in turn drives node 2 through a controlled current of transconductance gm=4 mSg_m=4\,\mathrm{mS} (an inverting stage of gain −gmR2-g_mR_2), where node 2 presents R2=80 kΩR_2=80\,\mathrm{k}\Omega in parallel with C2=6 pFC_2=6\,\mathrm{pF}. A 1.5 pF1.5\,\mathrm{pF} capacitor bridges node 1 to node 2 across this stage; with the other capacitors open, it sees a resistance R1+R2+gmR1R2R_1+R_2+g_mR_1R_2. ((a)) Find the open-loop gain magnitude A0=(gm,preR1)(gmR2)A_0 = (g_{m,\rm pre}R_1)(g_mR_2) and the dominant pole by the sum of the open-circuit time constants. ((b)) A ×10\times10 oscilloscope probe with 10 pF10\,\mathrm{pF} input capacitance (as characterised in Chapter 1) is then clipped onto node 1 to observe the waveform there, adding its capacitance in parallel with C1C_1. Recompute the dominant pole. ((c)) State, in one sentence, why the probe's capacitance alone barely moves the pole even though it is far larger than any single capacitor already at that node. ((d)) The same probe also presents 10 MΩ10\,\mathrm{M}\Omega to ground (Chapter 1). Recompute R1R_1, A0A_0 and the dominant pole with both the probe's resistance and its capacitance on node 1, and say which part of the probe actually disturbs this node.

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