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Could someone help to analyse the circuit above, which is a single common emitter amplifier. And for the purpose of larger voltage gain, a current mirror building block is applied as active load. But I could not analyse how it work for larger voltage gain, and even be confused with current variation when there is a input signal applied.

when there is no input signal Ic2 is approximately equal to IR, and will be constant also Ic1 = Ic2

when there a small input signal, say Δib, so will be Δic1, and the Q1 collector current Ic1 now will be Ic1 + Δic1 but Ic2 is still unchanged, that is to say when applied input signal, Ic2 will not be equal to Ic1, that is what has confused me. Would someone help to explain it ? Thank you!

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That difference in the currents between IC1 and IC2 goes into the load attached to Out.

Also the output impedance of a real transistor is not infinite (could be megohms) that will put a limit on the achievable gain.

That circuit is not practical without some DC feedback to set the DC voltage at the collector of Q1 and Q2, as drawn the voltage would either saturate Q1 or Q2.

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  • \$\begingroup\$ Thanks Kevin. That difference in the currents between IC1 and IC2 goes into the load attached to Out, So, when Δib is positive, then the current across load(i-load) flows from ground to Q1. when Δib is negative, then the current i-load flows from Q1 to ground . Is that correct? \$\endgroup\$ – John Lu Sep 7 '17 at 5:37
  • \$\begingroup\$ when Δib is negative, then the current i-load flows from Ic2(NOT Q1) to ground. Is that correct? \$\endgroup\$ – John Lu Sep 7 '17 at 5:43
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You need something like this

schematic

simulate this circuit – Schematic created using CircuitLab

Add R11, providing DC&AC feedback from output to constant-current-biasing, to have programmable-Rout of the Constant Current source; this allows programmable gain of the CommonEmitter transconductor.

Q5 minimizes Miller Capacitance, thus input signal energy gets used, usefully, in the transconductor and not wasted in charging Cob of Q1.

C3 provides high-frequency Power Supply Rejection.

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