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Comments on Help with differential to single ended voltage converter

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Help with differential to single ended voltage converter

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Hello I don't understand how this differential to single-ended voltage converter can work.

What I don't understand is why we place a current mirror(a current mirror copies the current flowing through one transistor to another transistor) and how is the output voltage affected by the input voltages.

SeeMe

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Question: What is the purpose of the current mirror?

Answer: It allows to combine the current changes in BOTH transistors (T1, T2).

Explanation In a classical differential amplifier (with separate collector resistances for T1 and T2) we must use the difference between the collector voltages or currents for further processing - when we want to use the full dynamic capabilities of the amplifier.

This can also be accomplished using the shown current mirror - with simultaneous double-ended into single-ended conversion.

Resulting from a differential input signal the collector currents of T1 and T2 change their values in the opposite direction. However, the current mirror does not "allow" such a change because it tries to keep the currents equal.

Therefore, the difference between the currents is available as an output current iout which drives the next stage, which makes the current-to-voltage conversion.

So we have:

ic1=ic3+2ib and ic2=-ic1=ic4+iout;

With ic4=ic3 this gives: iout=-2(ic1+ib)

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2 comment threads

You ave nothing to do than to solve for iout and set ic2=-ic1 and ic3=ic4 - that`s all. (1 comment)
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You ave nothing to do than to solve for iout and set ic2=-ic1 and ic3=ic4 - that`s all.
LvW‭ wrote almost 3 years ago

You ave nothing to do than to solve for iout and set ic2=-ic1 and ic3=ic4 - that`s all.