current-mirror amplifier

[jutek]

hello

i'm working on attached circuit. it enhances slew rate of amplifier.

could you tell me what's the role of the current mirror in the amplifier?
it converts differential output to single-ended and allows to achieve higher gain in one stage opamp. what else?

what if i don't want to use this current mirror cause it makes amplifier's output resistance high and deteriorates bandwidth. the opamp has two outputs without thic current mirror and i need one. when i use simple single-ended ota how should i connect slew rate enhancement circuit on the right side then?

regards

 

[ahmad_abdulghany]

what's the role of the current mirror in the amplifier?
it converts differential output to single-ended and allows to achieve higher gain in one stage opamp. what else?

Current mirror is used to supply current (or mirror current) to all other circuit pathes, and it doesn't affect how the system is ended (differential or single ended), you may use diode connected load to get unbalanced output (single ended) but the current mirror isn't responsible for that.

As well as current increases, in general, you will get better slew rate, higher BW, I think you can even sense it in mind, right?

Best Wishes,
Ahmad,

 

[jutek]

Current mirror is used to supply current (or mirror current) to all other circuit pathes, and it doesn't affect how the system is ended (differential or single ended), you may use diode connected load to get unbalanced output (single ended) but the current mirror isn't responsible for that.

As well as current increases, in general, you will get better slew rate, higher BW, I think you can even sense it in mind, right?

Best Wishes,
Ahmad,

what if i cancel the current mirror M5-M12 (i have to do that to lower output resistance). I need a single ended opamp to conect its output to Vout between md7 and md8. Do you mean i can simply connect m4's gate to m3's gate instead of its drain to achieve single ended output?? what are another possibilities?

regards

 

[tekno1]

read abdulghany's reply again, he is giving the answer.

The Op-Amp you are talking about is current-mirror type therefore your differential input stage (this is not op-amp by itself but an Gm stage) and current mirror stage together is an Op-Amp. current mirrored output provides necessary current drive at the output and is merely a current multiplier (yes it also reduces BW, because nothing is free).

You can (as you wished) connect input differential pairs output to Vout but than you may not have enough current drive (also to do this differential stage pmos loads should be either diode connected as it is now --lower gain -- or mirror connected 2x more gain). If you do this you will no longer able to talk about an Op-Amp but a Gm stage (if you remove current mirror arrangement). Than your output impedance will be even higher and you better do not have any DC current at the output (driving capacitor load is OK).

I hope it is clear.

 

[wpchan05]

Note the following points about this form of opamp:
1. The SR is high as long as the quesicent current in the output node is high (SR ≈ Ibias over capacitance)
2. The output resistance will not destory the BW. In fact, this opamp can be seen approximately as one pole device (the dominant pole occurs at the output node, other nodes are non-dominant and located at high frequency). To achieve high speed with stability, a large dominant pole (larger Rout or larger Cout) will yield higher phase margin.
3. The drawback of this opamp is that the dominant pole cannot be known in advance because we usually do not know the exact Cout. Therefore, a compensating cap is usually needed to ensure stability. In this case, SR will be degraded.