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nmos current mirror with differential pair

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leo47

New Member
Why is it better to use a differential pair to guide the current to the output node, rather then using just one simple switch?
Anyone an idea?
I made a current steering dac and somebody asked me that question but I just took the diff. pair because everbody uses it, i don't know why it is better.
 

dknguyen

Well-Known Member
Most Helpful Member
Pair

Mind providing a circuit? Some of us don't have the names of circuits fresh in our head as we should ;)

One of the first things I learned about differential pairs is that you don't need to use capacitors on the output (for amplifiers? I don't remember exactly) which is a big plus when designing on silicon...something like that anyways, I might have a few things mixed up. Not sure if that applies to what you are talking about.
 

Roff

Well-Known Member
This is a differential amplifier. The differential output voltage is proportional to the difference between V1 and V2. The common mode gain is very low, meaning that any voltage components of V1 and V2 that are the same (not differential) will have little effect on the single-ended and differential output voltages.
 

dknguyen

Well-Known Member
Most Helpful Member
Bias

Took me a little while to recognize that the current mirror was just a current source for the differential-amplifier. I agree with Ron, this entire circuit is just a basic differential amplifier.

Differential amplifiers are better than regular amplifiers because
-they reject noise better because they reject common signals on the input lines (and if the input lines are very close to each other, the noise present on both lines will be almost identical)
-they do not require you have to have large filtering capacitors on the amplifier output to filter out the DC-biasing which is a big plus when designing on silicon

I find that a bit odd that you knew about current mirrors and DACs without knowing about differential amplifiers. At least for me, I learned about diff-amps and a year before I learned about current mirrors.
 
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Roff

Well-Known Member
EDIT:
dknguyen, I see you changed your mind. I'll delete your original statement, which I had quoted.
The current mirror is a standard way of biasing the diff pair.
 
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dknguyen

Well-Known Member
Most Helpful Member
Yeah, it took me a while to recognize that the current mirror was just a current source for the diff-amp. Looking at the current source completely expanded out like that with a diff-amp looked a bit different to me. I kept looking at the circuit with the mindset that the current mirror was the dominant purposeful part of the circuit because it took up so much space. From that perspective, I kept thinking the differential inputs were an addition to the current mirror and that V1 and V2 were DC bias voltages. So I couldnt figure out why the differentials were there in the first place. I also go to wondering why the right branch of the mirror was connected to Vdd when it was supposed to be connected to the load of the mirror. THen it took me a bit longer to realize that the differentials ARE the load of the current mirror and that the differentials were meant to be an amplifier.
 
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