Question on single ended vs. fully differential OTA

[talon]

I am having trouble understanding gain in the single ended OTA case and its use in creating an op amp.

Say I have a first stage that is an OTA - diff pair with current source load. This diff pair has transistor M1 on the left and M2 on the right.

In small signal I input Vin at the left input of the diff pair and ground the right input. I take a single ended output at the drain of M2. Then the gain of this stage is positive(?)
A1 = gm1(ro4 || ro2)

I am unsure about the above result. The OTA should have output current Io = -GmVin with the output current Io point out towards the next stage but I am confused about where the negative sign comes from since increasing Vin causes current through M2 to decrease and the output node V1 to increase.

If my second stage consists of a common source amplifier with a current source at the source (implemented as a transistor) resulting in a gain of -gm6(ro6 || ro7) , then my total gain is
A1*A2 = -gm1(rop || ro2)gm6(ro6 || ro7)

Now if I disregard the necessary output stage for now (buffer), is what I have an op amp? Shouldn't the total gain be positive Vo=A(V+ - V-) where V+ is Vin and V- is grounded.

What is incorrect about my logic?

Thanks,

talon

 

[erikl]

In small signal I input Vin at the left input of the diff pair and ground the right input. I take a single ended output at the drain of M2. Then the gain of this stage is positive(?)

Yes. 2 voltage inversions.

A1 = gm1(ro4 || ro2)

Correct.

I am unsure about the above result. The OTA should have output current Io = -GmVin with the output current Io point out towards the next stage but I am confused about where the negative sign comes from since increasing Vin causes current through M2 to decrease

This is true for the current (transconductance).

... and the output node V1 to increase.

That's why your voltage gain is positive.

If my second stage consists of a common source amplifier with a current source at the source (implemented as a transistor) resulting in a gain of -gm6(ro6 || ro7) , then my total gain is A1*A2 = -gm1(rop || ro2)gm6(ro6 || ro7)

Ok.

Now if I disregard the necessary output stage for now (buffer), is what I have an op amp?

No, it's still an OTA. For an opAmp, you need the buffer.

Shouldn't the total gain be positive Vo=A(V+ - V-) where V+ is Vin and V- is grounded.

For feedback you must feed back (a part of) the output voltage to the inverting input, no matter where your Vin is connected to. Moreover, you can always reverse your input differential connection points.