Single-ended amplifier gain vs Fully differential amplifier gain

[CHL]

Hello

I wonder the gains of single-ended (typical opamp) and fully differential amps.

Let's assume that there is a simple differential amplifier structure.

In a fully differential amplifier, the gain is

(Vout+ - Vout-)/(Vin+ - Vin-)

On the other hand, in a single-ended amplifier, the gain is

(Vout+)/(Vin+ - Vin1-)

This means that if the input signal and the gain of each input common source amplifier are same, the single-ended amp gain is the half of the fully differential amp gain.

Also, if the positive input port of the singe-ended amplifier is connected to a bias voltage to make an inverting amplifier (let's assume it is GND), the amplifier gain is

(Vout+)/(- Vin1-)

and a single common source amplifier operates in this case, not both input amplifiers.

Does what I understand correct?

Although it looks simple in an opamp block level, I'm confused in integrated circuit level.

 

[D.A.(Tony)Stewart]

That seems correct,

Some differential structures are highest bandwidth Video amplifiers with low gain. ( eg x2) But I understand you were assuming they had the same structures.

 

[LvW]

This means that if the input signal and the gain of each input common source amplifier are same, the single-ended amp gain is the half of the fully differential amp gain.

The input signals (equal or not) do not influence the gain. For equal input signals it is only the common-mode gain that is effectiv.

Also, if the positive input port of the singe-ended amplifier is connected to a bias voltage to make an inverting amplifier (let's assume it is GND), the amplifier gain is
(Vout+)/(- Vin1-)
and a single common source amplifier operates in this case, not both input amplifiers.
Does what I understand correct?

I am not sure what you really mean. Are you using one transistor only - neglecting the other one? This could be "problematic" because it influences the feedback.
Please, show us a circuit diagram.

 

[CHL]

The input signals (equal or not) do not influence the gain. For equal input signals it is only the common-mode gain that is effectiv.


I am not sure what you really mean. Are you using one transistor only - neglecting the other one? This could be "problematic" because it influences the feedback.
Please, show us a circuit diagram.

Sorry for my unclear question.

This image is from "Design of Analog CMOS Integrated Circuits - Razavi".

I assumed the M1, M2 and Iss in both circuits are same respectively.

Also the 'input signal' doesn't mean the common mode voltage, it is a small signal input such as microphone signal.

If an inverting amplifier is made using the amplifier above (a), M1 gate should be biased at certain voltage and the output and M2 gate are connected through a feedback resistor. Also the input signal goes to the M2 gate through a resistor.

Therefore, I think only M2 operates as an amplifier while M1 is just biased.

 

[FvM]

If an inverting amplifier is made using the amplifier above (a), M1 gate should be biased at certain voltage and the output and M2 gate are connected through a feedback resistor. Also the input signal goes to the M2 gate through a resistor.

Therefore, I think only M2 operates as an amplifier while M1 is just biased.

If the circuit would be operated in the described way, there's still a signal path through M1, M3 and M4 which contributes half of the output signal. In so far it's not correct to say that M1 is "just biased". M1 and M2 form a differential pair, if M1 current is increased by a positive gate signal, M2 current is reduced by the same amount.

Another point to consider, a single stage amplifier can't be seriously used as OP, it neither has the output swing nor the required gain. It can be operated as inverted amplifier with a very limited voltage range.

 

[CHL]

If the circuit would be operated in the described way, there's still a signal path through M1, M3 and M4 which contributes half of the output signal. In so far it's not correct to say that M1 is "just biased". M1 and M2 form a differential pair, if M1 current is increased by a positive gate signal, M2 current is reduced by the same amount.

Another point to consider, a single stage amplifier can't be seriously used as OP, it neither has the output swing nor the required gain. It can be operated as inverted amplifier with a very limited voltage range.

Thanks for your explanation but I think you misunderstand what I meant.

That's true, a single stage amplifier isn't used as an opamp, but I just want to understand the operation of a single differential pair.

If the M1 gate is the positive input and M2 gate is the negative input, M1 gate is biased at a certain point when I want to make an inverting amplifier with two resistors at the negative input.

'M1 current is increased by a positive gate signal' the mention is not for this case.

 

[FvM]

You can easily exchange M1 and M2 in the statement. Things are the same. The transistor M1, which you assume to be inactive, is part of the signal path.