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# CMOS vs ideal op-amp

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#### boarensteinboars

##### Newbie
When we design an op-amp at the transistor level (CMOS), both inputs of the op-amp are biased at the same DC level. The output DC level, however, is not zero. The output is a certain level between Vdd and Vss.

What the op-amp really does is it amplifies the difference of changes in the + and - inputs into a large change/swing in the output.

vout=Aol(vinp-vinm)

and we use small v to signify AC or incremental signals in contrast with DC signals.

Then why do we use the following equation (with DC voltages):
Vout=Aol(Vinp-Vinm)
when designing op-amp circuits, say for ex.) a summing amplifier.
We use this equation together with KCL at the input nodes to calculate the closed loop gain. Then we let Aol tend to infinity.
This assumes that the output is normally zero with equal voltages at the input and output.

#### KlausST

##### Super Moderator
Staff member
Hi,

I hope I understand your question correctly.

This is a general formula true for DC and AC.
But
* Aol is considered infinite for (coarse) calculations in feedbacked configuration. For exact gain calculation you need to use true Aol at the given frequency
* Aol varies with frequency. There is Aol for DC, then usually at rather low frequencies there is the first order cutoff and with some higher frequency there is a second order LPF characteristic.

So instead of a general Aol you should use the frequency dependent Aol(f)

**
Output voltage at non feedbacked Opamp with inputs shorted:
--> you say "it's normally zero"
--> I say it as "undefined"

There are some reasons:
* with single power supply this means "saturated at GND". This may be, but there is no "must"
* dual supplied Opamps have to "GND" ... so how could they know what "0V" means?
* If Aol(DC) is considered infinite, then the formula is Vout = 0 x infinte --> which is undefined
* for real (standard) Opamps it is not important at all, because the input offset and it's drift will make this calculation useless.

Klaus

#### AMS012

##### Full Member level 4
The equation Vout=Aol*(Vinp-Vinm) is the general model of an ideal opamp, i.e for any values of Vinp and Vinm, the equation is valid. This is the definition of an ideal opamp.

However, the same equation is valid even for a real opamp. The difference is the range over which it is valid is smaller depending on the linearity of the devices used to build the opamp. So, over very small range (typically a few tens of mV), that equation can be used in case of real opamp. Therefore it is called small signal analysis when it comes to real opamp networks. The range over which the above equation is valid is called a small signal.

It is like, if you have a very non linear function say y(x), you can determine the adjacent values of the function over a small range around some x value say x0 by knowing its derivative around the point i.e y(x1)=y(x0)+(x1-x0)*dy/dx@x0+..... taylor series. This is valid with an assumption that x1-x0 is very small or the difference in dy/dx at the two points are very similar (similar to real opamp). If you knew that y(x) is linear, then you would be able to predict y(x) for any values no matter what (ideal opamp). It is the same with opamp equation.

#### boarensteinboars

##### Newbie
What I'm trying to say is when designing op-amp circuits we use the formula:

Vout=Aol(Vinp-Vinm)

where Vout, Vinp and Vinm.

When designing a single-ended differential amplifier at the transistor level, we bias both inputs to the same DC level. If the above formula was correct, the output would be zero

For a real op-amp, it senses the small differential change in the input and amplifies this as an output change/swing.

ΔVout = Aol(ΔVinp-ΔVinm) or vout = Aol(vinp-vinm)

We use small letters to denote incremental or ac signals.
However, ΔVinp-ΔVinm = (Vinp-Vbias) - (Vinm-Vbias) = Vinp - Vinm

So that:

ΔVout = Aol(Vinp-Vinm)

The output swing , ΔVout, is not necessarily equal to Vout. If Vout starts from a value of zero at the bias point then yes, but that is not strictly true. So I am thinking, shouldnt the formula be:

Vout = Aol(Vinp-Vinm) + Vout,bias

Vout,bias
is the output at the dc operating point

##### Super Moderator
Staff member
The upper and lower transistors resemble a class A amplifier. If you want Vout to center around 0V, the lower supply rail must be negative. (Customary bipolar supply setup for op amps.)

Furthermore you trim transistor bias so that the upper and lower transistors duplicate a resistive divider at the required proportion, so as to achieve Vout =0V.

By performing these adjustments back and forth, you can obtain 0V output with 0V input, that is, with the inputs connected to 0V ground.

#### AMS012

##### Full Member level 4
What I'm trying to say is when designing op-amp circuits we use the formula:

Vout=Aol(Vinp-Vinm)

where Vout, Vinp and Vinm.

When designing a single-ended differential amplifier at the transistor level, we bias both inputs to the same DC level. If the above formula was correct, the output would be zero
View attachment 172896

For a real op-amp, it senses the small differential change in the input and amplifies this as an output change/swing.

ΔVout = Aol(ΔVinp-ΔVinm) or vout = Aol(vinp-vinm)

We use small letters to denote incremental or ac signals.
However, ΔVinp-ΔVinm = (Vinp-Vbias) - (Vinm-Vbias) = Vinp - Vinm

So that:

ΔVout = Aol(Vinp-Vinm)

The output swing , ΔVout, is not necessarily equal to Vout. If Vout starts from a value of zero at the bias point then yes, but that is not strictly true. So I am thinking, shouldnt the formula be:

Vout = Aol(Vinp-Vinm) + Vout,bias

Vout,bias
is the output at the dc operating point
You are saying you will put Vinp and Vinm equal to zero, you want the output to be zero. That won't happen. The circuit is not linear when the inputs are zero. So the equation is not valid for absolute zero input. This is the very definition of linearity of a function or system. You can expect the validity over a small range defined as the small signal range.

#### boarensteinboars

##### Newbie
You are saying you will put Vinp and Vinm equal to zero, you want the output to be zero. That won't happen. The circuit is not linear when the inputs are zero. So the equation is not valid for absolute zero input. This is the very definition of linearity of a function or system. You can expect the validity over a small range defined as the small signal range.
No, i'm just saying if Vinp=Vinm=Vbias, where Vbias is not zero, the ideal opamp equation predicts an output of zero. However, when designing opamps this is not true. For the differential amp shown in the image I attached, the inputs are biased at Vbias, but the output is not zero. It should be around Vdd minus the vgs of the pmos at the top.

I'm just confused because there is a sudden leap in the definition of CMOS opamp and the opamp used in circuit level design.

In a cmos opamp, when both inputs are equal, there is zero swing or change in the output relative to the bias point. But the output voltage itself is not zero, it is at the bias point.

But when using the opamp in a circuit, we use
Vout =Aol(Vinp-Vinm) when it should be ( i think)
ΔVout = Aol( Vinp-Vinm)

#### boarensteinboars

##### Newbie
The upper and lower transistors resemble a class A amplifier. If you want Vout to center around 0V, the lower supply rail must be negative. (Customary bipolar supply setup for op amps.)

Furthermore you trim transistor bias so that the upper and lower transistors duplicate a resistive divider at the required proportion, so as to achieve Vout =0V.

By performing these adjustments back and forth, you can obtain 0V output with 0V input, that is, with the inputs connected to 0V ground.
I see. So we do have to design it so that the output is normally zero, when the inputs are equal?

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