# MOSFET Operational Amplifier

I am confused. Why we always say that the gain of an op amp should be as large as we can? Supposing that an op amp has a power supply of 1.8 V, the voltage gain is 1000 V/V. If we give a small signal 5 mV of sine wave, it is easy to swing to the triode region of MOSFET. However, it is useful that if small signal is under 0.5 µV. Why do we pursue a high voltage gain?

For almost all op-amps (whether bipolar or MOSFET technology) we pursue high open-loop gain. Then, when we connect up the feedback resistors we are (more or less) guaranteed a closed-loop gain that is low to moderate but, accurately defined by the resistor values.

For instance a typical op-amp might have an open loop gain of around 1 million at DC and, for this device, we could have a closed-loop gain (set by accurate resistors) of up to 1,000 (without any eyebrows being raised at all).

But, we can also use more negative feedback and engineer circuit gains of 1,000 down to unity (non-inverting configuration).

Also be aware that at higher frequencies, the open-loop gain naturally reduces and, as a result, we are not permitted the same levels of AC closed loop gain for the same accuracy as DC.

Maybe you are confusing open-loop gain and closed-loop gain?

• Yes, I think I am confusing open-loop gain and closed-loop gain. Thanks a lot Commented May 31 at 14:57

The gain of the op amp you mentioned is open loop gain. It may be high but the highest voltage you can see on output is upto the voltage on power supply minus .5 to 1 volt to see any output without distortion. The maximum limit of amplitude of an output signal is the value of voltage in its power supply on positive and negative rails. In your case, 1.7 or 1.8v is the maximum output you will see without distortion for dual power supplies with +1.8V and -1.8V. If you dont use feedback, the output might be at the same level as the power supplies.

If you use feedback, the maximum limit of amplitude of an output signal would be the same as voltage of the power supplies for positive and negative amplitudes of output. This is because the transfer function of the op amp is limited upto the power supply. Any amplification done which is higher than the power supply would be chopped with the output truncated in amplitude.

• Thanks for your explanation Commented May 31 at 15:00

# Need for high gain

More precisely, we need not so much amplifiers with a very high gain, but amplifiers with a moderate but stable gain. We do the latter through the former by reducing their gain many times through negative feedback, and it gets better the higher their gain.

I will illustrate the philosophy of high-gain (not only MOSFET) amplifiers with a series of CircuitLab schematics and graphs with some explanatory text to them.

# How to use high-gain amplifiers

The amplifier is an analog device whose output voltage is proportional to the input voltage (Vout = A.Vin). If the amplifier is "ideal", there is no problem in doing this for any value of the input voltage. Let's take one such amplifier (without power supply terminals) from the CircuitLab library and examine it.

# "Ideal" amplifier

Vin = 1 V, A = 10: So, if we apply a 1 V input voltage to a 10 x amplifier, the output voltage will be 10 V.

simulate this circuit – Schematic created using CircuitLab

We can see it graphically, if we sweep the gain from 1 to 10 (using the DC Sweep Simulation).

Vin = 1 V, A = 100 -> Vout = 100 V: If we increase the gain 10 times, the output voltage will be 100 V.

simulate this circuit

Vin = 1 V, A = 1000 -> Vout = 1000 V and so on...

simulate this circuit

Thus, for each value of the input voltage, a corresponding proportional value of the output voltage appears.

# Real amplifier

However, the output voltage of a real amplifier can vary only within the limits of its supply voltage; this is because it is a fraction of the supply voltage produced by two devices in voltage divider configuration. Let's take such an amplifier (with power supply terminals) from the CircuitLab library, supply it with +-10 V, and examine it.

## No clipping

Vin = 1 V, A = 10 -> Vout = 10 V: So if the output voltage is not required to exceed the supply voltage, proportionality is preserved like in Schematics 1.1 above. Let's take one such amplifier (without power supply terminals) from the CircuitLab library and examine it.

simulate this circuit

The results are identical to those obtained from schematic 1.1.

## Clipping

Vin = 1 V, A = 100 -> Vout = 10 V: But if it is necessary for the output voltage to exceed the supply voltage, this is impossible and it remains equal to the supply voltage.

simulate this circuit

We can see it graphically, if we sweep, as above, the gain from 1 to 100.

## Remedy 1: Decreasing Vin

... Inside the input sources: To avoid the limitations, we are forced to reduce the input voltage...

simulate this circuit

... e.g. by a voltage divider.

simulate this circuit

By the way, this attenuation is done for each device (e.g., voltmeter, ammeter, etc.) in order to extend its operating range.

## Remedy 2: Neutralizing Vin

The idea of first attenuating and then amplifying the input voltage does not seem very smart to us because errors accumulate. It is much trickier to reduce the input voltage, and in this way, the overall gain, by subtracting voltage from it which is the idea of the so-called negative feedback. Here is how it is implemented in a non-inverting amplifier.

simulate this circuit

Note that the 100x amplifier gain is reduced to 10x circuit gain…

Vin = 1 V, A = 1000 -> Vout = 10 V

simulate this circuit

simulate this circuit

… and the 1000x amplifier gain is reduced to 10x circuit gain (with a smaller error).

simulate this circuit

• Nice graphs. Thanks a lot Commented May 31 at 15:01