# Why is open loop gain important in an op amp?

Can someone tell me why open loop gain in an op-amp is important?

Why should someone measure or even care about the open loop gain when op-amps are always used in closed loop with feedback only?

Why should someone measure or even care about the open loop gain when op-amps are always used in closed loop with feedback only?

Some relevant stuff about op-amps: -

• The open-loop gain you are likely referring to is only at DC
• Above a few hertz or tens of Hz, the open loop gain falls at 20 dB/decade: -

This last point is important if we want an op-amp circuit to have non-trivial gains at a decent frequency. Consider the small red circle in the image above. It shows you the highest frequency that this op-amp can muster when it has a voltage gain of 200.

• If you wanted a gain of 1000, you'd get a 3 kHz frequency response and, no more
• If you wanted a gain of 100, you'd get a 30 kHz frequency response.
• If you only wanted to amplify by ten then, you'd get a 300 kHz bandwidth.

Hence, open-loop gain is one of the cornerstones of the graph above. It and the natural 3 dB point of the graph fully define the unity gain bandwidth and any-gain bandwidth of most regular op-amps.

Also, having a high open-loop gain gives you more accuracy when processing DC signals (even when the closed loop gain is low-ish).

• Thank you very much for the detailed answer, Much appreciate it! Jun 19, 2023 at 14:55
• Readers might want to know that "Gain-Bandwidth Product" is the usual term for what this answer describes. Jun 20, 2023 at 15:21
• @BenVoigt if I projected the sloping section of the open-loop gain to the y-axis (assuming the y-axis was at 1 Hz which on this occasion it is) then that would be the GBWP line. It would cross the 1 Hz y-axis at a magnitude equal to the frequency magnitude along the x-axis when the gain is unity. That would be the constant GBWP line. Jun 21, 2023 at 8:06

I take it you understand that the relationship between input potentials $$\V_{INV}\$$, $$\V_{NONINV}\$$ and output potential $$\V_{OUT}\$$ of an op-amp with open-loop gain $$\A\$$ is:

$$V_{OUT} = A\times(V_{NONINV} - V_{INV})$$

What many seem to overlook is that the op-amp always behaves this way, regardless of anything connected around it. Whether its part of an inverting amplifier configuration, or non-inverting, or differential amplifier, the op-amp still has that behaviour. In other words, even if you close the loop with feedback, the op-amp doesn't care, and still multiplies the difference between its two input potentials by its huge open loop gain.

An op-amp doesn't suddenly change its behaviour just because there are these resistors and those capacitors and diodes around it, in the same way your car doesn't suddenly change its relationship between steering wheel position and wheel angle just because there's a driver, providing negative feedback, sat in it.

When we talk of the gain of a "closed loop" system, we are referring to the behaviour of the whole system, which may indeed have very low "closed-loop" gain, much much lower than $$\A\$$. But, the op-amp itself still has gain $$\A\$$, and is still doing:

$$V_{OUT} = A\times(V_{NONINV} - V_{INV})$$

Importantly, when the op-amp output isn't stuck hard against one of the supply potentials (saturated), but instead roaming freely somewhere between the those supplies, then the difference $$\V_{NONINV} - V_{INV}\$$ must necessarily be very, very small, since A is so large:

$$V_{NONINV} - V_{INV} = \frac{V_{OUT}}{A} \approx 0$$

Therefore, if $$\A\$$ is huge, like 105 or one million, and the op-amp isn't saturated, then we are able to say:

$$V_{NONINV} \approx V_{INV}$$

During your studies, you have no doubt encountered this condition of $$\V_{NONINV}\$$ and $$\V_{INV}\$$ being equal, which is one of the consequences of negative feedback, and which makes all the calculations really trivial. It is due to this equality that we have really simple and linear input/output relationships in closed-loop systems, such as:

1. Inverting amplifier: $$\\frac{V_{OUT}}{V_{IN}}=-\frac{R_2}{R_1}\$$

2. Non-inverting amplifier: $$\\frac{V_{OUT}}{V_{IN}}=1+ \frac{R_2}{R_1}\$$

This (near) equality between $$\V_{NONINV}\$$ and $$\V_{INV}\$$, though, depends entirely on huge open-loop gain $$\A\$$, which the op-amp is still, and is always employing, regardless of what you connect around it. It is only because this open-loop gain is so large that we can have closed-loop behavior like $$\\frac{V_{OUT}}{V_{IN}}=1+ \frac{R_2}{R_1}\$$.

As $$\A\$$ gets smaller, the larger the difference between $$\V_{NONINV}\$$ and $$\V_{INV}\$$ becomes, and the less ideal becomes closed-loop behaviour, so that (for example, a non-inverting amplifier) $$\\frac{V_{OUT}}{V_{IN}}\$$ diverges farther and farther from the ideal $$\1+ \frac{R_2}{R_1}\$$.

Having argued for the highest open-loop gain possible, there are also arguments for reducing it.

In closed loops, there can be (and usually are) various phase shifts that occur around the loop, which can lead to instability, like unwanted ringing or oscillation.

In certain cases it may be necessary to reduce open-loop gain (either by using a different op-amp, or introducing signal attenuation some other way), to bring about stability.

Often, choosing open-loop gain is a balancing act between linearity (like $$\\frac{V_{OUT}}{V_{IN}}=1+\frac{R_2}{R_1}\$$) and stability.

• Thank you very much for the detailed answer, Much appreciate it! Jun 19, 2023 at 14:55

The answer is simple: Because only in case of a very large open-loop gain Ao (in practice, we often assume an infinite value for Ao), the feedback factor Hr determines the gain.

Example: Non-inverting opamp configuration.

Acl=Ao/(1+HrAo)=1/[(1/Ao)+Hr].

As you can see, for Ao approaching a very large value the closed-loop gain Acl can be determined with Acl=1/Hr.

In many cases, this approximation is good enough when the error (caused by this simplification) is smaller than the other neglections and simplifications:

• Neglecting opamps finite input and output impedances
• parts tolerances contributing to Hr.

The above given feedback formula is valid also for the inverting opamp configuration. However, in this case, we have to consider the fact that the input signal is NOT connected directly to the (inverting) opamp input terminal but through an input damping (forward) network (-Hf). Therefore, this factor Hf does appear in the numerator of the formula: Acl=-Hf/Hr

• Important remark: We should not forget, however, that for stability analyses, we always have to consider the frequency-dependent finite open-loop gain Ao=f(jw)
• Isn't the frequency dependent open loop gain used for stability analysis like Bode plots?
– sai
Jun 19, 2023 at 12:13
• Sorry - you are right. I have corrected this (bloody) typing error. For stability analyses we need the loop gain, which is nothing else than the product Hr(jw)*Ao(jw).
– LvW
Jun 19, 2023 at 14:11
• Analysis of inverting opamp configurations can follow the same procedure as non-inverting configurations if a Norton transformation is applied to the signal source and input resistor. Then you don't have to worry about sometimes accounting for an input network and sometimes not. Jun 22, 2023 at 16:31

If you shop for op-amps, you can expect that the open loop game is very high which let's you make some simplifying assumption for circuit design (see post by LvW).

Still, it is unlikely that the exact open loop gain of one op-amp vs. another notably affects circuit behavior. In fact, parts with rail-to-rail outputs (i.e. common emitter/source outputs) have strongly varying open-loop gain depending on the load impedance to complicate matters, and often, we don't notice or care.

Exceptions when you do care:

• precision (ppm level) regulators
• CMRR and PSRR are affected by open-loop gain