# Op Amps: how do they know whether the supply is split?

Most op amps can operate in single and split supply configurations. However, I cannot see how this is possible.

If I connect voltages $V_\text{SY}^+$ and $V_\text{SY}^-$ to the positive and negative supply pins of a op amp, and $V_+$ and $V_-$ to the non-inverting and inverting inputs, then the output is (if $A_\text{OL}$ is the open loop gain of the op amp): $$V_\text{out}=A_\text{OL}(V_+-V_-)+V_\text{SY}^-\tag{Single-supply}~\text{ }~~\text{ }~~\text{ }~~\text{ }~~\text{ }~~\text{ }$$ The above equation is from personal experience using op amps in single supply (i.e. $V_\text{SY}^- = \text{GND}, V_\text{SY}^+ > \text{GND}$).

In split supply, we can use the following diagram from MIT's 6.002 introduction to electronics lecture video: $$V_\text{out}=A_\text{OL}(V_+-V_-)+(V_\text{SY}^+-V_\text{SY}^-)/2\tag{Split-supply}$$

These two equations, however, are mutually inconsistent: If I take an op amp and hook up a single voltage source between its supply pins, how does the op amp know that the resulting voltage is not split between two sources?

It must know - but how does it know?

• When you are using some non-trivial subscripts/parameters you better first define them. – Eugene Sh. Jul 14 '15 at 16:59
• The voltages are ALL measured relative to the common zero. Go from there. – Eugene Sh. Jul 14 '15 at 17:13
• They don't "know" -- they have permitted common mode input voltage ranges and output voltage ranges that depend upon where the rails are. – Scott Seidman Jul 14 '15 at 17:13
• Yeah, I'm trying to use virtual ground, because I own only one voltage source. So "common zero" could be anything. – Dave Jul 14 '15 at 17:16
• Of course it can. But then all of the values should be reajusted to it. Your Vsy and the others are measured relative to something, right? – Eugene Sh. Jul 14 '15 at 17:18

Let's go back to $V_\text{out}=A_\text{OL}(V_+-V_-)+(V_\text{SY}^+-V_\text{SY}^-)/2$ for a minute. The equation holds in both cases.

Assuming a 5V single supply, $V_+ = 5V$, $V_- = 0V$, and $A_{OL} = 10^5$,

$$V_\text{out}=10^5(5V-0V)+(5V-0V)/2$$

In this case, the open loop gain dominates, and drives the output to the positive rail.

Assuming a +/-5V supply, $V_+ = 5V$, $V_- = 0V$, and $A_{OL} = 10^5$,

$$V_\text{out}=10^5(5V-0V)+(5V-5V)/2$$

The open loop gain still dominates, and the output goes to the rail.

The difference is, in the single supply case, 2.5V is the middle of the supply rails, and in the split supply case, 0V is the middle. You wouldn't know any different because the open loop gain forced the output to the positive rail. Let's repeat this process with the noninverting pin at $1\mu \text V$.

$$V_\text{out}=10^5(1\mu V-0V)+(5V-0V)/2 = 3.5V$$

$$V_\text{out}=10^5(1\mu V-0V)+(5V-5V)/2 = 1V$$

Negative feedback is what gives you control of the open loop gain. I'm sure it will be in the next video.

• Personal experience - well: all I can say is, if you take an op amp with negative feedback, hook the negative supply rail to ground, and the positive supply rail to, say, +5V, then the output will be measured above ground, NOT +2.5V. – Dave Jul 14 '15 at 17:22
• @Dave And it won't be zero if you do the same thing to +/- 5V. I don't see what you're getting hung up on here. – Matt Young Jul 14 '15 at 17:24
• No, in the second case, we have a circuit identical to that in the youtube image, with V_S = 5V. And in that image, V_out is measured relative to 0V. – Dave Jul 14 '15 at 17:28
• @Dave So in your split supply case you're tying the inverting pin to 0V still? If so you're comparing apples to bananas. – Matt Young Jul 14 '15 at 17:31
• It doesn't care about where your reference point is! It's maximum output would be the Vsy+, it's minimal - Vsy- in any case. But these values will be different if you are taking different references. – Eugene Sh. Jul 14 '15 at 17:34

Rules for most op-amps: -

1. Output can usually get to within a couple of volts of the power rails and then will hard-limit unless it is a rail-to-rail opamp (then it might get to within 50mV)
2. Inputs can't normally be taken to within a couple of volts of the power rails but there are some notable exceptions.
3. An op-amp has no-idea that the negative supply to the op-amp isn't ground - if the inputs are all above ground (but less than the +V rail as per above) and the output (by negative feedback) constrains itself to within the rails on the chip then it works fine.
• Op-amps can be rail-to-rail on the output, the input, or both. "Rail-to-rail" on a data sheet usually means on the output, but some amps are described on the features list as rail to rail on input and output – Scott Seidman Jul 14 '15 at 20:14

Your first equation is not correct. It should be identical to the second equation, assuming the bias point of the output is really halfway between the supplies. (In practice, there's a small DC offset.) An op amp is a differential amplifier with high gain. It doesn't "know" what kind of supplies it's hooked up to, and in fact there is no way to know. The choice of ground is arbitrary. If you have +/-12V supplies with an input referenced to ground, it's equally correct to say that you have a +24V/0V supply with an input referenced to +12V.

Adding negative feedback is what really references the output ground to the signal ground. For example, consider a voltage follower: simulate this circuit – Schematic created using CircuitLab

The op amp will change its output to make the inverting input voltage equal to the non-inverting input voltage (V1). Since V1 is referenced to ground, the output voltage will be as well. This is independent of the supply voltages

That's usually okay with split supplies, but with a single-supply the output can't go below zero volts. If you have a ground-referenced AC signal, that's problematic. For example, here's an inverting amplifier: simulate this circuit

When V1 is negative, the amplifier works as expected. But when V1 is positive, the output runs into the 0V rail. The solution is to add a negative offset to the signal. Because the op amp is a differential amplifier, that's the same as adding a positive offset to the inverting input. (Remember, the choice of which node to call ground is arbitrary!) simulate this circuit

This offset is called a virtual ground, and is usually half of your single supply voltage.

• "Adding negative feedback is what really references the output ground to the signal ground." - could you elaborate on that point? – Dave Jul 14 '15 at 17:32
• I've updated my answer. – Adam Haun Jul 14 '15 at 18:53

Consider a graph of $V_+-V_-$ vs $V_{out}$ (referenced to $V_{SY}^-$) for the two equations given. The opamp can't produce more than $V_{SY}^+$ or less than $V_{SY}^-$, of course. Here you can see that at 0V input, the output is 0V relative to $V_{SY}^-$. Below 0V input, the opamp can't produce less than $V_{SY}^-$. The linear range is from 0V to $\frac{V_{SY}^+}{A_{OL}}$. For the second equation, you can see the output (referenced to $V_{SY}^-$) at 0V input is 2.5V or $V_{SY}^+/2$. The linear range is from $-\frac{V_{SY}^+}{2A_{OL}}$ to $+\frac{V_{SY}^+}{2A_{OL}}$.

This is the equation that is actually used, for both single supply and split supply circuits. The linear range is where the opamp 'works', and the most natural place to define its properties is at the middle, not at the edge where it's about to stop working.

Now, real opamps have an input offset voltage $V_{OS}$ which dominates this concern. An equation that incorporates it would be $V_{out} = A_{OL}(V_{OS}+V_+-V_-) + V_{SY}^+/2$. Notice how the input offset gets multiplied by the open loop gain. A cheap opamp like the LM324 would have an input offset voltage randomly distributed between say -5mV and 5mV, and an open loop gain of 100k, making this equation something like $V_{out} = A_{OL}(V_+-V_-) - 500V + V_{SY}^+/2$! 500V is a lot bigger than $V_{SY}^+/2$. Graphically, it looks like this: 