There's an AC signal, 1 V amplitude, which is the input to an op-amp, in differential mode:

figure 1

However, the output gets clipped and the amplitude isn't unity for some reason, even though gain is set to be 1.

figure 2

To "fix" this clipping, one needs to "bias" the op-amp by putting a reference midpoint (resistor divider network for example) at the positive terminal of op-amp, like this:

figure 3

An "explanation" is that:

for the opamp to swing high, your input signal has to go below the ground, so you set up "virtual ground" somewhere between rails so that your opamp has some reference against which to invert the input signal

Or the "op-amp can't go beyond its output rails" etc. These explanations do not help me understand the why.

To understand why, I'd like to know how the output of the OP284 changes with changing inputs to the OP284.

Is this approximately what the OP284 looks like?

figure 4

Let's assume it is, which is just a bunch of BJTs.

V1 = negative terminal input to U1 V2 = positive terminal input to U1.

Output of U1 is: \$ V_{out} = Gain * (V_2- V1) \$

Gain is unity, so it's just: \$(V_2- V1)\$

Let's assume V5 (input source AC signal) goes like this:

At t = 1, Vin (positive terminal of V5) -> V1 is 0.1 V, V2 = 0 V.
t = 2, Vin = 0.2 V, V2 = 0 V
t = 3, Vin = 0.3 V, V2 = 0 V

For these three time instances, what's Vout?

Well, from figure 4, Q2 is closed off pretty much all the time, right? Q1 is slightly on, so Vo1 is about 0.1 V, while Vo2 is the maximum 5 V? (VCC5 rail).

So what about Vout when:

t = 10, Vin = 1 V, V2 = 0 V
t = 11, Vin = 0.9 V, V2 = 0 V.
t = 20, Vin = 0 V, V2 = 0 V
t = 21, Vin = -0.1 V, V2 = 0 V


  • 3
    \$\begingroup\$ A very simple answer is that there is nothing inside of an opamp that can generate new voltages, just resistors and transistors that can switch and divide the voltage that you provide. By raising your signal to be above your negative rail and below your positive rail, you hopefully avoid trying to produce voltages outside of that range (at least if your input signal amplitude doesn't get too large). \$\endgroup\$ Jan 12, 2022 at 5:13
  • \$\begingroup\$ So is the question about why an op-amp, or any circuit, can't output voltages that are beyond the supply voltages? \$\endgroup\$
    – Justme
    Jan 12, 2022 at 5:17
  • \$\begingroup\$ @user1850479 how am I "raising" my signal above my negative rail???? Input source AC signal swings fine between +1V and -1V, and I give to op-amp +5V and the same -1V for the negative supply. So the negative supply side of U1 is same node as negative source terminal V5. \$\endgroup\$
    – user234571
    Jan 12, 2022 at 5:27
  • \$\begingroup\$ @Justme the questions have question marks in them \$\endgroup\$
    – user234571
    Jan 12, 2022 at 5:29
  • \$\begingroup\$ The input signal to the opamp in your second diagram has no negative voltages at all, the addition of the positive voltage to the input raises it above zero so everything is positive. Since you got rid of the negative voltages, you don't clip. \$\endgroup\$ Jan 12, 2022 at 5:34

2 Answers 2


Why does differential op-amp require mid point biasing at its input to prevent clipping?

It's not just a differential op-amp circuit, it's any linear op-amp circuit...

So, you are confusing yourself by not recognizing the basic problem. The confusion arises because you have chosen an overcomplicated circuit to learn a very basic thing: -

enter image description here

The above is not a differential op-amp; it's a single-ended input and the op-amp's negative rail is at ground potential. It's an inverting amplifier with a gain of unity; virtually as simple as it gets.

So, when the input signal goes below 0 volts, in order to keep Vin- at the same value as Vin+ (which happens to be 0 volts), the op-amp output rises to a positive value. That positive value enforces Vin- == Vin+. Not a millivolt higher nor a millivolt lower; that's what the op-amp is conditioned to do; it must makes Vin- == Vin+.

It has no other task.

However, if the input signal has a positive value then the op-amp output cannot make its output go below 0 volts and thus it clips. The output is limited to values within its power supply range; some are better and some are worse of course.

However, the output gets clipped and the amplitude isn't unity for some reason, even though gain is set up to be 1.

No, of course it isn't unity; it's clipped; it's shortened; it's reduced.

To understand why, I'd like to know how output of OP284 changes with changing inputs

It's got nothing to do with the op-amp model number; all op-amps will do the same.

Back to the basic circuit but with an input capacitor

enter image description here

It's a the same story; the op-amp can't fight against the input signal rising positively above 0 volts but, because we have added a capacitor there's a chance we can shift the output up a few volts and avoid it clipping: -

enter image description here

So now, the op-amp has a fight it can win; it's trying to make Vin- have a DC value of +2.5 volts; the op-amp can pull and push that voltage around by moving its output towards the positive supply rail or by moving towards ground. It's a symmetrical situation and, unless the input has an amplitude that is close to and beyond the power rails, the op-amp will produce a relatively undistorted output signal.

  • \$\begingroup\$ >"When the input signal goes below 0 volts." Before that, I'd assume the positive rail of V5 was positive, and the negative rail, well at 0V, which is a reference point for everything aka "gnd". So when Vin, with respect to GND (which is also the positive terminal input to op-amp) becomes 0, and then current direction changes, and V5 source's polarity changes, thus "GND" rail becomes positive and Vin side negative, right? Is this the moment you're talking about. \$\endgroup\$
    – user234571
    Jan 12, 2022 at 8:40
  • 1
    \$\begingroup\$ Try and order your thoughts and speak in simpler sentences. What you wrote in your comment is far too loaded for me to unpick. Assume the input signal peaks at +1 volt and -1 volt - don't overcomplicate things with it having a peak value that equals the op-amp power rail. \$\endgroup\$
    – Andy aka
    Jan 12, 2022 at 8:44
  • \$\begingroup\$ Here's circuit with labeled nodes: i.imgur.com/o54Y6dK.jpg. It's comment section, so have to make it short not detailed, but I'll try. "When the input signal goes below 0 volts, in order to keep Vin- the same as Vin+, the op-amp output rises positively." How does input signal go below "0V"? Vin1 starts out positive, rises, then decreases, this is all with respect to Vin2(gnd). When Vin1 =0V with respect to Vin2, Vin2 becomes positive and Vin1 is now negative (current direction change). But I think that the supply rails for OP-amp are fixed (VCC5 and Vin2/GND) and can't be swapped. 1/2 \$\endgroup\$
    – user234571
    Jan 12, 2022 at 8:50
  • \$\begingroup\$ But if you do swap, then op-amp would be able to output the input signal, right? Like when Vin1 is -1V with respect to Vin2, and U1's negative supply side is connected to Vin1? 2/2 \$\endgroup\$
    – user234571
    Jan 12, 2022 at 8:51
  • 2
    \$\begingroup\$ How does input signal go below "0V"? the input signal is a sinewave hence it goes positive then negative etc.. Just ask one question at a time because you are building on ideas that might be wrong. One question at a time please. Don't overload a comment with too much info because it won't work. \$\endgroup\$
    – Andy aka
    Jan 12, 2022 at 8:56

Is this how approximately OP284 looks like?

No. Your circuit won't work close to the negative rail because the input transistors don't get enough bias voltage, but it can work close to the positive supply rail.

The OP284 has rail-to-rail inputs and outputs. To handle signals close to the negative rail it has an inverted form of your circuit wired in parallel, with PNP input transistors. The outputs of the NPN and PNP circuits are then combined in the following stage. Here's the 'simplified' schematic:-

enter image description here

But you don't need to know all this. Just remember that input voltages must stay between Vcc and Vee for correct operation, and the output voltage is constrained between Vcc and Vee.

In your circuit where the non-inverting input is referenced to ground (via R4) an AC input signal goes below ground during negative half cycles. This violates the common mode input voltage range, but even if it didn't the output voltage cannot go below ground so the bottom half of the waveform must be clipped.

There are two obvious solutions to this problem:-

  1. Supply a negative voltage to Vee that exceeds the maximum negative voltage swing on +IN. The output will then swing below ground on negative halves of the AC signal.

  2. Bias +IN with a positive voltage so when the input goes negative the voltage at +IN is above ground (Vee). Since the DC gain of the op amp is 1 in this circuit, the output will be 'centered' at the bias voltage.

  • \$\begingroup\$ Even if I connect negative terminal of V5 to U1's negative supply rail, making them the same node, there's still clipping: i.imgur.com/5dEBfTg.jpg Why can't output just simply repeat what the input is? I'm guessing the issue is that difference of inputs is outputted. I provided examples, and I still don't know how output behaves at t=10, Vin = 1V, V2=0V t=11, Vin=0.9V, V2=0V. ... t=20, Vin=0V, V2=0V t=21, Vin=-0.1V, V2=0V. Also I don't need output to go below "ground". Input source doesn't "go below" gnd to show -1V. \$\endgroup\$
    – user234571
    Jan 12, 2022 at 6:30
  • \$\begingroup\$ Because the op amp is biased at 0V (ie. ground, through R4), so the output is centered on 0V and positive input signal on -IN will try to make the output go below ground - which it cannot do (unless Vee is negative). \$\endgroup\$ Jan 12, 2022 at 6:38
  • \$\begingroup\$ yeah I read all that already, I don't get it, I just need to know how output behaves at different time instances \$\endgroup\$
    – user234571
    Jan 12, 2022 at 6:59
  • \$\begingroup\$ Not sure I can explain it any better. perhaps you should try just considering the voltage on each input and calculating what the output would be with an 'ideal' op amp. Then when the answer is a negative voltage (which is impossible)... \$\endgroup\$ Jan 12, 2022 at 7:14
  • \$\begingroup\$ I kinda tried in the op-post, but nobody commented about those time instances. \$\endgroup\$
    – user234571
    Jan 12, 2022 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.