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

Question

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

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

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:

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?

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

?

Answer 1

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: -

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

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: -

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.

Answer 2

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:-

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.