Choosing the right opamp for biasing a square wave

Question

I am trying to bias a 1MHz, Vpp = 1V, bipolar square wave signal coming from an opamp.

I want to bias it by 1V, the 1V bias (Vcm) is available at an ADC's output port.

I am thinking about using this circuit with a rail to rail opamp:

I know I will get an inverted signal, but I can simply invert it again.

1. How do I choose the correct capacitor value?
2. What voltage do I need to apply to the rails?
3. Do I need a resistor between the ADC's Vcm output and the opamp's Vcm input?
4. I will need an opamp with a gain bandwidth product of 10 times the square wave frequency, right?

Answer 1

Technically, you don't need the opamp at all; your capacitor, feeding it to a higher-than-source impedance resistor to your reference voltage does exactly that.

So, if the original source of your square wave can drive your load that you plan to attach to Vout, this simple CR high pass filter totally biases the square wave to whatever you want.

The capacitor value would hence be chosen to be large enough to put your cutoff frequency below 1 MHz. That depends on the resistor.

Answer 2

You could bias it with a resistor to a +1V source and coupling capacitor, then buffer it with a voltage follower (no inversion). Your circuit is not that.

However to get a good fidelity square wave you need a high slew-rate amplifier with a large GBW product. Maybe 50MHz and 20 or 30V/us minimum. Also you cannot expect the amplifier to perform well right at the limits of output, so it would be better to give it +2V if you expect an output voltage of 1.5V peak. Even if it is advertised as rail-to-rail input and output (since it's a voltage follower, the input also has to have a common-mode range that approaches the positive rail).