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