Electronic – If positively biasing amplifier does minimum voltage on input pins apply

Agreed with the answer given by @KyranF, but I would also like to add that capacitors resist change in voltage. This means that when the AC voltage swings between -2.5V and 2.5V, the capacitor will draw/produce (Depending how you look at it) whatever current it needs to maintain this voltage. So that causes a voltage/current swing in the 1MΩ resistor.

I would also argue that the circuit given by @Black Emperor is not entirely wrong.

Ideally (Note ideally) an ADC pin would not draw any current, so we can treat the ADC node as an open circuit. Let's say we apply zero volts at the AC signal pin for a very long time. The cap would also be an open circuit and there wouldn't be any current flowing through the resistor, thus the ADC pin sees 2.5V, this is the known bias.

If the AC signal starts to fluctuate, the cap would force the voltage across it's terminal to remain the same, causing current to flow and changing the voltage across the resistor which can be read by the ADC.

The only reason why I wouldn't suggest to implement that exact circuit is because a cheap ADC would have a low input impedance. This low input impedance might be on the same scale as the 1MΩ resistor (causing it to appear less like an open circuit) and current would start to sink into the ADC and possibly cause damage.

Hope I was clear, good luck!

• Josh

Well, this is exactly what should happen ! The reason is as follows:

Your input signal is a sinewave between 0 - 2.5 V

I could also say that your input signal is a 1.25 V (2.5 V /2) DC voltage + a sinewave with an amplitude of 1.25 V.

The thing that confuses you is the 1.25 V DC ! It is converted to differential as well, resulting in the DC voltages you see at the outputs.

If you want to prevent this you need to make that DC voltage at the input 0 V. This might be impractical.

Fortunately what you can also do is shift the DC reference voltage of the single to differential converter. I see that is has a Vocm input, if this is a common-mode level input then you could apply 1.25 V DC to it (the DC voltage I mentioned above) then the single to differential converter will consider 1.25 V as the commonmode level and the outputs will get the same DC voltage.