When using an op amp that is ac coupled I see a dc path to ground at the input has to exist. The simplest example I can think of is a voltage follower. A dc path can be provided on the + terminal but what about the negative terminal? Rather am I supposed to assume a dc path is provided through the output to the power supply to ground? I using a set up where the op amp's power supply has a path to ground.

Don Lancaster points out it is essential for both inputs to have a path. Lancaster says a dc return path to ground or some other stable dc bias point. I usually think of an output as varying. I suppose it works for the voltage follower.

 

A dc path can be provided on the + terminal but what about the negative terminal? Rather am I supposed to assume a dc path is provided through the output to the power supply to ground?

Yes, that is correct. The amplifier's input bias current has to have someplace to flow to/from.

Lancaster says a dc return path to ground or some other stable dc bias point. I usually think of an output as varying. I suppose it works for the voltage follower.

Doesn't matter if it varies or not. It just has to be a DC path.

EDIT: The attached application note from Analog Devices discusses this.

 

Yes, that is correct. The amplifier's input bias current has to have someplace to flow to/from.


Doesn't matter if it varies or not. It just has to be a DC path.

EDIT: The attached application note from Analog Devices discuss

 

Well, thanks, it's nice to see my thinking confirmed. I've been experimenting with filters and adding a path to ground with a resistor sure can change the cutoff frequency. Values up around 390k doesn't. There so much of great math shown with filters I'm surprised it hasn't been taken into account.

 

I've been experimenting with filters and adding a path to ground with a resistor sure can change the cutoff frequency.

Don't understand that.
Show an example.

There so much of great math shown with filters I'm surprised it hasn't been taken into account.

All filters I've seen using op amps do take that into account.
They always have a return path for the input bias current.

 

If I connect a low pass RC to an input where the input picks off across C and ground I get a transfer function of 1/(1+sRC). Assuming it was ac coupled I would add a resistor R1 across the cap to ground. I get for a result a transfer function of 1/(sCRR1 +R+1). It's not the same. Sure, a high value resistance R! would look like a light load on the output of the original RC filter and probably could be ignored. I think the original RC is the Butterworth filter type. The math calculated would be different.

 

If I connect a low pass RC to an input where the input picks off across C and ground I get a transfer function of 1/(1+sRC). Assuming it was ac coupled I would add a resistor R1 across the cap to ground. I get for a result a transfer function of 1/(sCRR1 +R+1). It's not the same. Sure, a high value resistance R! would look like a light load on the output of the original RC filter and probably could be ignored. I think the original RC is the Butterworth filter type. The math calculated would be different.

You already have a path from the input to ground through the R in the RC filter; you don't need a second resistor.

 

I don't quite understand. The R would come from some output to the input of the op amp. If it were dc coupled such an output would likely have a path. If it were ac coupled though, I would have a coupling cap, in series with the R and then the op amp input, with a cap C to ground. I would think the coupling cap would block the dc return path. I don't see a return path to ground.

 

I don't quite understand. The R would come from some output to the input of the op amp. If it were dc coupled such an output would likely have a path.

Not "likely"; it definitely would.

If it were ac coupled though, I would have a coupling cap, in series with the R and then the op amp input, with a cap C to ground. I would think the coupling cap would block the dc return path. I don't see a return path to ground.

Correct. In that case you would have to provide a DC path, via a resistor or by some other means.

 

Most sources, even if AC coupled, have a resistive path to ground at the output of the capacitor.

Any resistive path to a voltage is a path for the bias current to flow. The path doesn't have to be directly to ground (and often isn't).