DC Offset after a Voltage Follower

[aw_rw1]

I have a MiniGen from SparkFun. I am trying to get 1-100Hz out of it. So far, the MiniGen can produce this, but the output is offset by a Vcc/2, about 1.6VDC. I suppose the offset is due to the power being 3.3VDC. To remove the offset, I put an AC coupling capacitor in series with after the MiniGen.

The problem is after the capacitor when I pass the MiniGen signal through a voltage follower. I get another offset. I don't see an offset after the capacitor, but I see an offset after the voltage follower.

Oddly, the offset after the follower goes away by increasing the coupling capacitance to 1uF. Another way to get the offset after the voltage follower is to load the cap with a large resistor about 1MOhm.

Does anyone know why there is an offset after the voltage follower? The follower should just be 1-to-1 with its input.

[Slh]

Sounds like the opamp input bias current.  Opamps are sadly not as ideal as all the tutorials pretend.

I think I get what your circuit is but if you put your circuit with part numbers then I think I could be more help.

[aw_rw1]

The opamp that I am using for the voltage follower is a TL071.

[wasedadoc]

The opamp that I am using for the voltage follower is a LT071.

Is that a typo?  LT071 or TL071?

[magic]

Oddly, the offset after the follower goes away by increasing the coupling capacitance to 1uF. Another way to get the offset after the voltage follower is to load the cap with a large resistor about 1MOhm.

What do you mean by "loading the cap with 1MΩ"?
There must be a resistor to ground after every AC coupling capacitor if you want the output to be centered around ground.

Measuring the output voltage of the capacitor with a typical DMM connects ~10MΩ resistance to ground, which may act as the missing resistor and temporarily bring the voltage towards ground.

[shapirus]

It would be helpful to see schematics, also indicating the points at which you take measurements.

[Bud]

Does the buffer follow the gen's signal excursions? Then it *is* a follower. Local DC bias does not account for following action.

[Bud]

And if you power the follower from a unipolar power supply you cannot expect the follower output to go negative.

[Bud]

By the way, 1uF is awfully small for 1Hz (even 100Hz) signal, you should put more like 100uF

[aw_rw1]

Sorry, there was a typo. The opamp is a TL071.

[aw_rw1]

Attached is the schematic of the circuit that I am working with.

Again, opps, typo again. Opamp is TL071.

[aw_rw1]

By the way, 1uF is awfully small for 1Hz (even 100Hz) signal, you should put more like 100uF

I suppose the capacitance is small. I saw on this site common capacitor sizes:

http://www.learningaboutelectronics.com/Articles/What-is-a-coupling-capacitor

"For coupling a 100Hz signal, a 10μF capacitor can be used."

However, when I used a 10uF without the voltage follower, I saw a huge offset. I tried 10uF with the voltage follower and RHP just now. The offset is nearly 0 VDC, about 0.05VDC.

[gf]

C1 and RHP form a high-pass filter. Therefore the value of C1 depends on the the value of RHP and the desired cut-off frequency.

[timenutgoblin]

Attached is the schematic of the circuit that I am working with.

Again, opps, typo again. Opamp is TL071.

I'm not sure I understand exactly what your circuit is doing.

I think that your schematic has a typo.



https://www.electronics-tutorials.ws/opamp/opamp_2.html

For a non-inverting opamp circuit, the input signal goes to the non-inverting input and the output is connected to the inverting input.

https://www.electronics-tutorials.ws/opamp/opamp_3.html

[Bud]

The output of the op amp has to be connected to the Inverting input. And inpug signal should go to non-inverting pin. Your wiring of the op amp is wrong.

[aw_rw1]

C1 and RHP form a high-pass filter.

Yes, the two components do form a high-pass filter. Ideally, RHP shouldn't be there if C1 was an AC coupling capacitor. RHP is there since we are talking about this thread.

I think that your schematic has a typo.

The output of the op amp has to be connected to the Inverting input. And inpug signal should go to non-inverting pin. Your wiring of the op amp is wrong.

Yes, there is a typo in the schematic. I draw it in a rush. Attached a revised version. The device name and wiring of feedback were corrected.

[magic]

Okay, so what are the values of C1 and RHP?
What's the voltage at the input of the opamp?
And at the output, preferably measured simultaneously with a second meter?

[shapirus]

Yes, the two components do form a high-pass filter. Ideally, RHP shouldn't be there if C1 was an AC coupling capacitor. RHP is there since we are talking about this thread.

Check this: https://electronics.stackexchange.com/questions/206275/do-i-need-to-make-a-dc-path-at-the-input-of-an-op-amp

...or, a little more fundamentally, this: https://www.analog.com/media/en/technical-documentation/application-notes/an-937.pdf

[aw_rw1]

so what are the values of C1 and RHP?

This values are in the spec out phase. At the moment, 10uF and 1uF for C1 seems to work, and 1MOhm and 0.5MOhm for RHP seems to work. I think RHP should be lower to lower any Johnson noise.

What's the voltage at the input of the opamp?

Input voltage ranges from 0.7 to 1.1 Vpk-pk, and the frequency ranges from 1 to 100Hz.

...or, a little more fundamentally, this: https://www.analog.com/media/en/technical-documentation/application-notes/an-937.pdf

This app note is spot-on.
 

[shapirus]

Input voltage ranges from 0.7 to 1.1 Vpk-pk, and the frequency ranges from 1 to 100Hz.

Keep in mind your high-pass filter cut-off frequency when you consider 1 Hz. However if you take 10uF and 1MOhm, it shouldn't noticeably affect your input signal even at 1Hz.