Hi all,
I have attached an image of the circuit I have built - it is a simple non-inverting amplifier used as a preamp for an electret microphone. I tested the circuit once on a bread board and everything functioned properly. However now that I have built it, the output of the amplifier drifts to either the +12V or -12V supply. This DC offset occurs when there is no input to the opamp and also when I have the microphones attached.

The opamp in use is an LME49740, and the 500k resistor (R1 in the image) is a trim pot. If I turn the gain down via the trim pot, the dc offset in the output seems to decrease. Any ideas as to what could be causing this?

 

I would check the voltage at the op amp (+) input and the connections of R1, R2, and R3.

 

The DC gain seems awfully high. A small amount of drift at the input of the op-amp will cause a level shift. You can try putting a capacitor in series with the 280 ohm resistor to see if the drift goes away.

Unfortunately, this has to be a very large value to have an impedance small in relation to the 280 ohms. At 20 hertz, I calculate that a 30 uF cap would be needed. Because of the bipolar power the capacitor cannot be polarized. For testing you can try two electrolytic caps in series with the negative ends tied together.

 

I would check the voltage at the op amp (+) input and the connections of R1, R2, and R3.

Agreed. I think there is an open between +input and R3 to ground.

Mark

 

Agreed. I think there is an open between +input and R3 to ground.

Yes. Better than my theory and a lot easier to fix.

 

Checked the resistors, they all seem to be wired correctly and are making a proper connection.

When I probe the (+) input with my multimeter I read a DC voltage of ~-7 mV. So I would assume the DC voltage is somehow present on the input is getting amplified to the point of reaching the supply rail voltage, causing the DC offset that is at -12V.

Could the DC voltage at the input be present because of the input offset voltage? The capacitor should block any DC from entering the amplifier, and this is the only thing I could think of that would cause a DC voltage at the input.

 

One thing you might try is to make the 1 meg a 100k.

 

It is very unusual to see a good working opamp circuit with a gain as high as this in a single stage, and I don't think I have ever seen an audio amplifier with this much gain in a single stage.

You would do well to reduce the DC gain such as with RichardO's excellent suggestion AND reduce the input resistance as suggested by ronv.

It would be better to also reduce the signal gain in the first stage otherwise your amplifier will be come virtually open loop at audio frequencies, if that's important to your application. The conventional approach is to use a preamp with low value resistors followed by one or more gain stages (again, depending upon your application), but

By the way, what are you using as an opamp?

 

The OP said he successfully bread boarded this circuit. He has now built it into a assembly and is having some problems with it. Other than the feedback trim pot being too high I see no reason why this circuit should not work. (I doubt he is running at a gain of >1700.)

 

The input offset current of the amplifier is 11na typical across 1 meg is 11mv, X gain of say 1000 = 11 volts. It could be much worse. Throw in some leakage current for the cap and some offset voltage and I could see it drifting up with the IC temperature. Both the cap and the IC would benefit from a lower input resistor with very little loss of amplitude from the microphone.

 

ronv, I agree with your math. But, I don't agree that he is running a gain of 1000. I pulled out some electret mic elements and powered one up. Normal speaking voice at 12 inches away produced 30 Milli-volts output. To amplify this up to a line level of 1 volt only requires a gain of 33.

- Would the op-amp work better with a much lower input resistor? Absolutely
- Should the circuit work, as is, at gains of less than 100? Absolutely
- Is it possible the OP used a 500K pot instead of a 20K pot because that's all he had? That's my bet.

Mark

 

Most of them I have used only output a couple of mv from a foot or 2 away, but having said that the gain is out of sight. But maybe it's a leak detector or something.

 

So I replaced the 1 meg resistor with a 100k resistor per everyone's suggestion and that seemed to lower the dc offset of the output to ~2V. I then tried a 10k resistor in place of the 100k resistor and it lowered the offset even more. Anybody have an explanation as to why that is?

And just to clarify I have the gain set to roughly 300 with the trim pot - I only used the 500k pot because that was all I had.

 

There are 2 possible culprits.
The op amp you are using has an input bias current of 72 nano amps max. That current will flow thru the resistor and develop a voltage - .0072 volts. That gets amplified times your gain - 300 to 2.16 volts max.
The other possibility is the cap between the mic and the circuit. It will also have some leakage current that will put a positive voltage on the input that also gets amplified.
You can reduce the effects by using a FET input op amp - they have very low bias current.
You can also use film capacitors instead of electrolytic to reduce the capacitor problem.

 

ronv made some excellent suggestions. I would like to expand on one of them. You changed the input resistor from 1meg to 100K and the output drift dropped to 2 volts. I suggest that you drop the input resistor again to 10K. This should drop the effect of the input current to .2 volts on the output. You can do this with very little effect on gain because the impedance of electret mics are around 1K or less. (I think ronv was implying this change in his comment.)

With a 10K input resistor I would change C1 to a 10uf metal film capacitor. With the original 1uf and a 10K resistor you will have a 100Hz roll off. 10uF will bring that down to 10Hz.

The op-amp you are using is an excellent audio amplifier with very very low noise. It has lower noise density than the legendary 5534a. It can also drive 600 ohm loads. I would try my best to make this op-amp work.

Mark

 

Thank you for the responses, you all have been very helpful. Dropping the input resistor to 10k appears to have resolved my issue.

One last thing I have noticed - it appears that the drift on the output seems to slowly move toward -12V once I turn the gain up to unusually high levels (a gain greater than 1000). I am concerned that my output might be slowly drifting toward the negative supply rail, and that the huge increase in gain might be making it more noticeable. Would this be a result of instability due to such a high gain, or possibly some sort of feedback issue where a small voltage is making its way back to the inverting input? My only experience with op amps is with the "ideal" case in class, so any help with these "real world" issues is again greatly appreciated.

 

NonLineon
Generally any op-amp will exhibit some instabilities with extremely high gains. The data sheet for this op-amp does not have any applications that have a gain over 100. Running at a gain of 1000 will amplify every bad feature of the op-amp to the point of non-functionality.

Why do you need a gain of 1000 with an electret mic? At a gain of 1000, even if the op-amp were to function perfectly, the output would be dominated by the thermal noise of the mic preamp.

Mark

 

I do not intend to use the opamp with a gain of 1000, the drift was just something that I noticed when I would adjust the pot to find a usable gain. I just wanted to make sure that the same drift was not also occurring when I was using a lower gain, and that the higher gain made it much more noticeable.

Based on your answer, Lestraveled, it seems like the drift was most likely caused by some form of high gain instability which is fine as I don't intend to use the opamp at such a high level. Thanks again for your help, I have really learned a lot

 

As has been memtioned above, the DC gain is excessive.
You can easily overcome that. In the attached circuit simulation, DC gain is reduced to 2 by use of resistor R6. This is bypassed by C2, so the AC gain is effectively determined just by R1, R2. Here, AC gain is ~100 and is flat from ~10Hz - 20kHz. Note the deliberate introduction of a 5mV input offset voltage. This results in a ~10mV output offset, which is stable with temperature.

 

Alec_t,
I appreciate your response. Seems like a very simple solution to the problem I was having. I have been using the circuit successfully by lowering the input resistor to 10k, however I have come up with an alternate circuit is based off of your suggestion of lowering the DC gain.

The difference is that in the attached circuit there is really no DC gain, as there is no path for the DC to take that is parallel to capacitor C4. I have breadboarded the circuit successfully, however I am curious if there are any reasons to chose the layout you suggested (where R6 and C2 are in parallel) vs my layout (where there is no "DC path" parallel to C4). Thanks again for the help.