DC on op-amp input, audio amplifier

Question

I have built the following audio amplifier circuit on a breadboard. The circuit works. However the circuit makes a DC offset on the non-inverting input of the op-amp.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

I get the input from a volume control circuit. If I measure the volume control output without connecting it to the amplifier, i measure a DC offset < 10 mV (Still after C1 and with R1).

If i connect the volume control to the amplifier, i am able to measure a DC offset of 350 mV on the non-inverting input of the amplifier.

Naturally if i use the op-amp as a buffer, and use the amplifier without feedback, i will have some DC-offset on the output of the amplifier. I would like to use the feedback to correct this DC-offset. This is not possible, with DC on the input.

Any hint on what i am doing wrong, and where i should start to debug?

Ne5534 is not unity gain stable, have you got a compensation capacitor? — Colin, Jun 25 '19 at 21:15
No i did not, and i did not have any of that size. I tried replacing it by a NE5532, and the seems much better now! — keffe, Jun 25 '19 at 21:55
I'd have supplied the source to the (-) input via a managed resistor value and used another managed resistor value for the NFB from the output (also back to the (-) input.) And I'd have just left your (+) input grounded via \$R_1\$ as shown. This way I can also set a fixed voltage gain greater than one. (I'd probably also add a zero in the NFB.) — jonk, Jun 25 '19 at 21:56
Just for testing (I know what it will do to the frequency response): could you replace C1 for a 220n (or larger) polyester capacitor? — devnull, Jun 25 '19 at 23:03
@keffe, that makes sense, the 5532 is internally compensated for unity gain, I suspect it was oscillating before. — Colin, Jun 26 '19 at 07:28

TimWescott · Accepted Answer · 2019-06-25T22:03:28.547

3

Always read the datasheet. The NE5534 has an input bias current of 500-2000nA. (500nA)(22k\$\Omega\$) = 11mV. So you're getting just about exactly the expected voltage across R1. If you put a resistor between your output and the inverting input of the op-amp, then it'll have the same bias current, \$\pm\$ the input offset current (which is 20nA up to a fairly hefty 300nA, so you may not win if you need better DC offset than that).

Here's a multi-purpose canonical connection for an op-amp with bias current. You need to choose R1, R2 and R3 so that R3 is equal to the parallel combination of R1 and R2. An alternate way of thinking of it is that the DC resistance to ground from the inverting and non-inverting inputs needs to be equal -- and you need to take the output of the amp, and the outputs of any pure voltage sources as "ground" for the purposes of the calculation.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

edited Jun 25 '19 at 22:03

answered Jun 25 '19 at 21:23

TimWescott

44,867
1
41
104

1

This would explain a small offset, but what i measure is far larger. I tried replacing the opamp with a NE5532, and it seems better now. The NE5532 has a smaller input bias current, but not that much smaller. – keffe Jun 25 '19 at 21:59
@keffe I agree. Although the explanation is perfectly valid it doesn't seem to explain the difference you see when you connect the volume control. – devnull Jun 25 '19 at 23:11
1

I missed the part about the offset rise with the volume control. I'm going to leave the answer here because a lot of people get this wrong. However -- the real answer is probably that the part isn't unity-gain stable, and when you change the load on the non-inverting input it starts oscillating, probably at a high enough frequency to not affect the other parts of the circuit. The solution to that is to either use a compensation capacitor, or to replace it with a unity-gain stable part. – TimWescott Jun 25 '19 at 23:49

score 2 · Answer 2 · answered Jun 26 '19 at 05:36

The following is a template you might consider:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Your current source is replaced by \$R_{15}\$ and the bootstrapping capacitor \$C_2\$. This is a common approach that is worth some study. If you have any questions about how it works, just ask. But think about it, first.

I've included a similar \$V_\text{BE}\$ multiplier. But I've included a way to adjust it for a quiescent current you can live with. Also, the \$V_\text{BE}\$ multiplier includes a resistor, \$R_{11}\$, designed to compensate for the Early Effect.

Instead of Darlington arrangements, I've chosen Sziklai. This choice improves the thermal behavior. (Keep \$Q_3\$ and \$Q_4\$ thermally isolated from the main driver BJTs, \$Q_1\$ and \$Q_2\$.)

\$C_3\$ and \$R_{17}\$ provide a useful zero in the feedback loop; which includes also \$R_{16}\$ and \$R_{18}\$ (setting the fixed voltage gain.) I've set the voltage gain to be about \$A_v\approx 12\$. But feel free to adjust that, per your own requirements.

There are a few more components (not shown) that may help out. But this is already complex enough. So I stopped here.

score 0 · Answer 3 · answered Jul 06 '21 at 05:34

I have found a very simple solution to boostered opamp audio amplifiers:

the output stage is a common emitter amplifier stage which is controlled by the currents through the power rails of the opamp. This allows for nearly rail-to-rail output voltag swing. R8 is a fine tuning trimmer of 10 Ohm, to be set that the bias current through the output stage is 50 mA.

DC on op-amp input, audio amplifier

3 Answers3