Offset voltage

What is the recomended minimum DC offset voltage into a loudspeaker element? Both into bass and into treble.

The reason I am asking is that I am building a DAC with digital X-over (more on the particulars later). I will also add volume control to this. I'd like to keep it DC coupled, and there is no guarantee that whatever power amp I connect is AC coupled.

Thanks,
Børge
 
Only the bass element is affected of the offset. All other elements has a cap in series.

I'll guess you'll never get an exact answer but according to me 100-200 mV is nothing to worry about even though I prefer lower than 50 mV.

This is very dependent of how big the bass element is and the power handling.

If the power amp also is DC connected and can amplify DC you must take care of the offset and an easy to eliminate the offset is to use a DC servo.
 
I've seen amps that trip the protection circuit at 800mV :bigeyes: so that designer thinks that's not too high. But I agree with peranders, 50mV or less should be good.

I know you want it DC coupled by why not use 1 cap in the input and it would solve a lot of your problems. It's not like our running several GHz through it. :hot: (or are you?)🙄
 
opamp input offset voltage measurement

One approach you could try is diagrammed below. You might want to also include a ceramic bypass capacitor between pins 4 and 7, about 100 nanofarads (0.1 microfarads).

The output at pin 6, equals the opamp's input offset voltage, times the gain of the overall circuit including negative feedback. This overall circuit has a gain of +1.000, so its output equals the input offset voltage of the opamp. Simple. It'll be somewhere in the vicinity of 500 microvolts (0.5 mV), according to the OPA134 datasheet.

If your meter is unable to measure microvolts accurately, you might need to change the circuit design. For example, if the circuit gain were 100X, then the output at pin 6 (where your meter is connected) would be somewhere in the vicinity of 50 millivolts. The opamp's input offset voltage is (meter_value / circuit_gain) = (50 mV / 100) = 0.5 millivolts. As before. But now the meter is being asked to measure a voltage 100 times larger.

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