This might be a noob question, but here goes anyway.
I've built a simple transistor headphone amp, (DIY IRF610 MOSFET Class-A Headphone Amplifier Project) with output coupling caps. When powered up for testing I measure 4V DC offset on the outputs AFTER the coupling caps. I thought the caps were supposed to block DC and only allow the AC audio signal through?
I've triple checked the circuit and my soldering and everything is fine
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I've built a simple transistor headphone amp, (DIY IRF610 MOSFET Class-A Headphone Amplifier Project) with output coupling caps. When powered up for testing I measure 4V DC offset on the outputs AFTER the coupling caps. I thought the caps were supposed to block DC and only allow the AC audio signal through?
I've triple checked the circuit and my soldering and everything is fine
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The voltage is perfectly normal. Even if the cap is fully charged there could still be sufficient leakage current (normal) to show a substantial reading on a DVM given the meters high input impedance.
Do as Scott suggests and use a resistor in place of the load. That will fully charge the caps and in practice using any resistor value below around 100k as a load will show no voltage as being present.
The omission of such a resistor is actually quite a serious oversight and one should ideally be permanently connected (a 1k would be fine and would quickly charge the cap)
Do as Scott suggests and use a resistor in place of the load. That will fully charge the caps and in practice using any resistor value below around 100k as a load will show no voltage as being present.
The omission of such a resistor is actually quite a serious oversight and one should ideally be permanently connected (a 1k would be fine and would quickly charge the cap)
Thanks a lot guys. I followed your instructions and sure enough, the DC offset disappeared. The ~1k resistor in parallel with the headphone load makes a lot of sense. Now that I know what to look for, it seems a large number of schematics out there with AC coupled outputs lack this resistor. I wonder how many people have damaged their headphones when the accumulated charge suddenly dumps when they're connected.
That resistor is not optional. It is necessary for capacitor coupled line level inputs and outputs too.
Now you know what that resistor does because you observed and tested it for yourself. That's the best way to learn.
The output resistor is needed to charge up the coupling capacitor at start up and to discharge the coupling capacitor at switch off.
Don't use your headphones to charge or discharge that coupling capacitor.
Insert after switch on and remove before switch off. Let the resistor do it's job.
If you can wait a few seconds longer then you can adopt a resistor that is 100times the nominal rated load impedance, i.e. for 8ohms to 16ohms speakers/headphones use a 1k resistor.
For 300 to 600ohms headphones use 20k
Don't use your headphones to charge or discharge that coupling capacitor.
Insert after switch on and remove before switch off. Let the resistor do it's job.
If you can wait a few seconds longer then you can adopt a resistor that is 100times the nominal rated load impedance, i.e. for 8ohms to 16ohms speakers/headphones use a 1k resistor.
For 300 to 600ohms headphones use 20k
Probably 0; I don't think the energy is large enough to actually blow up a headphone. At least I don't believe it will cause damage due to overheating, maybe mechanically something might go wrong with very feeble headphones.
With a bias voltage of 10 V and 680 uF the energy dissipated per channel is (1/2) * C * V^2 = 34 mJ. The power rating of headphones is usually between 50 mW and 5 W per channel, so this energy is less than what a headphone dissipates in a second of normal use at high volumes. I guess the thermal time constant of the voice coil is much larger than that, at least a couple of seconds (but that's just a guess).
Nonetheless, it is definitely good practice to include a bias resistor.
I built this amplifier and did include the 1k resistor as suggested to get rid of the 4V DC offset after the output cap. I have included the exact circuit I built below. But the I still measure 40mV DC offset. Why am I still having DC offset ?
An externally hosted image should be here but it was not working when we last tested it.
Assuming that you have waited for at least ten seconds before taking the measurement, so the RC network has had time to settle, it could be electrolytic capacitor leakage current. Does it decrease slowly over the first hour after turn-on? If so, when you try it again the next day, does it immediately start with a low value?
When I immediately measured the DC offset after turn on, it started at 135mV and slowly started decreasing. Then after 5 mins the DC offset is stable at 40mV.
The electrolytic is passing 135uA initially reducing to ~40uA after some time to show an offset of 40mVdc across the 1k0 resistor.
You need to reform the electrolytic to reduce the leakage.
You may find that the electro needs to be reformed periodically (one every couple of years) to keep the offset manageable.
Edit:
forget this next bit, the circuit is single polarity.
You can trim the output offset in this CCS loaded single ended amplifier.
The 10r resistor is passing ~125mA when the mosFET loads it.
You can add a VR across the 10r. Try a 200r VR and see how the offset responds.
If it goes the wrong way, then change the 10r to 11r and remeasure offset.
You need to reform the electrolytic to reduce the leakage.
You may find that the electro needs to be reformed periodically (one every couple of years) to keep the offset manageable.
Edit:
forget this next bit, the circuit is single polarity.
You can trim the output offset in this CCS loaded single ended amplifier.
The 10r resistor is passing ~125mA when the mosFET loads it.
You can add a VR across the 10r. Try a 200r VR and see how the offset responds.
If it goes the wrong way, then change the 10r to 11r and remeasure offset.
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indeed it's not excessive.
But one can probably reduce that to <10% by reforming the electro.
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