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Supply rail imbalance and DC offset

SinglePoint

SinglePoint

Active Member
I am trying to understand how DC offset is introduced on the output terminal (speaker connection) of a power amplifier when a supply rail imbalance develops. For example, referencing the attached schematic, DC offset is nearly zero, but if the (-) rail fuse F2 blows or is removed DC offset goes to 25vdc.

Can someone help me understand the basics of how that happens, or point me to a resource so I can learn more? I have searched the Web and read some, but still need some help. Thanks.

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Q17 and/or Q18 possibly shorted. Isolate from circuit and check for shorted junctions.
 
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Your amplifier output is your voltage balance point. + 25 - 25 = 0 .. Zero volts is your ideal DC offset.
 
avionic said:
Q17 and Q18 possibly shorted. Isolate from circuit and check for shorted junctions.
Click to expand...
Thanks for your response. Speakers were reportedly buzzing and when I saw the large DC offset I pulled all output transistors and they all tested OK, with no shorts. They could be leaking at operating levels though. It behaves ok (very low offset) with DMMs across the fuse clips watching currents. When the fuse and DMM is removed I see the high DC offset. That is now with all outputs removed.

Compensation circuits are probably just doing their thang, but I do not understand fully how they work and I can't explain the high offset.
 
That is ridiculous. How can you imagine that your amp has any chance to work with only one rail supply???
 
what @avionic said about how its supposed to be zero. Add the positive and negative rail and it should equal zero.

single rail supply amps do exist, but the point the output comes from sits at 1/2 the supply voltage, so on a +70 volt supply, the output is at +35 volts. There is a big output coupling cap to block the DC and make it all work.
 
In a single supply rail amplifier, the "lower rail" is connected to ground. If F2 is removed in this amp, there is no current source for the "upper" differential stage, the collector currents of the "lower" differential stage has no where to go, as well as the "lower" VAS stage, the "lower" driver and power transistors. It's like trying to run a car on one wheel....
 
This amp does not work with one of the fuses removed/blown, of course. The purpose of this thread is to learn why a large DC offset develops at the output. What, in general, in the circuits is happening to cause the offset. I believe the remainder of the amp may be functioning fine, except for a single blown fuse on the channel. I haven't found any other failed components so far... Will keep looking.
 
the offset with one fuse out has nothing to do with component failure, its just how this type of amplifier works. This is totally normal behavior.
 
gadget73 said:
the offset with one fuse out has nothing to do with component failure, its just how this type of amplifier works. This is totally normal behavior.
Click to expand...
Yeah, I am seeing that. With both rail fuses in place, the amp seems to behave just fine and offset is very low. So, without one fuse, the current on that side has nowhere to go. I follow that, but how does the offset end up on the output? Maybe it's obvious to y'all, but I dont get that part. Thanks.
 
current flow through the output transistors pulls the output up or down, depending which rail the transistor is connected to. When both rails are powered, it pulls equally in both directions so its at zero. When there is no supply on one side, there is nothing to pull against whichever side is powered, so it ends up close to rail voltage.

think of it like a tug of war. If you have equally strong people on either end, the rope doesn't move. Take all of the people away on one side and the rope just goes towards whoever is still there.
 
gadget73 said:
current flow through the output transistors pulls the output up or down, depending which rail the transistor is connected to. When both rails are powered, it pulls equally in both directions so its at zero. When there is no supply on one side, there is nothing to pull against whichever side is powered, so it ends up close to rail voltage.

think of it like a tug of war. If you have equally strong people on either end, the rope doesn't move. Take all of the people away on one side and the rope just goes towards whoever is still there.
Click to expand...
Okay, got it. And, thanks for the explanation. Since all of the output transistors have been removed and I see the same result (high offset at the output), I will guess that the lopsided push-pull is also happening in earlier stages and hitting the output through the feedback circuit?

Thanks again. I am going through this thing as an exercise to help understand and learn how it works, and refreshing it along the way. So, this info is very useful to me. I appreciate you sharing.
 
the driver stage of Q14/Q17 also ties to that mid point, so does D10 and D11 so those will pull the output line one way or the other as well, just not with a whole bunch of current capacity.
 
Transistors are switches, and mainly amplify current. Audio transistors are designed to have a long and fairly even switching pattern from full off to full on to allow for volume changes in amplification - at least those in vintage audio equipment. Class D is different, so this explanation isn't useful there.

What happens in a complimentary or quasi complimentary system is that the transistors on each "side" (plus and minus) are operated with some minimum current flow -- in the mA range -- for a couple of reasons. First, it takes time to switch a transistor on from zero, and secondarily it is necessary to have enough current flow through the complimentary pair at the output will transition from positive sine wave to negative sine wave froms without a "step" as the slower transistor (NPN I think, but I'm dylexic and confuse them all the time) switches on. Manufacturers specify this bias current, usually as mV across the emitter resistors of the output. It can also be set with a sine wave generator and an oscilloscope, which is probably better.

What that means when a rail goes off (blown fuse or something similar) is that the pair is STILL conducting, only with one side at ground potential, hence with no connection between the speaker plus and minus (ground) terminals, rail voltage will "leak" through at very low current flow. The speaker protection circuit will open the speaker relay or ground the input to the amplifier to shut off current flow as DC at any significant level and current will certainly overheat voice coils. The drivers and outputs with functioning rail will still have 1V or so on the base and will conduct, but at low current flow. Which is good, else they would fry.

Center voltage (or offset) is set by the differential pair resistors at the first stage in the amplifier. Some amps are adjustable, some are not and rely on carefully matched transistors in the differential pair for offset control. Offset adjustments will have no effect on an amp with one rail open,

Note that output transistors and usually the drivers are connected between the rail and the output, usually with only emitter resistors in the circuit. The emitter of the driver will be connected to the base of the output transistor. With both rails hot, the potential of the center of the output emitter resistors should be zero volts with some mA level current flowing from the positive to the negative rail (or actually negative to positive since that's the actual electron path I think). Take away one rail voltage, and the current flow will be zero, but the opposite rail voltage will appear at the center of the resistors!
 
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HPMguy said:
Transistors are switches, and mainly amplify current. Audio transistors are designed to have a long and fairly even switching pattern from full off to full on to allow for volume changes in amplification - at least those in vintage audio equipment. Class D is different, so this explanation isn't useful there.

What happens in a complimentary or quasi complimentary system is that the transistors on each "side" (plus and minus) are operated with some minimum current flow -- in the mA range -- for a couple of reasons. First, it takes time to switch a transistor on from zero, and secondarily it is necessary to have enough current flow through the complimentary pair at the output will transition from positive sine wave to negative sine wave froms without a "step" as the slower transistor (NPN I think, but I'm dylexic and confuse them all the time) switches on. Manufacturers specify this bias current, usually as mV across the emitter resistors of the output. It can also be set with a sine wave generator and an oscilloscope, which is probably better.

What that means when a rail goes off (blown fuse or something similar) is that the pair is STILL conducting, only with one side at ground potential, hence with no connection between the speaker plus and minus (ground) terminals, rail voltage will "leak" through at very low current flow. The speaker protection circuit will open the speaker relay or ground the input to the amplifier to shut off current flow as DC at any significant level and current will certainly overheat voice coils. The drivers and outputs with functioning rail will still have 1V or so on the base and will conduct, but at low current flow. Which is good, else they would fry.

Center voltage (or offset) is set by the differential pair resistors at the first stage in the amplifier. Some amps are adjustable, some are not and rely on carefully matched transistors in the differential pair for offset control. Offset adjustments will have no effect on an amp with one rail open,

Note that output transistors and usually the drivers are connected between the rail and the output, usually with only emitter resistors in the circuit. The emitter of the driver will be connected to the base of the output transistor. With both rails hot, the potential of the center of the output emitter resistors should be zero volts with some mA level current flowing from the positive to the negative rail (or actually negative to positive since that's the actual electron path I think). Take away one rail voltage, and the current flow will be zero, but the opposite rail voltage will appear at the center of the resistors!
Click to expand...
Thank you for adding much clarity and detail and taking the time to post that response. I have learned a few lessons working through this amp (Acurus A100), including that there can be large DC offset even with well-matched differential pairs and fully functioning components, if one rail drops out or I presume if any substantive asymmetry develops between the rails. Also, a large DC offset can endanger speakers, but does not necessarily kill/smoke the driver/output stage, I presume thanks to the low currents involved. And, I learned that I do not like desoldering components from a multilayer, double-sided board.
:beerchug:
 
A lot of things have speaker protection circuits to keep any DC from hitting the speakers. Its more there in case an output shorts and gives full power supply current but any DC will trip them out. Not everything has it though. There are add-on circuits to add this feature though.
 
gadget73 said:
A lot of things have speaker protection circuits to keep any DC from hitting the speakers. Its more there in case an output shorts and gives full power supply current but any DC will trip them out. Not everything has it though. There are add-on circuits to add this feature though.
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I would like to add a proper speaker pro circuit, but would probably need to come up with a lower voltage supply for it (~12v?). Not sure I would be up for that, but I will look into it. I imagine there are various options available. Are there any you would recommend? I believe my BK ST-140 simply has inline fuses... Do you have an opinion about that approach?
 
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