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Sherwood Power Amp Crossover Distortion

KilgoreCemetery

Apr 12, 2017
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This has been an ongoing project for quite a while, but long story short, I had to replace all of the output transistors (2SK1058 & 2SJ162) with modern ones (IRFP9240PBF & IRFP240PBF). The pinouts are different, so I had to twist the legs to get them to connect properly and was very careful to make sure that none of the legs touched anything they weren't supposed to.

Now, it powers on and comes out of protect mode fine, but there's crossover distortion on both channels. I'm guessing that the new transistors are just different enough that I need to adjust the bias, but there are only two trim pots in the whole unit and, while I can adjust them to get rid of the crossover distortion, it also raises the DC voltage at the speaker connects to the point where it would damage speakers (upwards of 500+mV).

How do I go about correcting the crossover distortion?
 

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duke37

Jan 9, 2011
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There is a diode D107R? which I think controls the voltage between the gates of the output transistors. You could try another diode in series to raise this voltage but check dissipation.
 

KilgoreCemetery

Apr 12, 2017
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It took me a while to find it, but I did add another diode to one side only. Unfortunately, I can't tell any difference between the two channels. The original is a 1n4148. Do I need to have another one of those, or is there something similar that I can use. I do have some 1n4007's on hand, but I'm guessing they aren't gonna work quite the same.

From what I've been reading, the goal is to keep the output transistors in a state of "almost on" so that they can switch on/off quicker. I think that the on voltage for the new mosfets is supposed to be between 2v - 4v and I'm showing 2.55v and 3.28v on the gates. Do I need to try to balance the voltage between the push/pull better, or do I need to increase it slightly, or both?
 

BobK

Jan 5, 2010
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With BJTs 1 diode per transistor gives you about the right drop (2 for single output transistors, 4 for darlingtons). With MOSFETs it is anybody's guess how much voltage you need between the two gates. Most MOFSET output stages I have seen incorporate a pot to adjust that voltage, since the gate thresholds are not well controlled. The idea is to get the two output transistors to be slightly on (say 1 to 10 mA) so that there is no dead zone when crossing zero. Adding resistance is OK with MOSFETs. The reason diodes was typically used with BJTs was so that they could be placed on the same heatsink with the output transistors and thus kept at the same temperature so that the forward drops tracked each other as they change with temperature. With MOSFETs that is unnecessary since the gate threshold is not so dependent on temperature.

Bob
 

KilgoreCemetery

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That makes sense, BobK. I did notice a non-populated spot on the circuit board right around D107R that was for a trim pot. I suppose the earlier releases of this model must have had them, then they started using diodes to save cost. Either that, or maybe the board was used in higher models that needed the trim pots. Any ballpark guess as to what resistance trim pot might work?
 

duke37

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I do not think that the difference in the gate/source voltage is important. It will affect the static output voltage but you should be able to adjust this to zero.

Any silicon diode will do, IN4148 or 1N4007. Rather than a string of diodes you could use a zener or has been suggested a resistor. Try a variable 10k, starting with zero resistance and gently working up. Measure the voltage across an output source resistor to get the current.
 

BobK

Jan 5, 2010
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I think it would be in the range of a few hundred ohms, but this is only a guess. Have you compared the gate threshold voltage of the original MOSFETs as opposed to the new ones you installed? If my theory is correct, the new ones should have a higher threshold.

Bob
 

Ylli

Jun 19, 2018
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Got a model number on this Sherwood Power Amp?



Oh-oh. Did a little searching of spice models.

2SK1058 Vto = 0.4 volts
2SJ162 Vto = -0.08 volts

IRFP9240 Vto = -3.8 volts
IRFP240 Vto = 4.0 volts

Not a good choice for replacements.
 
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BobK

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Now the curcuit with a single diode makes sense. One diode drop will have both MOSFETs on. You need seven volts for your replacements.

I would look for something else at this point.

Bob
 

Ylli

Jun 19, 2018
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Still curious about the model number of this amp. I see no feedback, either AC or DC. That's a bit unusual.

The VAS is probably running in the 3 - 5 mA range. You could try adding a 2K or 5K pot in series with D107. Start with the lowest resistance and increase it until you see 10 mV or so across any of the output stage source resistors. If things start to get hot, back off.
 
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KilgoreCemetery

Apr 12, 2017
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Sorry for the slow replies, I was out of town for a few days. This is a Sherwood AM-7040 power amplifier. I'll be going down the troubleshooting list as soon as I can, but this part definitely stands out:
2SK1058 Vto = 0.4 volts
2SJ162 Vto = -0.08 volts

IRFP9240 Vto = -3.8 volts
IRFP240 Vto = 4.0 volts

Not a good choice for replacements.

*sigh* This has been a learning project and I remember asking before I bought them whether they would be adequate replacements and was evidently given some bad information. Live and learn, I suppose.

So, is Vto the same as VGS(TH) then? That would make sense with what you mentioned about comparing them, BobK.

You could try adding a 2K or 5K pot in series with D107. Start with the lowest resistance and increase it until you see 10 mV or so across any of the output stage source resistors. If things start to get hot, back off.
I would have to look around for a 2K or 5K pot, but I do have a brand new 10K one I could use for testing. Would that work, or do you think it would step up too quickly?

I would look for something else at this point.
I kinda hate to dump more money into this, but frankly, a simpler solution would be best. I'll see what I can find that would be a better match. Any recommendations?
 

Ylli

Jun 19, 2018
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So, is Vto the same as VGS(TH) then?
There may be some subtle differences in the way they are measured, but basically yes.

I would have to look around for a 2K or 5K pot, but I do have a brand new 10K one I could use for testing. Would that work
Yes, just be sure to start at the lowest resistance - and it would probably be a good idea to power the unit though your trusty DBT.

better match. Any recommendations?
Suggesting replacement parts is out of my range of knowledge.
 
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