HK620 balance pot bypass

Hi
Can anyone help? I have an old Harman Kardon HK620 which has duff balance control. Can someone explain how to cut this out of the system as I never used it anyway. Simple terms please as I'm no expert.
Thanks.

If there is something that is stopping one channel from working, or makes it a lot quieter than the other channel, it might not be the balance control.



This is the circuit diagram that I found online.

The balance control seems to be a linear stereo 100 kOhm potentiometer. If you can find any linear, stereo 100 kOhm potentiometer that fits, you could use that. Although the diagram has two separate part numbers for VR501 and VR502, the parts list says they are one item. A stereo potentiometer is just two potentiometers controlled by a common shaft.

If you want to bypass the control, you could just fit four 50 kOhm resistors. You would need one from each slider to each end.


This seems to be the layout of the balance control. The stereo potentiometer has two banks, one is the three terminal closer to the shaft, and the other is the three terminals further from the shaft.

Each bank has the slider in the middle. If you remove the potentiometer and put four fixed resistors where the red lines are, you will have bypassed the balance control with it set to centre.

Electrically it's quite simple. It's not going to be that easy to get to, and if you take out the potentiometer, there may be nothing to hold the knob on.

Diver300 said:

If there is something that is stopping one channel from working, or makes it a lot quieter than the other channel, it might not be the balance control.

View attachment 124900

This is the circuit diagram that I found online.

The balance control seems to be a linear stereo 100 kOhm potentiometer. If you can find any linear, stereo 100 kOhm potentiometer that fits, you could use that. Although the diagram has two separate part numbers for VR501 and VR502, the parts list says they are one item. A stereo potentiometer is just two potentiometers controlled by a common shaft.

If you want to bypass the control, you could just fit four 50 kOhm resistors. You would need one from each slider to each end.

Click to expand...

No need to be that 'technical' - simply put a blob of solder between the top and slider of each half of the pot, and leave the pot in place.

Or, if he wants to remove the pot, do the same but place a 100K resistor between top and bottom where each pot was - this is simply to provide a DC path for polarizing any potential electrolytics on either side of the pot. If there are no electrolytics then the 100K's wouldn't be needed.

Thanks
There are six pins. Do the resisters go from the four outer pins to the two middle pins. I want to be sure I do it right.

Arny1109 said:

Thanks
There are six pins. Do the resisters go from the four outer pins to the two middle pins. I want to be sure I do it right.

Click to expand...

Yes.

Each middle pin will have two resistors connecting to it. Do not have any connection or resistors between the front row of 3 pins and the rear row of 3 pins.

Nigel Goodwin said:

No need to be that 'technical' - simply put a blob of solder between the top and slider of each half of the pot, and leave the pot in place.

Or, if he wants to remove the pot, do the same but place a 100K resistor between top and bottom where each pot was - this is simply to provide a DC path for polarizing any potential electrolytics on either side of the pot. If there are no electrolytics then the 100K's wouldn't be needed.

Click to expand...

It may not be obvious which end the "top" of the potentiometer is, as the top is clockwise for the right channel and anticlockwise for the left channel. Blobbing the wrong end with solder would result in no signal at all.

Diver300 said:

It may not be obvious which end the "top" of the potentiometer is, as the top is clockwise for the right channel and anticlockwise for the left channel. Blobbing the wrong end with solder would result in no signal at all.

Click to expand...

But easily spotted, and obvious what was wrong - and it's no more difficult to spot than deciding which is the slider. It's also dead simple to tell, as the two ground pins will be shorted together, plus they go to ground

Great, thanks a lot for your advice everyone.

Great, thanks a lot for your advice everyone.

Hello again,
I did as instructed, just put it back together. Both channels are equal but the overall volume is about 60% less than before. This is what I have done. Any ideas?

Did you remove the pot?.

Yes.
Could the wrong value resistors cause this?

These are the resistors I was supplied.

Those are only 100 ohm, one thousand times too small - hence the low volume. You want Brown, Black, Black, Orange, Brown

As far as I can tell from the circuit diagram, I don't think that you have to have the resistors at all if you are putting both sides to maximum. You can just cut the resistors and it will probably be at full volume.

My suggestion with four 50 kOhm resistors was to simulate the balance control being in the middle. With the links and the 100 kOhm resistor, that would give the maximum volume on both sides.

Thank you again, Diver and Nigel. I will get the correct parts on Monday.
Regards....Brian

Diver300 said:

As far as I can tell from the circuit diagram, I don't think that you have to have the resistors at all if you are putting both sides to maximum. You can just cut the resistors and it will probably be at full volume.

My suggestion with four 50 kOhm resistors was to simulate the balance control being in the middle. With the links and the 100 kOhm resistor, that would give the maximum volume on both sides.

Click to expand...

That's only an incredibly small difference, too small to be noticeable (as it was a linear pot).

But as I mentioned earlier, he 'might' need the 100K to replace the pot if it's used to bias electrolytics - we don't have the rest of the circuit to tell.

I found the circuit diagram here:- https://elektrotanya.com/the_harman_kardon_hk620.pdf/download.html

It's not that easy to read, but I don't think that the resistance of the pot will be needed for bias.

One odd thing is that the 100 kOhm balance pots feed 30 kOhm volume pots, which have 1.5 kOhm load resistors. Those loads are so much lower resistance than the potentiometers that the potentiometers are being used more like variable resistors than voltage dividers.

OK, I've downloaded the circuit now, and as you say the resistors aren't required - so OP, just cut the resistors out, and all will be fine.

The 1.5K resistors are there to make the linear pot into a log one - linear pots are far more accurate across the two halves of the stereo pot, and doing this makes for a better more accurate control.

I didn't see your last posts so I got the correct resistors a couple of days ago, replaced the 100 ohm ones and it's now fine so thanks again for your help. I'd love to know why increasing the resistance increases the volume and taking them out altogether has the same effect but I doubt I'd understand anyway.