# Help understanding balance circuit in Kenwood KR-5600, and how I can replace the pot

#### bdondo

I have a Kenwood KR-5600 receiver, which was my dad's while I was growing up. I'm trying to fix it up as it's one of the few things I have of his. The balance pot's shaft was broken off a long time ago, probably by me as a kid, and it's proving to be the hardest replacement to find.

It's 200K, 17mm wide, 30mm shaft length, knurled, and linear taper. Going by the part number description from Bourns, I believe the theoretical replacement part number would be PDB181-K230K-204B (or PDB181-K430K-204B would work too). As far as I can tell, this doesn't exist, and I haven't found anything like it online anywhere.

So I thought maybe I could use a 100K pot instead. My circuit analysis skills faded long ago, but I tried to figure out what effect changing the pot would have on the amplitude of the signal reaching the first transistor in the power amp. I'm hoping someone can chime in and tell me if using a 100K pot instead of 200K pot would be ok. The relevant part of the circuit is below:

I tried drawing a simplified schematic of one channel here, but I'm not totally sure I got it right. In this sketch I'm replacing the 100K volume control with two 50K resistors, and setting the balance to the center, so it's 100K on each side.

I roughly calculated the resistance of all the resistors to the right of the first 50K resistor (from the volume pot) to be about 25KOhms (I assumed the capacitors could be replaced with shorts for this calculation, based on fuzzy recollection from school 25 years ago). If I replace the balance pot with a 100K pot, and set it to center so that only 50K is used, then this resistance becomes 20KOhms.

I also simulated the circuit on circuitlab (before my free trial ran out) and using 1KHz, 1V (peak) signal in, the signal at the base of the transistor is 317mV (peak) using the 200K pot. With the 100K pot, it drops 14.5% to 271mV (peak).

I'm not very confident that my simplified circuit is correct, or if these differences in resistance and voltage I've calculated/simulated are significant. I was thinking they aren't a big deal, but then I read about balance controls in Douglas Self's Small Signal Audio Design book, and now I'm not so sure. It seems that these values could be important in maintaining the right amount of noise, dB loss at center position, etc., but I'm also not sure I fully understand what I just read in that book.

Any help would be greatly appreciated!

#### russc

1 - It would appear to be a pot with a centre tap - am I correct? Count the terminals - are there 4 rather than the usual 3?
Note that connection 3 (ground) goes to the centre of the track AND to the wiper.

At centre position, the wiper matches the position of the centre tap.
Each channel signal has a 100k path to ground.
Move wiper up and lower channel still sees same 100k to ground so is unchanged.
Upper channel sees lower resistance to ground so is progressively muted.

The idea being that rather than raising one channel and reducing the other, one channel is reduced while the other remains constant volume.

2 - is the pot mounted on a pcb or wired? Obviously pcb mount reduces your options although you may be able to hard wire a different pot to the pcb.

3 - centre tapped pots are rare as hens teeth.
It probably also has a centre detent.

Whatever you choose to do, trial it first by linking the pot to the pcb by just dobbing some wires on the track side of the pcb so you are not repeatedly soldering & removing the pot from the pcb which may damage the tracks.
Just wiring a normal linear pot will work, just a little differently.
A 200K pot in its middle position would be electrically identical as the current pot.
In service I expect you will not move it far from centre for subtle changes therefore the difference in operation will be neither here nor there.
30mm spindle appears to be the limiting factor, can you fit an extension to a shorter shaft?

Using the 100K pot, the loss would be greater - you would turn up the volume to compensate.
Add 50K in series with each end and centre position gives the same as original loading. Turning would reduce one channel while raising the other but you would not completely mute one channel.
Adding less than 50K increases the loading at mid position. Worth experimenting. Remember what the remote wiring was for?

What might be worth trying is using another pot for a donor shaft to repair the original pot.
Now there's a thought!

#### bdondo

Thanks russc. Yeah, the pot has a center tap (4 terminals total) and a center detent. It's pcb mounted too. I forgot to attach a pic of the pcb/pot. I'll do that now.
At centre position, the wiper matches the position of the centre tap.
Each channel signal has a 100k path to ground.
Move wiper up and lower channel still sees same 100k to ground so is unchanged.
Upper channel sees lower resistance to ground so is progressively muted.
Thanks, this is what I was thinking here too.

Using the 100K pot, the loss would be greater - you would turn up the volume to compensate.
Yah, that's probably fine. I suppose this extra attenuation would theoretically increase noise that gets amplified later, if my understanding of that stuff is correct.

Just wiring a normal linear pot will work, just a little differently.
A 200K pot in its middle position would be electrically identical as the current pot.
The 100K pot I ordered doesn't have the center tap, so I guess I'd have to do without anyway.

30mm spindle appears to be the limiting factor, can you fit an extension to a shorter shaft?
Oh, I didn't know there was such a thing...

Add 50K in series with each end and centre position gives the same as original loading. Turning would reduce one channel while raising the other but you would not completely mute one channel.
I was thinking of this as well, and it would probably be fine since I never need to completely mute one channel anyway.

As for experimenting, it's going to be difficult because it's very time-consuming to remove this board from the receiver. It also involves several wire-wraps (which I've never done before but have a tool on the way), and the wires are a bit brittle now. I'm hoping I can figure out a good solution without having to try it in the circuit multiple times. That might not be realistic though.

#### bdondo

As for getting a donor shaft, that's a good idea. I've never tried to take a potentiometer apart before, but I suppose it's possible.