Replacing 50K pot with 10K pot

Hello,

I am replacing a 50K pot in my integrated amplifier with a DACT 10K pot.

The pot I am replacing is one of those fake Chinese Alps stepped pots that was on E-bay a couple of years ago, the ones with individual resistors.

So, I imagine I will not have the range I need with the new pot and my music will be too loud even on the first couple of clicks of the new DACT pot.

Now to correct this, could I just add a 40K ohm resistor on the output side of the new pot? Is it that easy?

There is already a resistor on the pcb at the output of the pot, it feeds a capacitor. Should I series the 40K resistor with this resistor? Or just replace this resistor with one of its value plus 40K ohm?

Any advice would be appreciated.

Also is there anyone that has some good quality resistors in the 40K range for sale?

Thanks

Kevin
 
KevinLee said:
Hello,

I am replacing a 50K pot in my integrated amplifier with a DACT 10K pot.

Kevin

Someone smarter than me propably have a better explanation than i can give, but Your pot is just a voltage divider.
Then, a 50K and a 10K will ( theoretically ) give the same results.

But... There might be an issue with the impedance at the place Your pot belongs. I guess there's a reason the manufacturer put in this value.

best regards
Ebbe
 
Why did you remove "fake" Alps pot? It is better than original Alps and almost same as DACT.

If you put a 10K instead of 50K, you will probably have a lower input impedance of integrated amp becouse integrated amps are almost always designed as power amps with selector and passive pre (atenuator) in front of power section.

Very small number have a buffer stage before attenuator.
 
KevinLee said:
So, I have been doing some more thinking on this, It would probably make the most sense to put a resistor at both the input and output of the pot?

Any thoughts?

It will work in a way that you will increase input resistance, but you will loose gain. Adding 10k resistor before the 10k pot will lower signal by 6dB.

Depending on your source, you may not need any adjustements at all and 10k pot will work equally well as your 50k.
 
Peter Daniel said:


It will work in a way that you will increase input resistance, but you will loose gain. Adding 10k resistor before the 10k pot will lower signal by 6dB.

Depending on your source, you may not need any adjustements at all and 10k pot will work equally well as your 50k.


So, I put the 10K DACT into the amplifier and everything seems OK. I have to play a few more CDs to confirm that the sound did not get worse. I also have to re-bias and then re-listen.