changing biasing pot in rod elliotts p3a

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Hi all!

just a quic question really.
now, i have been trying to simmulate this as well as figuring out for myself:
what effect will changing the biasing pot, VR1, have on the amp?
will it work, or will it screw something up?

i plan on using 10k multi turn pots.
i figure it'l work, since the "bulk" of resistance will be paralelled by the center conductor as a 0 ohm load just leaving "the bottom" of the pot to be relevant resistance..

if you understand what i'm trying to say..

I have found 2k pots to be hard to source here, and i have some other pots lying around, and using them will not only save me from having to purchase pots with cash i dont have, but save me some time too, letting me finnish tonight.

thanks for any help.
Marius
 

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Hi,
go ahead and change the pot to 10k.
I suggest 2 alternatives.
either
1. Change r16 to 4k7, disadvantage is the current through the resistor string is too low and should be kept above 10 times the base current.
or
2. add a parallel resistor to the pot of a value about 2k0, worth trying even lower - down to 1k0, disadvantage is finding a position to add it neatly to the PCB, maybe on the track side?
 
AndrewT said:
Hi,
go ahead and change the pot to 10k.
I suggest 2 alternatives.
either
1. Change r16 to 4k7, disadvantage is the current through the resistor string is too low and should be kept above 10 times the base current.
or
2. add a parallel resistor to the pot of a value about 2k0, worth trying even lower - down to 1k0, disadvantage is finding a position to add it neatly to the PCB, maybe on the track side?

:xeye:
uhh.. well, i'm the jackass here, but i must ask why?
now, the paralelling resistor i can understand, if the 8kohms ekstra would have to be part of the resistor string, but it's not, is it?
I figure the center lead on the pot, the variable one, acts as a straight piece of wire into the remaining 2k's of resistance right?
then why do anything else?

know i'm the noob here, but i just dont see the reason..

thanks
Marius
 
Hi,
I see two reasons for the parallel resistor.

1. The pot will now be acting as a fine adjust on the 2k0 resistor and you will be able to use a wider range giving a less sensitive feel to the adjustment. That was why I thought you might prefer an even lower value of lower leg resistor.

2. The power the pot can carry is probably not an issue here but I will explain anyway.
P=I*I*R. so 6mA through the 10k pot is about 360mW. The power rule for pots is that the track must never see more than the max current that would generate the max power through the whole track. So for a 10k 360mW pot irrespective of what setting it is adjusted to, the maximum current is 6mA.
For long life, stability and tolerance I would recommend a much lower current, say 2 to 3mA.

Are you OK with that?
 
AndrewT said:
Hi,
I see two reasons for the parallel resistor.

1. The pot will now be acting as a fine adjust on the 2k0 resistor and you will be able to use a wider range giving a less sensitive feel to the adjustment. That was why I thought you might prefer an even lower value of lower leg resistor.

2. The power the pot can carry is probably not an issue here but I will explain anyway.
P=I*I*R. so 6mA through the 10k pot is about 360mW. The power rule for pots is that the track must never see more than the max current that would generate the max power through the whole track. So for a 10k 360mW pot irrespective of what setting it is adjusted to, the maximum current is 6mA.
For long life, stability and tolerance I would recommend a much lower current, say 2 to 3mA.

Are you OK with that?

Ah, get it.

concerning fine regulation, the pot in question is a 15 turn one, so no stress there. if however the problem should arise, i'l add a resistor. there's plenty lying around here anyways..

well, i see the benefit of not trying to push to much amperage through a pot, especially not a multiturn one, since tracs are marginal, and heat dispassion becomes a problem possibly leading to a fracture in the carbon resistor\conductor. correct?

and as you say, i dont think either that this will be a problem here, considering the relative low I flowing through the resistor chain.

thanks a lot andrew, cheers!
-Marius
 
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