Ohm's Law application: choosing a resistor

[centervolume]

Hello all-

Can someone demonstrate the way to solve this with Ohm's Law?

I have a 56K resistor between 2 power supply caps. On the downstream side, the voltage is 330 volts. What value is needed to get the downstream voltage about 350?

please show work, I'll need to apply this again

 

[rayma]

Hello all-
Can someone demonstrate the way to solve this with Ohm's Law?
I have a 56K resistor between 2 power supply caps. On the downstream side, the voltage is 330 volts.
What value is needed to get the downstream voltage about 350?

You need the upstream voltage also, to get an answer.
The load circuit will usually draw a different current at a different voltage, so the answer will be approximate,
but probably close enough.

Say the upstream voltage is 400V. Then the resistor current is (400V-330V) / 56k, or 1.25 mA.
If we assume that this current draw by the load circuit stays about the same when the voltage changes a little,
then to get 350V, we have: (400V-350V) / R = 1.25 mA. Then R = (50V / 1.25 mA) or 40k.

Can you post your circuit?

 

[centervolume]

apologies, I just realized that. The voltage drop is approximately 100 volts. So 430 to 330 across the 56K resistor.

I did a ratio type approach, took the target voltage (350) as 80% of the total drop.

Divided the resistor in 5ths = approx 11K (x5=55K), then took 4 of those 5ths to get the 80%: 4 x 11 = 44K

But I don't know how well this squares with Ohm's law

 

[rayma]

apologies, I just realized that. The voltage drop is approximately 100 volts. So 430 to 330 across the 56K resistor.

I did a ratio type approach, took the target voltage (350) as 80% of the total drop.

Divided the resistor in 5ths = approx 11K (x5=55K), then took 4 of those 5ths to get the 80%: 4 x 11 = 44K

But I don't know how well this squares with Ohm's law

(430V-330V) / 56k = 1.79 mA current through the 56k

(430V-350V) / R = 1.79 mA what you want by changing the 56k to a different value

R = (80V / 1.79 mA) = 44.7k for the new resistor value

 

[centervolume]

Re the circuit: it is a Fender Super Amp 5G4 (1960).

http://ampwares.com/schematics/super_6g4.pdf

It has a replaced PT that is not delivering enough b+, so I am modifying the power supply to get a few of the stages what they are looking for.

 

[centervolume]

thank you very much!

 

[rayma]

Re the circuit: it is a Fender Super Amp 5G4 (1960).
http://ampwares.com/schematics/super_6g4.pdf
It has a replaced PT that is not delivering enough b+, so I am modifying the power supply to get a few of the stages what they are looking for.

Should be ok. Check and make sure that your AC line voltage is equal to the rated voltage by Fender,
any difference there can throw things off by this much easily.

 

[centervolume]

yes, indeed. Between the higher wall voltages in 2015 versus 1958 and the less powerful PT, I am in a position to get most of the stages within 5% of the original 1958 voltage specs, so that's kind of promising.

 

[rayma]

yes, indeed. Between the higher wall voltages in 2015 versus 1958 and the less powerful PT, I am in a position to get most of the stages within 5% of the original 1958 voltage specs, so that's kind of promising.

Also, a weak 5AR4 could drop some extra volts. Check the related caps near the rectifier as well.
The one right after the rectifier, before the inductor, is hit the hardest.

 

[centervolume]

I have a silicon rectifier plug in there now.

I see the 2x 20@600 between the rectifier and the choke... they are all new electrolytics in the power supply now. Someone had removed the 2nd one of those 2 caps and just put one in there but 40uf instead, which I suppose is the same. I put in 2 x 30uf instead - not liking to see the empty eyelets under there.