
Home  Forums  Rules  Articles  diyAudio Store  Blogs  Gallery  Wiki  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
13th May 2015, 01:50 AM  #1 
diyAudio Member
Join Date: Aug 2010

Ohm's Law application: choosing a resistor
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 
13th May 2015, 02:05 AM  #2  
diyAudio Member
Join Date: Apr 2011

Quote:
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 (400V330V) / 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: (400V350V) / R = 1.25 mA. Then R = (50V / 1.25 mA) or 40k. Can you post your circuit? Last edited by rayma; 13th May 2015 at 02:16 AM. 

13th May 2015, 02:24 AM  #3 
diyAudio Member
Join Date: Aug 2010

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 
13th May 2015, 02:29 AM  #4  
diyAudio Member
Join Date: Apr 2011

Quote:
(430V350V) / 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 Last edited by rayma; 13th May 2015 at 02:32 AM. 

13th May 2015, 02:33 AM  #5 
diyAudio Member
Join Date: Aug 2010

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. 
13th May 2015, 02:33 AM  #6 
diyAudio Member
Join Date: Aug 2010

thank you very much!

13th May 2015, 02:39 AM  #7  
diyAudio Member
Join Date: Apr 2011

Quote:
any difference there can throw things off by this much easily. 

13th May 2015, 02:54 AM  #8 
diyAudio Member
Join Date: Aug 2010

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.

13th May 2015, 03:05 AM  #9  
diyAudio Member
Join Date: Apr 2011

Quote:
The one right after the rectifier, before the inductor, is hit the hardest. Last edited by rayma; 13th May 2015 at 03:08 AM. 

13th May 2015, 03:22 AM  #10 
diyAudio Member
Join Date: Aug 2010

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. 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Help understanding a few Ohm's law calculations  jmillerdoc  Tubes / Valves  6  2nd January 2010 05:42 PM 
Where is the error with this ohm's law calc?  riotubes  Parts  19  24th November 2006 05:33 PM 
need some help with Ohm's lawfast!  mr mojo  Tubes / Valves  4  30th September 2005 10:55 PM 
Ohm's Law 101  Steve Eddy  Everything Else  49  21st November 2003 12:11 PM 
Calling for help with ohm’s law  chris ma  Everything Else  2  26th May 2003 07:12 PM 
New To Site?  Need Help? 