Interesting math conundrum. - diyAudio
 Interesting math conundrum.
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 11th July 2004, 05:00 PM #1 diyAudio Member     Join Date: Apr 2002 Location: Belleville, IL. Interesting math conundrum. Hi all, I have been playing around with trying to come up with a formula to set the biasing of an Ixys current regulator found here: http://www.ixys.com/98704.pdf The ratio of Rk to current is a linear one but it is an inverse proportion. By this I mean that you get 10mA with a 300 ohm resistor and 5mA with a 600 ohm resistor.........etc. Does anyone have a handle on algebra well enough to write out this relationship as a formula? I'm getting a headache. __________________ Gavin
 11th July 2004, 05:21 PM #2 diyAudio Member     Join Date: May 2002 Location: albuquerque nm it only looks linear looks like a straight line on a log-log plot... rt
 11th July 2004, 05:29 PM #3 diyAudio Member   Join Date: Apr 2004 Location: Monroe Township, NJ Gavin, Would you believe the equation you're looking for is Ohm's Law (V = IR)? When there is an inversely proportional relationship between any 2 entities, the equation takes the form of xy = k. Assuming the examples you gave are correct, the constant of proportionality is 3. In this case, the units of k are Volts. __________________ Eli D.
diyAudio Member

Join Date: Apr 2002
Location: Belleville, IL.
Quote:
 Originally posted by Eli Duttman Gavin, Would you believe the equation you're looking for is Ohm's Law (V = IR)? When there is an inversely proportional relationship between any 2 entities, the equation takes the form of xy = k. Assuming the examples you gave are correct, the constant of proportionality is 3. In this case, the units of k are Volts.
It can't be that easy! So I guess that the voltage dropped across the regulator divided by the desired current will give you the value for the resistor. My god. I must be having a worse day than I thought if that's what you are saying. No. That's not working. Have mercy Eli. Put it in the form of an equation so I can make my feeble mind digest ( and hopefully absorb ) what you are saying.
__________________
Gavin

 11th July 2004, 07:00 PM #5 diyAudio Member   Join Date: Jun 2004 Location: San Diego, California i=15 - (R/60) or R = 900-60i i is current; R is resistance
 11th July 2004, 08:09 PM #6 diyAudio Member   Join Date: Apr 2004 Location: Monroe Township, NJ Gavin, Both of the examples you provided satisfy Ohm's Law for a fixed 3 V. You can calculate the true fixed Voltage (constant of proportionality [k]) from a known IR pair that works with the current regulator. Once you know V (k), you can calculate R for any given I by using R = V/I. __________________ Eli D.
diyAudio Member

Join Date: Apr 2002
Location: Belleville, IL.
Quote:
 Originally posted by Eli Duttman Gavin, Both of the examples you provided satisfy Ohm's Law for a fixed 3 V. You can calculate the true fixed Voltage (constant of proportionality [k]) from a known IR pair that works with the current regulator. Once you know V (k), you can calculate R for any given I by using R = V/I.

I think I'm beginning to see light at the end of the tunnel. Since 3 is the constant then all I have to do is divide 3 by the desired current value and I will have the value of my resistor. What was throwing me was the voltage across the device. That essentially plays no role in the equation and is controlled by the voltage drop across the device being regulated. If that makes sense. I will have to bias the tube to the voltage drop across it that I want and then good ole Kirchoff will take it from there. So for this device R=3/I. If that is true then Peter drew the schematic wrong for the EL34 test.:

He shows an Rk of 15 for 65mA. That should be 3/.065=46 ohms or so. Either that or I'm still missing something.
__________________
Gavin

 12th July 2004, 03:00 PM #8 diyAudio Member     Join Date: Apr 2002 Location: Belleville, IL. After thinking about it I think the best way to do it is just to put a 1K resistor in parallel with a 1K pot and dial in the current that I want. A 500 ohm pot in // with a 500 ohm resistor will do for the biasing on the output stage. I will have to find a 2 watt carbon pot though for this application. It should work ok. __________________ Gavin
 13th July 2004, 03:09 AM #9 diyAudio Member   Join Date: Jun 2004 Location: San Diego, California In my previous post I didn't realize you were working with a log- log plot. Since that is the case this equation should help you. log(i) = 3.2 -0.9log(R) where log is to the base 10, i is in ma, and R is in ohms. The decimal place is only one significant figure since I had a difficult time reading values on the log-log plot you referenced.

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