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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.
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Gavin 
11th July 2004, 05:21 PM  #2 
diyAudio Member

it only looks linear
looks like a straight line on a loglog plot...
rt 
11th July 2004, 05:29 PM  #3 
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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.
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Eli D. 
11th July 2004, 06:14 PM  #4  
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Join Date: Apr 2002
Location: Belleville, IL.

Quote:
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Gavin 

11th July 2004, 07:00 PM  #5 
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Join Date: Jun 2004
Location: San Diego, California

i=15  (R/60) or
R = 90060i i is current; R is resistance 
11th July 2004, 08:09 PM  #6 
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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.
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Eli D. 
11th July 2004, 08:32 PM  #7  
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Join Date: Apr 2002
Location: Belleville, IL.

Quote:
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.: http://www.pmillett.addr.com/current_source.htm 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.
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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.
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Gavin 
13th July 2004, 03:09 AM  #9 
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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 loglog plot you referenced. 
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