Hi All,
I'm putting together a quick and dirty pre using a ECC99 with IXYS current regulators as the plate loads. I've gandered at the PDF for the IXYS regulator and the "graph" for adjusting the current is as clear as mud: http://www.google.com/url?sa=U&start=1&q=http://www.ixys.com/98704.pdf&e=7317
What I would like to do is put a 2K resistor in parallel with a 2K pot on the "cathode" of the IC and use that to adjust the current through the tube. The problem is I would like to use a pot with a power dissipation of at least 3W. Unfortunately the only pots that fit the bill for a reasonable price are wirewound pots. Will the miniscule inductance of a wirewound pot cause a problem in this application?
I'm putting together a quick and dirty pre using a ECC99 with IXYS current regulators as the plate loads. I've gandered at the PDF for the IXYS regulator and the "graph" for adjusting the current is as clear as mud: http://www.google.com/url?sa=U&start=1&q=http://www.ixys.com/98704.pdf&e=7317
What I would like to do is put a 2K resistor in parallel with a 2K pot on the "cathode" of the IC and use that to adjust the current through the tube. The problem is I would like to use a pot with a power dissipation of at least 3W. Unfortunately the only pots that fit the bill for a reasonable price are wirewound pots. Will the miniscule inductance of a wirewound pot cause a problem in this application?
I don't see your problem - just read the required resistor value for the current you want from graph Fig 2. and wire the anode load according to Fig 1. except that the -ve supply side is the anode connection of the tube. Example 1K to give about 3.3mA etc. Don't be concerned about the "Plateau Current" graph axis name. When used as an anode load it will always be at the "Plateau" current.
You can fine adjust the resistor value later if you don't get exactly the current you wanted.
A pot is not necessary - I think high voltage on a pot is a seriously bad idea, I would be too "s**t" scared to grab the shaft to adjust it. I also don't see where you got the 3 Watt power figure from. 2K resistor parallel 2K pot would give 1K - gives total current of 3.3mA, 1/2 of this through the pot. Current Squared times Resistance gives 5.3 milli-Watts.
Stick to a single fixed resistor and adjust value on test if required.
Cheers,
Ian
You can fine adjust the resistor value later if you don't get exactly the current you wanted.
A pot is not necessary - I think high voltage on a pot is a seriously bad idea, I would be too "s**t" scared to grab the shaft to adjust it. I also don't see where you got the 3 Watt power figure from. 2K resistor parallel 2K pot would give 1K - gives total current of 3.3mA, 1/2 of this through the pot. Current Squared times Resistance gives 5.3 milli-Watts.
Stick to a single fixed resistor and adjust value on test if required.
Cheers,
Ian
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