using voltage drop to calculate current
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 1st September 2007, 04:21 PM #1 cbutterworth   diyAudio Member   Join Date: Oct 2006 using voltage drop to calculate current I am sure that this is an easy calculation, but right now, I cannot think of how to do it. I have a tubed pre-amp with ss PSU that delivers around 330 VDC unloaded (well there is 20K of bleeder), BUT WITHOUT tubes. When I insert four 6SN7s, the voltage drops to around 290 VDC. So, can I use the voltage drop to calculate the current draw of the tubes? If so, how? Thanks, Charlie
 1st September 2007, 08:29 PM #2 cbutterworth   diyAudio Member   Join Date: Oct 2006 OK, How does this sound? If I measure from B+ to the GND, I get a resistance of 16.9K. I can plug this number into Ohm's Law. If I use the sans tube VDC of 330V, I get 19mA of current. If I use the voltage after drop due to current-draw by the tubes (287.5), I get 17mA. Which of these figures to use? I expect that I would use 330V for the calculation as it is the voltage available to the tubes. Now, I already have 20K of bleed across the final cap of the PSU, which itself draws around 18mA. This means that I get a total current draw of 37mA. So, if I am using 10H chokes, I should be pulling more than minimum voltage to prevent them from ringing. Is this correct? Charlie
 3rd September 2007, 01:21 AM #3 KSTR   diyAudio Member     Join Date: Jul 2007 Location: Central Berlin, Germany Seems you are on a wrong track, Charlie. All that you know from measuring the B+ impedance in cold condition is that there is some other leakage beside the 20k bleeder (some 110k in total). Why not measure the current directly? Add, say, 10 ohms in series to the supply after the bleeder (preferably on GND side of the supply, if that is possible) and check the Voltage drop across which reads 10mV per mA. Both unloaded and loaded, of course, then take the diff as the total tube quiescent current. Four 6SN7s (eight triodes) should give you way more than 20mA, something like ~100..150mA. Of course you could also hook up an ammeter directly. With these values and the B+ voltages (loaded and unloaded) you can get the impedance of the supply, which might be useful to know. Regards, Klaus

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