boyblue wrote:Hi Andy, I,ve never seen a 5V rec heater winding with a CT, you usually take the HT from either heater pin. I don't think it matters which htr pin you use. I,ve always thought that this setup would introduce some hum but its always been ok for me.
Michael Watterson wrote:... No HT or AC needed to test a Signal or Power Valve rectifier. Only heater supply and Low voltage variable PSU.
Do NOT use HT supplies, AC Transformers or Load resistors to test a rectifier emission. ...
There's a general principle that the best way to test a valve to see if it really will work properly in the circuit for which it's intended is to put it into the circuit for which it's intended. The 5U4G is intended to be an AC rectifier. I'm afraid I wouldn't trust any low-voltage DC test quite as much as I would a high-voltage AC one.
Michael Watterson wrote:[... The 3 things on a rectifier:
1) PIV (leakage / flashover) (Capacitor / Leakage tester etc)
2) Forward V/I curve . Ordinary PSU ...
But most valve testers have poor instructions for testing them and essentially tell you nothing more (and maybe less) than a leakage tester, Low voltage adjustable PSU and heater/Filament supply.
boyblue wrote:Hi All, it would be nice to try taking the HT from each end of the heater in turn to find out if the ripple varies, on load of course. Any takers.?
Michael Watterson wrote:With Low Voltage bench PSU you can make a valve rectifier hotter than most valve testers.
P = V x I It makes no difference if DC or 100Hz short pulses charging a cap. Average Power = RMS volts x RMS current.
Valvebloke wrote:P = V x I It makes no difference if DC or 100Hz short pulses charging a cap. Average Power = RMS volts x RMS current.
The RMS value of 100mA average current supplied as DC is 100mA. The RMS value of 100mA average current supplied as a 20% duty cycle pulse train (so 500mA peak) is 224mA. If the loss mechanism was Ohmic (I know it isn't in a thermionic diode, but indulge me for a moment) then the RMS voltage across the diode would also be raised by the same factor. So the average power being dissipated in the diode would be 2.24 x 2.24 = 5 times higher. I would call a factor of 5 'a difference'.
The RMS value of 100mA average current supplied as a 20% duty cycle pulse train (so 500mA peak) is 224mA.
Michael Watterson wrote:The RMS value of 100mA average current supplied as a 20% duty cycle pulse train (so 500mA peak) is 224mA.
This means nothing to me. Either the RMS is 100mA or 224mA, it can't be both. Average current we don't care about, its meaningless. Average power though is RMS volts x RMS current.
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