[AMRadio] Solid State 575 mercury vaporRectifierReplacementsneeded |
Jim Wilhite
w5jo at brightok.net
Mon Oct 5 10:20:13 EDT 2009
Thanks for the info Steve, but not quite what I had in mind. Is the resistor you use a wirewound (given the wattage, it must be)? Something many people overlook is the voltage value of resistors. Most carbon comp or film are good only up to ~500 volts and many of them only to 350. So in this particular application if one wants to build the rectifier circuit in a tube base, you should have room for a 15 watt resistor. Then on top of that you will have the heat dissipated by the resistor to contend with which is probably more than the tube. So the best situation would be to ignore the voltage drop and just put in modern diodes which have tighter specifications that don't require equalizing resistors and caps, then just work around the higher voltage in the circuit. Most tubes will withstand a much higher plate voltage than their specifications. Just off the top of my head I forget how the manufacturer determines maximum plate voltage, but tubes will withstand much more and operate fine. So if you add the resistor you have the heat but lower voltage, without it you have higher voltage but not the heat. So the easiest thing would be look at the capacitors in the circuit and see if you need to raise the voltage value of them and just leave the voltage dropping resistor out of the circuit. And given the price of ceramic coated wirewound resistors, much cheaper too. 73 Jim/W5JO ----- Original Message ----- > > I wrote: > >To calculate the value of resistor, use Ohms law R=E/I and P=IxE where > R is in ohms, > is the desired drop in volts, and I is the current in amps, and P in > watts > For example, lets say your B+ is 50 volts too high. The load draws > 0.250 amps. > R= 50/.25 = 200 ohms. > P= 0.25x50=12.5 watts. > > > Steve WD8DAS
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