[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 

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.



----- 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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