[AMRadio] Solid State 575 mercury vaporRectifierReplacementsneeded

sbjohnston at aol.com sbjohnston at aol.com
Tue Oct 6 09:34:19 EDT 2009

W5JO wrote:

>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)?

Yes, perhaps my message did not come thru intact.  Here's the pertinent 

I prefer cylindrical wirewound power resistors for this job (they seem 
to be able to take more surge current than the sandy rectangular ones 
in case of a fault downstream).

If the resistor ends up to be hard to find or expensive, it might be 
easier to eliminate the reasons why the higher B+ is a problem. And 
more efficient too. Upgrading some capacitors could be a reasonable 

>Then on top of that you will have the heat dissipated by the resistor 
>contend with which is probably more than the tube.

If the resistor is simulating the internal resistance of the tube 
rectifier, that dissipation is roughly the same.  And without the 
filament there's a bunch less heat at the tube, and in the filament 
power supply.

I've not put the resistor on the tube base - I'd mount it on insulators 
nearby in series with the positive lead coming from the stack.  I can 
see why you'd like to put it with the tube - I just couldn't work out a 
solid mounting arrangement.

Steve WD8DAS

sbjohnston at aol.com
Radio is your best entertainment value.

-----Original Message-----
From: Jim Wilhite <w5jo at brightok.net>
To: Discussion of AM Radio in the Amateur Service 
<amradio at mailman.qth.net>
Sent: Mon, Oct 5, 2009 9:20 am
Subject: Re: [AMRadio] Solid State 575 mercury 

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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