[Boatanchors] Dropping Resistor Question

[Jacques Fortin]

Hi Mark,

If you use a dropping resistor in series with each diode, you are right (~
130 ohms).
Despite this will be only a peak current value in practice, not an average
one.
The point here is to use a resistor to mimic the hollow state rectifier
voltage drop at any current, right ?
But graphically, that will give a straight line, which is not the case for a
rectifier tube.
A tube will drop more than the resistor at currents lower than 175mA, for
ex.
Some experimentation can be done, however.
Another trick that works very well and give a better voltage regulation than
an added resistor is to use a Zener diode in series with each rectifier
diode.
I once had a "solid stated" supply that gave me 195V at full load when the
"target" was 175V.
I just used two 20V, 5W zeners, and everything is fine since.

73, Jacques, VE2JFE

-----Message d'origine-----
De : boatanchors-bounces at mailman.qth.net
[mailto:boatanchors-bounces at mailman.qth.net] De la part de Mark K3MSB
Envoyé : 5 octobre 2017 14:36
À : Boatanchors List <boatanchors at mailman.qth.net>
Objet : [Boatanchors] Dropping Resistor Question

I'm solid stating my Johnson Valiant as I work on it.

For the 5V4,  my Tung-Sol data sheet shows 23V drop @ 175 ma per plate.

I see some one web site using 2x175 mA for the dropping resistor current.

As I see it even though both plates will be used,  it will be only one side
at a time so the current through the single dropping resistor should be 175
mA.

Comments?

73 Mark K3MSB
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