[Boatanchors] Dropping Resistor Question

[Mark K3MSB]

Got it;  I had to sit down and draw out the waveforms and look up the
formulas for the average of a fully rectified and half wave rectified sine
wave.....

Current through each leg (Half rectified sine wave):  1/pi * Ip =  0.318  Ip

Current through load (Fully rectified sine wave):  2/pi * Ip = 0.637 Ip

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

If I put the dropping resistor (or zener) after the two rectifier output
sides join I need to use 2x175 (as the one web site referenced),  but If I
put the dropping resistor (or zener) before they join,  it's 175 mA.

73 Mark K3MSB



On Fri, Oct 6, 2017 at 3:36 PM, WA5CAB--- via Boatanchors <
boatanchors at mailman.qth.net> wrote:

> In a power supply with a full wave vacuum tube rectifier or with two
> semiconductor diodes replacing the tube, the transformer secondary is
> center-tapped.  The rectifier plates or anodes are connected to each end
> of the
> secondary winding.  The cathode or cathodes tied together is/are the
> positive source
> and the center-tap is the negative.  The average current that flows in the
> leads between the ends of the transformer winding and each plate or anode
> is
> one half of the average current flowing in the external circuit.
>
> So if you are using resistors or zener diodes to try to simulate the
> characteristics of the vaccum rectifier, if placed in the leads between the
> transformer and the rectifiers, they will have to dissipate one-half the
> average
> power that one placed in the output lead would have to dissipate
>
> Robert Downs - Houston
> wa5cab dot com (Web Store)
> MVPA 9480
>
> In a message dated 10/06/2017 13:47:07 PM Central Daylight Time,
> mark.k3msb at gmail.com writes:
> >
> >
> > I still don’t see why it matters if you put a dropping resistor (or
> zener)
> > in each leg, or one after the legs are combined.
> >
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