[Boatanchors] Dropping Resistor Question

[WA5CAB at cs.com]

No.  Ignore the max current per plate rating.  The maximum rated output 
current of a 5V4 is 175 mA DC.  So assuming that you are going to add 
resistance to reduce the ouput voltage from a solid state full wave rectifier to 
approximate the 5V4, and that to do that, you need a voltage drop of 23 Volts, 
the resistor value is the same at all three locations.  During the positive 
half cycle during which the upstream resistor  is in use,, it is in series 
with the output.  But during the negative half cycle, the resistor on that 
side doesn't have to do anything but cool off.   So the resistance regardless 
of where you put the resistor(s) is 23/0.175 or 131 ohms.  However, although 
the wattage being dissipated at full load from the single resistor is 
23*0.175, or abour 4 watts, that dissipated by the two upstream resistors is only 
2 watts each because on alternate half cycles, each of those resistors does 
nothing but cool off.

Robert Downs - Houston
wa5cab dot com (Web Store)
MVPA 9480

In a message dated 10/06/2017 18:00:06 PM Central Daylight Time, 
mark.k3msb at gmail.com writes: 
> 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
>