From: "Gary Schafer" <garyschafer@comcast.net>
I have seen a few older commercial amps with a resistor on the transformer
side of the filter capacitors. I have also seen it in older handbook power
supply projects.
### Installing a resistor in one / both legs, between sec of plate xfmr
and FWB/ FWD is fubar, doesn't work.
I haven't seen that done on anything in a long time.
I have an old TMC amplifier that had dual 25 ohm resistors between the
transformer secondary and the diodes. The regulation was terrible. I shorted
them out and it works much better.
## agreed. the pulses every 8.3 msec, coming from the plate xfmr
are typ 1.5 msec duration, and 4-7 times the plate current. You can easily
model that on PSUD. A lot depends on plate xfmr pri and sec dc resistance
and Z back to the street, including all wiring. I didn't pay Dahl huge
bux, for my .009 ohm pri.. and 6 ohm sec... then lose it all, with a resistor
in the sec.
## You end up with a massive V drop. If they wanted a resistor
in there, only one leg required [either leg]. If you want really lousy
V regulation, a resistor will do it. So will 240 vac wiring back to main
200A panel. It has to be sized big enough to handle the huge pulses
every 8.3 msec. [ which also end up the 240 v line] . Ditto with
contactor resistance, connections, etc. I use cool-amp on all
contacts, and also a tiny bit of 'conducto-lube'. Contactor
contacts need to be over rated as well. This is the drawback for
using a high C filter.
If the supply delivers say 1 amp of DC to the tubes and you have a 25 ohm
resistor on the transformer side of the filter capacitors the voltage drop
is not just 25 volts. The problem is that you have to be concerned with the
PEAK charging current which is many times the DC output current.
## agreed, see above. And it won't matter which side of the FWB
/FWD, the resistor is located.
The filters charge up to peak AC voltage. As current is drawn from the
supply the current that recharges the capacitors is drawn from the
transformer in very narrow pulses as the diodes only conduct near peak
voltage. That makes conduction time very short. In order to supply the
amount of energy needed with these short pulses the current has to be much
higher. So voltage drop across the series resistor (or transformer
resistance) is much higher than the DC current out of the supply.
I agree with the fellow that says it is a waste of expensive transformer if
you put a resistor on that side of the filter.
## agreed, it's a no brainer. And easy to prove. Just add a 25 ohm
resistor in one or both legs.. and watch the regulation go to hell.
The reason for a glitch resistor AFTER the filter capacitors (between
filters and the tube) is to absorb the energy held in the filter capacitors
in the event of a short. Even if the primary is disconnected, when there is
a fault the filter capacitors store enough energy to do damage. The glitch
resistor limits the amount of energy available.
## the 50 ohm glitch resistor limits the fault current. Total energy
delivered, depends on how fast the HV fuse, preceding the glitch
R open up.
#### 3500 vdc / 50 ohms = 70 A of fault current. 70A will open
a 1 A rated sand fuse very fast.
Using too large a filter capacitor presents a problem too as the amount of
glitch energy will be higher.
### Who cares. Install a Buss sand filled HV fuse, PRIOR to the glitch
R... and the B+ HV fuse will blow in < 2 msec every time. Filter cap
size doesn't enter into the equation. I have been experimenting with
using surplus lytics [ caseloads of em]. 2000/2300/2500/3900/5400
uf caps have been used to make a series string. Also, multiple,
identical strings have been paralleled . A 2nd HV sand filled fuse
is installed between sec of plate xfmr and FWB [ either leg]..
+ a mag- hydraulic breaker installed in the 240 line.
## with this set up, anything, any where can be cro-barred repeatedly,
and nothing happens, except blown HV fuses.
## We tried 8 x 2500 uf caps in series = 312 uf = 1400 joules. [3 kv]
[The stock 25 uf in a SB-220 is only 136 joules.. 3.3 kv]
## If you want say 2.5 kw DC input [1.5 kw out]... you won't get it
with caps that only store 136 joules of energy. 1 joule= '1 x watt
second'
## End result ? Ripple dropped from 3%.. down to .24%
Dynamic regulation on ssb is superb. Pep output is up
>140 w . I'm convinced imd is reduced as well.
The 2500 uf CD caps were 2" diam x 4" long. [ 9 x oz] You end up with
a quasi, brute force regulated B+ supply. It was a fun experiment,
and a real eye opener. Everybody said it wouldn't work, but it does.
### PSUD is only for static loads, but it will still provide an estimate of
level of the pulses..every 8.3 msecs, both in duration, and peak level.
## PSUD also showed the static regulation was improved a little,
for modes like FM /RTTY. What surprised me was the static regulation
improves a whole bunch, when plate current is very high.
later... Jim VE7RF
73
Gary K4FMX
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