[Amps] Power Supply loaded DC output
craxd
craxd1 at ezwv.com
Sat Oct 2 12:54:16 EDT 2004
Manuel,
It's according to what type of rectifier you'll be using, the
capacitance of the filters, and the frequency (50 Hz or 60 Hz). A full
wave rectifier is approximately a 5% drop and a half wave 10% with the
usual 36.7 uF of filtering. I listed both formulas but yours will be a
full wave. If the total capacitance is below say 36.7 uF, it will be
lower than the usual 5% on a full wave circuit. 36.7 uF comes from using
six (6) 220 uF @ 450 Vdc capacitors in series. The drop is due to the
fixed resistance of the supply compared to the resistance of the load.
The formulas take this into account. However, for figuring plate
resistance, the peak voltage and current is used.
There's two formulas for this drop;
Full Wave Vavg = Vpeak - I / 4 x C x f
Half Wave Vavg = Vpeak - I / 2 x C x f
I = Plate current on meter
C = Capacitance in Farads
f = Frequency in Hz
Will Matney
Hello Again,
Trying to establish the loaded voltage output of the plate transformer that
I am going to use with full wave bridge rectification and capacitor
filtering. It is the one that Ten-Tec buys from MCI Limited for the Titan
III. I tried to get some "real life" numbers from somebody on the reflector
but have had no responses.
The transformer itself has a label with the following ratings, two 117 V
primaries and a 2015 V secondary at 1.5A
Tentec claims a 3000 V "no-load" plate voltage on the Titan III.
My calculations are as follows:
Actual Primary Voltage 240 V
Actual Secondary Voltage w/ 240V on primary 2067VAC
2067 x 1.414 = 2923 VDC NO-LOAD
2923 less 10% = 2631 VDC @ 1.5 AMP
Are these calculations correct?
Thanks
Manuel R. Alonso
KC4MNE
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