[Amps] Power Supply loaded DC output

[craxd <craxd1@ezwv.com>]

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


_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps