Power supply question

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In simulating a power supply using Dunkan's Amp Tools I get a result I don't understand. Say you use a 600v center tap transformer with a full wave rectifier using a pair of 1000v piv diodes and a 220 uf cap. It reports that the
rectifier's PIV may be exceeded with 1008 volts. My rough calculation said the
PIV would be ~ 846 volts why the big error ?
 
600Veff gives 600x1.4= 840Vtop, that is a swing from -840V to +840V and back.
The cap is charged to +840V and is on the cathode of the rectifier.In the mean time the anode swings back to -840V giving a voltage across the diode of 2x840=1680V.
End of story (for the diode).
Mona
 
In simulating a power supply using Dunkan's Amp Tools I get a result I don't understand. Say you use a 600v center tap transformer with a full wave rectifier using a pair of 1000v piv diodes and a 220 uf cap. It reports that the
rectifier's PIV may be exceeded with 1008 volts. My rough calculation said the
PIV would be ~ 846 volts why the big error ?
IIRC, PSUD wants the transformer spec showing the voltage between the CT and the end of the winding - so I think (check this) that your 600v CT PT should be entered as 300v. (Rt click and edit)
I'd expect a 300-0-300 transformer to give about 400v rectified DC, and entering the PT as '300v' does that.

BTW, using the 'soft start' and also 'Delay' options in PSUD can reduce those warnings.....
 
I had originaly entered 600v then when I changed to a full wave rectifier the program changed the transformer to a 600v ct. That is 300v cent tap then 300v and i did get
about 423 V and as I understand it the maximum PIV the diodes should then see is twice this or 846 .
 
In simulating a power supply using Dunkan's Amp Tools I get a result I don't understand. Say you use a 600v center tap transformer with a full wave rectifier using a pair of 1000v piv diodes and a 220 uf cap. It reports that the
rectifier's PIV may be exceeded with 1008 volts. My rough calculation said the
PIV would be ~ 846 volts why the big error ?

rms volts x 3.1 is the absolute value of voltage possible when diodes are not conducting, so that the 1800volt piv recomended is okay, more wouldn't hurt...

also consider what the maximum line voltage in your area is like, what is it up to?
what does the 600v ct secondary become at such high line voltage?
so i will prepare for that...
 
Mains supply voltage tolerance above "normal" could be +10%, i.e. 1.1times.
Then multiply by sqrt(2) to get the peak voltage of the sinewave.
Then double for the peak INVERSE voltage.
i.e. 3.1 (= 1.1 * 1.41 * 2)

In the UK where "normal" is 240Vac and maximum used to be 254Vac (now reduced to 253Vac, under Harmonisation rules) we arrive at 2.99 times.

i.e. PIV is 3 times AC voltage.
 
I give up. I think that program I used has a small bug. A 600v center tap transformer using a bridge, 220uf and a 5k load does show an output of ~ 423 vdc and it does show the PIV of 826 as expected but it also shows a warning that the diode's PIV has been exceed with a value of 1008 at 16 seconds.
 
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