Is a 600v diode suitable?

JPB33

Senior Member
Cannot get my head around this one! Am I right in thinking that as they are in series 600piv diodes will be suitable for use with this voltage doubler circuit with 250v output from the transformer?


Thanks Peter

Delon Use .jpg
 

oracacle

Senior Member
that looks like a half wave rectifier to me, but I am pretty far from being any form of expert.
Don't forget that transformers don't work on DC (generally speaking anyway) so your input voltage would need to be 125V AC and no 125V DC
 

PhilHornby

Senior Member
I thought I'd found a design document but this GOOGLE BOOK excerpt contradicts itself: "PIV rating of each Diode is only Vm = Vp " followed by "Diodes have to withstand 2Vm" (Where Vm is voltage across each cap).

Clear as mud :confused:
 

fernando_g

Senior Member
That is what happens whenever the description in the sentence is incomplete.

The Peak Inverse Voltage (PIV) of a diode in a full wave voltage doubler , which is what you are showing here, is the sum of the stored voltage in the capacitor (Vm) plus the AC's negative peak voltage (Vpk-).
Since for such a high voltage, the diode's forward voltage drop can be disregarded, then one can safely simplify Vm = Vpk-
We further know that in an ideal sinewave Vpk = sqrt(2)*Vrms.

Thus for your 250 volt AC source, the total inverse voltage the diode will see is 2*1.41*250 = 705 volts.

Applying some safety margin, you should use 800 PIV diodes.
 

Reloadron

Senior Member
The pictured circuit is a basic voltage doubler circuit. With a 250 VAC output from the transformer here is what you will get:
Voltage Doubler.png

Capacitor C1 is charged through D1 on the positive half cycles. C1 will charge to 1.414 times the input voltage of the input or 1.414 X 250 Volts = 353.5 Volts. C2 is charged through D2 on each negative half cycle to 353.5 Volts. The capacitors C1 and C2 are in series so we add the voltages 353.5 + 353.5 = 707 Volts. C1 and C2 should be about 220 uF and rated for at least 450 VDC. Diodes D1 and D2 should be around 1 KV PRV using a 1N4007 or similar diode. You could likey get by with a lower PRV diode but a 1 KV 1N4007 is a cheap and plentiful diode. The value of the caps will be a function of the output current I picked the capacitor working voltage based on common values. You could likely get by with a lower working voltage like 400 VDC but I like 450 to be on the safe side. If you build this doubler keep in mind if you get across the output it will be a memorable and shocking experience. Not quite the 5.0 VDC or 3.3 VDC those who use micro-controllers are accustomed to so be careful.

Also not shown I would place a 100 Ohm 1 Watt resistor in the input line going to the junction of C1 & C2 to act as a current limiter for any surges.

Ron
 

JPB33

Senior Member
Many thanks everyone for the help--I read similar documents and got more confused by the minute! I intend using surge limiting as I also read that each cap is fully charged by the first cycle. Also fitting bleed resistors. A lower voltage version has been working fine for years.

I have a load of 600piv diodes, can I just use 2 in series? Remember seeing diodes in series in power supplies for valved power amps which had 1.5kv or more on the anodes during amateur radio days.
 

eggdweather

Senior Member
Yes, like resistors in series, the total applied voltage is divided between the devices so two 600v devices in series can survive across a 1200v circuit. The impedance of the devices are assumed to be equal for this to work and the probability of considerable impedance mismatch is minimal, which you could counter by using 3 diodes in series without affecting unduly the final output voltage.
 

techElder

Well-known member
Also, with diodes in series, watch that you don't try to draw too much current at the output. At minimum, you could be surprised at the voltage drop.
 
Top