I have a 240v 2kva variac , I would like to rectify and smooth its output so i can have a choice of an ac or dc output, what would the best diodes to use for a full wave bridge rectifier to use? ,i thought of old microwave oven hv diodes?.
I have no idea at what the microwave oven diodes are rated at. I would just use one of the 1N400x series of diodes. They're made by a number of different companies and are plentiful and cheap and you can get them from any of the standard electronics suppliers such as Mouser, Digkey or Farnell. They all have data sheets that you can look at online. Just be sure that the ones you pick are rated for the current and voltage that you plan on using.
Go look up "30A bridge rectifier"
A microwave diode is high voltage, low current and not appropriate for the application.
A commonly available 1000 volt 50 amp bridge rectifier is a good choice. Hard to kill, bolt down to the chassis, nice 1/4 inch push on connections or of course solder them on.
How much ripple voltage do want to allow? At the rated ~8 amp output, you’ll have ~41.5 volts of ripple voltage per every 1000 uf of capacitance added on the output. It’s all linear. Half the current, you’ll have half the ripple.
Example: 10 volts ripple voltage desired at 5 amps.
(41.5 / 10) * (5 / 8 ) * 1000 = 2600 uf required.
Don’t forget an appropriate bleeder resistor to discharge the cap on power down.
Keep in mind that if the variac does not have an isolated output, neither will the output of a simple rectifier conversion to DC. Mike
Calzap and ejeffrey are both correct:
1) Your rectified output will be very hazardous because it will not be isolated with respect to ground. If you attempt to connect a grounded oscilloscope to either end of the rectified output you will get smoke or worse. You need an isolation transformer.
2) Rectifier current ratings will assume a very hot rectifier when operating at full rating, and THAT maybe with a heatsink. Much better to be conservative. Go ahead and get a bridge that is rated for 20A or more and at least 600V.
You will need aluminum electroytic output filter capacitors. These will need to be big honkers.
will this be ok , Its from my parts bin
nope, minimum is the 04 version
OP: do you realise how dangerous this is? 320V DC, not isolated from the mains. That is serious shit. Especially as it will flow enough amps to ruin your whole day.
What do you want it for?
You might argue that 240V AC isn't much different, but let's be honest: mains AC circuitry is usually "managed": properly rated cables, wires, connectors and insulation. If you are creating 320V DC you won't be feeding it into the mains socket on anything. So presumably you'll have it sitting on the stripped end of a piece of wire. This is dodgy.
And scary. I'd be nervous.
That's more than 320V DC!
I see also an different result regarding the bridge/diodes voltage standing. Usually an variac puts 10..15% higher max voltage at out, than its input.
Lets assume, it delivers 260V - and for safety an additional overhead of 10% = 286V AC.
Reverse voltage at diodes is 286V x 1.414 = 404.4V. Multiplied by 2 = 808.8V - so the 08 version is too low. Should be the 10 (1000V).
And lets having an 4.700µF capacitor at output. The stored energy is 1/2 CU².
1/2(4.700x10^-6 x 404²) = (767.1)/2J = 383.5J
For comparison:
the preset energy of an defibrillator usually is 200J
Did you ever touch a fully charged input capacitor? Like 220 to 470 µF ones in computer PSUs? Not pleasant, right? This transformer will be giving the same kick… 100 times a second. With your hand unable to let the wire go. I hope this is much easier to comprehend than a defibrillator.
Also, with an autotransformer any point after the rectifier is at a lethal voltage with respect to anything around you. That’s unlike with separated windings, where typically you must either touch two points to complete the circuit, or only one side has elevated voltage.
Your rectified output will be very hazardous because it will not be isolated with respect to ground. If you attempt to connect a grounded oscilloscope to either end of the rectified output you will get smoke or worse. You need an isolation transformer.
I get grounding connections such as this confused when dealing with isolated and non-isolated. If the transformer is not isolated, then neutral on the primary and secondary will be common which (in my mind) seems would be electrically safe for a scope (ignoring the shock hazard issue).
Also, I assume you mean a DC powered scope since the OP wants to convert to DC.
A bridge (4-diode rectifier) following a transformer where one end of the secondary (or common for an autotransformer) is neutral (i.e. close to ground) gives a DC voltage that is not grounded nor connected to neutral at one end.
Normal bridge rectifiers follow a secondary that is not grounded, or else one with a grounded center tap to provide a split DC voltage.
If the transformer is not isolated, then neutral on the primary and secondary will be common which (in my mind) seems would be electrically safe for a scope (ignoring the shock hazard issue).
One end of the secondary in the transformer will be neutral. Right. But
the rectified output will be not! Both positive and negative end of the rectifier are at high voltage with respect to neutral.
That’s because on one half-cycle negative end is bound to zero and positive sees upper part of the sine, but on the other half-cycle negative end is bound to the lower part of the sine, while the positive is sees ground (that is now above the sine).
See
interactive presentation: voltages in a rectified autotransformer output (Paul Falstad’s circuit simulator).
--- edit (2025-12-19): a clarification on what may apparently confuse some people in the linked interactive presentation. The simulation depicts rectifier’s behavior
only. The inductors are included not to model an autotransformer, but to visually show its presence. It could as well be an AC source representing the secondary winding, but then the layout would be less obvious. To be clear and avoid any misunderstanding: the two inductors are not accurately modeling an autotransformer. The picture (both in simulation and here) is valid, because the assumption is the tap is at the topmost position.