Hi,

I'm a little new, so bear with me please:

I'm currently trying to create a full wave rectifier to convert the output from an AC power supply to run into a DC electronic load I have. I'm trying to run the power supply at full current/voltage which is roughly 4.7 AAC and 190 VAC. I'm trying to pick the right components for the rectifier, but am a little confused to diode specs because I want to be very careful not to blow up any components, and the store I'm going to has very little diode specification information online.

First off, would it be wise to buy around a 200 V diode at 5 watts? I am trying to calculate the max power, but I can't tell if the diode drop is just going to be 200 V, in which case the would the power dissipation be P = I*V = 4.7*200 = 940? Or is this voltage rating the max voltage allowed to run over the diode?

Second, if that is the case and I need to attenuate my AC signal a little bit (keep in mind I'm just trying to run full power out of the power supply for experimentation sake so I may run it through the electronic load) is it possible for me to decrease the amplitude with an RC filter using high wattage resistors and a high voltage cap, then run it through a rectifier, and then run the DC signal through the DC electronic load I have?

Sorry if this is written in a confusing way.

 

We need more information. You specify 4.7 AAC. Is that supposed to mean 4.7 amperes AC at 190 volts? That's a LOT. You would need diodes capable of 5 amp each. As for reduced voltage in the output, you can use heavy duty resistors to drop the voltage, but at 4.7 amperes, they would have to be rated at 50 watts, roughly.

The two critical ratings for diodes are PIV (peak inverse voltage) and current handling capability. I urge you to read up on power supplies, real soon now.
Best regards ....

 

While this thread breaks the TOS, This type of circuit is a common one in commercial products. That said, what you should use is called a "bridge rectifier" like this- http://www.google.com/products/cata...a=X&ei=AVdqT-2YA4nTgQfN3OjNCQ&ved=0CHgQ8gIwAg#

For a high amperage use a bridge rectifier is better in price and heat dissipation.

 

I think the original poster didn't give enough accurate information. The crucial missing point is how much current the load will take at 190 volts. I'm suspecting that it is on the order of 100 ma, in which case the answers are easily obtained from E=IR and W=EI formulas.

 

The power supply max is 4.7 Amps and 190 Volts at 60 Hz. The electronic load has specifications of max 130 Volts, max 15 Amps, and a power dissipation of 750 watts. Its a variable load. I figured I would put two small high wattage resistors before the rectifier (4.5 ohm, 8.5 ohm and each also capable of 750 watts power dissipation) to dissipate some of the power so that the electronic load may handle it. Mainly I just wanted a rectifier that would allow me to use the AC power supply on a DC electronic load.

I was only hoping to see if I could get the power supply to max out, but need adequate loading so the power supply will output at max voltage.

By the way do those rectifiers need a heat sink attached or anything?

 

If the rectifiers have to handle 4.7 amp, they would probably need a heat sink. I would put the dropping resistors between the rectifier and the load. 4.7 amp at 190 volts gives 893 watts.
If the output load is to vary as much as you seem to indicate, how about putting a variac in the circuit, between the AC mains and the transformer primary? You won't get anything near perfect regulation, but that's what the knob on the variac is for.