I need to run 12 amps through a 5-ohm electromagnet which means 60 volts must fall across it. I am building a power supply bridge rectifier circuit to fit my needs. I have a transformer capable of handling the wattage requirements but I am having trouble figuring out the amount of turns on the transformer to have on both the primary and secondary. The primary will be connected to an American outlet and I just don't understand how to get 60 volts DC out since the AC RMS voltage will be different than the rectified DC voltage.
If the rectified voltage isn't filtered the RMS voltage will be the same as the AC RMS going into it, minius the diode drop which will be anywhere from 1.4V-2V(assuming a full-wave bridge rectifier is being used which will have 2 diodes in the current path at any given time) depending on the diode and the load current. If the electromagnet needs 60VDC then the AC RMS going into the rectifier should be about 60V. So the turns ratio on the transformer will be roughly 2:1 with 120VAC on the input. I say roughly because there will be some loading effect on the transformer secondary and some drop in the rectifier diodes which you may want to compensate for by adding extra secondary turns. If you filter the rectified voltage, meaning you put a capacitor on it, the output voltage will be closer to the peak voltage(RMSx√2). How close it is to the peak will be determined by the load current and how big the capacitor is. There are multiple formulas you can grab online that compute 60Hz ripple voltage.
The transformer's primary should not disturbed, if possible. If not, then replace the primary wire using the same wire gauge and number of turns. The secondary winding needs a wire that handles about twice as much current as the primary's winding, which would require a wire that has twice the cross sectional area. I don't think that capacitor filtering of the bridge rectifier's output is necessary, but it would cause a lot of hum. If the capacitor is not used, the current and voltage feeding the magnet will be relative the average value (.63 of peak value), rather that the RMS value (.707 of peak value), Therefore, the peak value of the secondary winding would be 60/0.63 plus bridge rectifier drop or about 95 volts. Its RMS value is then 95*.707 67 volts and the turns ratio would be 67/102 .56 to 1.
If you want a different amplitude without using a resistor network, you probably need to transform before you rectify. You need a little more than half of the voltage you are starting with so not quite a 2:1 turns ratio. Make sure you have big enough wires for 60 A or they will get too hot.
OK If you want to use a transformer, you need to step down from 120V to 60V this is a turns ratio of 2:1 the AC RMS voltage will be different than the rectified DC voltage you are getting mixed up with rms and peak voltages. the input is 120V rms. (RMS tells you the work the electricity can do) if you take 60V and full wave rectify it you get a double hump - the shape is different but the rms is the same. skillbank .uk/psu/trfrec.ht I dont know why you are worrying about the turns ratio if you already have a transformer? Alternatively: use two electromagnets in series, so the will run directly from the rectified 120V supply; OR just use a single diode to half-wave rectify the mains supply, which will give you an output of 120RMS / 2 60V RMS. The electromagnet will get hot and buzz a bit but will work fine. LG, have you tried to source 1 FARAD capacitors with a rating of 60V and ripple current of 30 AMPS? skillbank .uk/psu/ripple.ht
