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#1
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H i I have access to some control transformers 500 watts to 1500 watts they are 480 primary to 120 secondary this is a 4/1 turns ratio .could I put 120 in on the primary and get 30 out on the secondary then rectify it to get the 42 volts dc? required for the board that I'm contemplating on using would this work or could I not get enough amperage out of it ? Bob |
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#2
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| These things are specified to work on and deliver the specified voltages. If you reverse your theory (120V connected to 480V) you know what would happen. Reversed again you have an underpowered transformer with decreased efficiency (=Heat). You cannot use everything you stumble on. Carel |
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#3
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| 060513-0955 EST USA bobleecnc: Yes. Any voltage lower than the rated voltage at the same frequency can be applied to the primary. Your output current rating will be whatever it was for the original specification. 1500 VA at 120 ouput is 12.5 A. So at your 30 V AC RMS output your rated output current is still 12.5 A which means you are good for 12.5 * 0.707 = 8.84 A from a full wave bridge with capacitor input filter. . |
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#4
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| great these are free so I guess I'll Try it What have i got to loose right? now if I wanted to get more amperage could I parallel two or three of these together through the same bridge and cap (assuming that they were big enough ) and get the desired amperage that I may need? |
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#7
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| If you are in the U.S. just spend the $18 and buy one of these. Actually, buy two and sell one to me.... They have a stupid $100 minimum order for international (Canadian) buyers! http://www.mpja.com/productview.asp?product=7846+TR |
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#8
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| You likely will have a heating problem. Power transformers are wound using a "turns per volt" design constant and you will be far from the design. It isn't as simple as a turns reatio. I have tried this and had filament transformers melt - down. I'd be looking for overheating |
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#9
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| 060519-2013 EST USA bobleecnc: You are getting all sorts of wrong answers to your question. This means the respondents do not understand the basic theory. A transformer rated 10 A RMS with a resitive load for continuous duty will not provide this same current magnitude for continuous duty from a full wave bridge rectifier with capacitor input filter, but will provide about 0.707 (that is 1/2 the sq-root of 2) or less times the above AC RMS to a resistive load. In other words your continuous duty load would be approximately 7.07 A or less. Are you sure you need 10 A continuous times 4 or 40 Amps? I would take one of the transformers connect a bridge rated greater than 20 A and 200 PIV, and a suitable capacitor. And run an experiment with one motor at mechanical loads and speeds you expect. For the present calculate the capacitor assuming a 10 A load. For round number convenience assume your output voltage is 40 V, then the effective load resistance is 40/10 = 4 Ohms. The time between charge pulses to the capacitor is 8.3 milliseconds. For one time constant the discharge is 63%. Thus, we would need about 10 time constants for 6% ripple, or a one time constant of 83 milliseconds. One time constant is R x C. Thus 0.083/4 = 0.021 Farads = 21,000 microfards. . |
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