amplifieroutputthree phasetransformervoltage
I've got huge three-phase transformer, primaries are connected in Y configuration, secondaries in Delta, then goes to 6 diode bridge rectifier. Output is about 200 amp 52 VDC.
In short, how many ohm resistor do I need to reduce current from 200 amp to 100 amp while dead-shorted ?
This is an old thread but I thought I would add something as there seems to be a widespread misunderstanding of this and there is not much information on the web. I think the designers and manufacturers of these devices want to keep the general population in the dark. I think what most people are missing is that these "fixed field" permanent magnet alternators have an inherently poor load regulation curve. With no load the output voltage can vary from 20V at idle to well over 100V at high revolutions per minute. However as the load current is increased a magnetic field is created in the stator which is in opposition to the field of the permanent magnets. Thus even short circuited it will not produce more than 40A or so. When the short is made by an SCR with a voltage drop of 1-2V this is only about 80W. The SCR is only turned on for part of the a.c cycle as it only turns on when the battery voltage exceeds 14.4V. Before it turns on the current is being delivered to the battery and all the other loads, the SCR only turns on when the other loads cannot absorb the current. When the SCR does short the windings the current is so high the the field in the stator almost cancels out the permanent magnet field so the torque load on the engine is reduced. This may seem inefficient but remember even if the field was produced by a wound rotor with slip-rings and brushes like in a car alternator that would also require power and there would also be power lost in the regulator so it isn't as bad as would first appear.
First of all the 7812 (positive voltage) and 7912 (negative voltage) regulators need about 2v higher input than the output in order to operate properly (that is at least 14v for the 12v regulator and -14v for the -12v regulator) .
The AC voltage in the input is rectified by two diodes to generate a symmetric unregulated supply (that is a positive and a negative voltage with reference to the ground) which feeds the regulators inputs and you get +12v and -12v regulated rails at the output. If you feed the input with 12v DC there is no way for the circuit to generate the negative rail voltage so this can't work.
Here is a link that explains the operation of the diodes in the circuit you show
http://metroamp.com/wiki/index.php/Half_Wave_Voltage_Doubler
I've got a 12V AC-DC adapter that delivers 200mA
Depending on the internal implementation of the AC-DC adapter it may be possible to use it.
If it is the classic transformer type rather than a switching type (you can easily judge it by size and weight) then you can omit the output circuitry (you'll need to open the adapter case to do that) and connect the AC output of the transformer to the AC input of the O2 power supply circuit.
Best Answer
Simple Ohms law can answer this. R = V / I
52V / 100A = .52Ω
It would also explain why this would be a pretty bad idea.
P = VI
52V * 100A = 5200 W. 5.2 KILOWATTS. Of pure waste heat. And a few hundred dollars.