How to calculate peak current in transformer secondary - Need specialist advice - Page 1
Forum section: electronics / projects

author: opusensemble
author_rank: Contributor
timestamp: October 06, 2015, 03:18:29 pm
content: I have a MOT transformer with the following specs:Primary Voltage: 220V Single phaseSecondary Voltage: 2280V (RMS)Output power: 2000WThe secondary is being rectified by a bridge rectifier.I would like to determine the maximum instantaneous peak current on the bridge output.I am assuming these calculations, but not sure if I'm doing it right:Secondary current (RMS) = 2000W/2280V (RMS) = 0.877APeak secondary voltage = Secondary current (RMS) x sqrt(2) = 1.24ADoes this make any sense at all?

author: bktemp
author_rank: Super Contributor
timestamp: October 06, 2015, 03:26:59 pm
content: The peak current is determined by the impedance formed by the load connected to the output + the impedance of the ouput of the transformer. A MOT has a magnetic shunt between the primary and secondary winding, therefore it has a rather high impedance.

author: opusensemble
author_rank: Contributor
timestamp: October 06, 2015, 04:00:07 pm
content: Thank you bktemp.My test circuit is attached. The transformer load is a 10uF/4kV capacitor.I would like to determine what would be its maximum charge (capacitor voltage) at 10ms, assuming the transformer has 2000W power output.I have dimensioned R2 around 2.4kOhm assuming the calcs above for a 1.24A peak current, but guess the problem is reduced to knowing if 2000W can output a 1.24A peak current, right?

author: opusensemble
author_rank: Contributor
timestamp: October 06, 2015, 04:12:53 pm
content: The transformer specs are attached. It's mentioned that the output resistance is 30 Ohm. Is that the impedance you were referring to?How to incorporate it in the simulation? As a series resistor only?

author: bktemp
author_rank: Super Contributor
timestamp: October 06, 2015, 04:22:32 pm
content: 30 Ohm is only the DC resistance. The AC impedance is much higher, typically around 1kOhm, so the short circuit current is about 2A.1.24A peak current is no problem for this transformer. The current flows only for a short time period. The resistor needs to be able to handle the high peak power and the high voltage.

author: opusensemble
author_rank: Contributor
timestamp: October 06, 2015, 06:33:08 pm
content: Thank you for your help bktemp. That really helps clarifying my doubt.I don't have the 2000W transformer yet, but was checking an old 1000W one and measured secondary inductance of 65mH (image attached)Calculating the inductive reactance value it comes XL = 2*PI*f*L, i.e. XL = 6.28*50*0.065 = 20.4 Ohm. Does this value make sense for 50Hz ?It's much smaller than the 1k you mentioned. Maybe you were considering a higher frequency or something?(My value must not make much sense as I'm measuring a higher DC resistance of about 100 Ohm. Do you have any clues for this?)

author: opusensemble
author_rank: Contributor
timestamp: October 06, 2015, 07:10:38 pm
content: I've read online people saying that the secondary inductance of a MOT with shorted primary, is around 5H, which would give a value close to the 1k Ohm "AC ohms" you've mentioned.Ref.1:http://www.overunity.com/13474/dc-power-tesla-coil/10/wap2/Ref.2:http://www.pupman.com/listarchives/2001/December/msg00766.htmlThis makes me wonder if I measured mine incorrectly. The secondary measurements should be made between the HV terminal, and the chassis, like in my picture, right?