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Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Finally get around to this, purchased a ac arc welder for $110 takening it to bits to get the inductor. Below is the inductor messured at 0.3Ohm with a saturation current of 130 Amps. Will be getting a muliter meter to get the inductance value. Ive left the high voltage winding incase I use them for the discharge.
Type of messured the inductance, it should be 1:37.5 compared to a 800 watt Mot transfomer, if a Mots 300mH it should be 8mH.
Thinking about get a 2volt 60amp psu, with a 4volt 60amp psu in series with two instead of one, and 6volt supply for three of the inductors...well the gernal theory.
Edit Did a four wire test, the resisance should be 0.175ohms A 12volt supply 5.5ohm current limiter and messured 0.079 volts across the inductor, with dc supply.
Tryed with ac and got 12.6 volt from a 24.4volt supply with 5.5ohm resistor.
Edit, thanks for the catch :) 12v / 5.5ohm = 2.18 0.079 / 2.18 = 0.036ohms
Registered Member #4465
Joined: Wed Apr 18 2012, 08:37AM
Location: Bucharest, Romania
Posts: 145
How did you measure the inductor resistance? In my opinion the 0.3 ohm value is too much for that wire. You need a four-wire connection method for measuring low resistance. It eliminates the resistance of the test leads from the measurement.
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Andy wrote ... Did a four wire test, the resisance should be 0.175ohms A 12volt supply 5.5ohm current limiter and messured 0.079 volts across the inductor, with dc supply.
Andy, it's always good to see you actually putting things together and measuring stuff. But your arithmetic answer is wrong. I bet you rushed it. Calm down and try again.
Andy wrote ... Edit, thanks for the catch :) 12v / 5.5ohm = 2.18 0.079 / 2.18 = 0.036ohms
That sounds much more like it. Now at 100 amperes RMS, that inductor winding is also a 360 watt heating element.
Care to compute the inductance from your 4-wire AC impedance measurement?
Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Lets simplify :
To estimate the primary inductance you may connect the primary on 220 or 110 V AC line and measure the the primary current. No load at secondary. A ac hook electrican ammeter is nice to measure currents.
The Reactance will be the voltage/current , in ohms
Then the inductance will be calculated as H=Reactance /(2*pi*f*). The results is in henryes.
And the inductance of such secondary could be calculated measuring the current using a 24 - 48 volts ac. and the above calculations.
As the secondary has 1/4 of the primary turns the secondary inductance will be 1/16 of primary inductance
The position of movable iron core can change the mutual inductance between the primary and secondary .
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Newton Messured the amps and got 2.36 with the voltage 10.9(adjusted the shunt) 10.9/2.36 = 4.61 4.61/314.2(6.48*50) = 0.014H Or 314.2*4.61 = 1448.46 1/1448.46 = 0.00069H
Got the Xl/2*pie*f from practical electronics book, but they are showing different values?
I think you got capactive and inducvtive foumla backtofront.
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
Andy wrote ...
Newton Messured the amps and got 2.36 with the voltage 10.9(adjusted the shunt) 10.9/2.36 = 4.61 4.61/314.2(6.48*50) = 0.014H Or 314.2*4.61 = 1448.46 1/1448.46 = 0.00069H
Got the Xl/2*pie*f from practical electronics book, but they are showing different values?
I think you got capactive and inducvtive foumla backtofront.
speaking from experience , the higher one looks most likely.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Thanks. Made a delayed switch circuit, it takes one RC of one msec to turn off, messured 1.87 amp with the ac reading on the multimeter getting 0.060 volt when the switch is turned off.
Just wanted to get a figure of the foward voltage of the inductor for a 1mS turn off time.
Tryed two test one is 1.87amp 2.4amp
0.060volt 0.10volt With the 100nF and 10kOhm keeping the switch on for time that the voltage drops below 3.35v level of the hex
Expanding the values to 100amp would get 90.6 volt for that charge up time, blowing up ifp540n fet, would like to use them just will always have low resistance path across the inductor to stop the voltage rising.
Registered Member #4465
Joined: Wed Apr 18 2012, 08:37AM
Location: Bucharest, Romania
Posts: 145
That's not so simple to test with only an multimeter. You need to use an oscilloscope in order to measure the time constants of the rise voltage across the inductor. However, for this experiment, if we are using the MOSFET as the main switch, we increase the resistance value by parasitic on resistance of the MOSFET, so the transient time and therefore the time constant of the circuit becomes shorter. This is because as the total resistance increases, and the circuit becomes more and more resistive. If the value of the resistance is increased sufficiently large compared to the inductance the transient time would effectively be reduced to almost zero.
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