how to create a high current source?

Question

I'm looking to observe magnetizing offsets of various current sensors. The idea is that when you pass a high current pulse (close to the rating of the device) through a current sensor and then return the current to zero, there is some small offset in the device. This is due to some polar magnetization of the core inside of the sensor.

My issue is that some of the devices are rated for 200 amps. I would like to somehow create a current source circuit that is able it apply 50 amps for short duration pulses (20 milliseconds or so). I would fool the sensor into thinking its 200 Amps by wrapping the wire in loops so that it passes through the center of the sensor 4 times.

Note: the reason I cannot do 2 amps and 100 turns (example) is because the wire size begins to stack up and will not fit inside the sensor.

Does anyone have any ideas on this, or can redirect me to a post about this?

Answer 1

A capacitor can deliver a large current for a short time. You don't say how much voltage you need to create the 50 A, but let's say 1 V. That means your wire is 20 mΩ, which is doable. Let's say 2 V to account for other drops in the system.

Let's further say that you want the current not to deviate more than 20% at any one time from the 50 A average. This will allow the current to start at 60 A and drop to 40 A during your 20 ms test time. That would mean 2.4 V down to 1.6 V, for a drop of 800 mV. From that, just do the math:

(50 A)(20 ms) / (800 mV) = 1.25 F

That's one whopping big capacitor, or more likely, quite a bank of capacitors in parallel, although they only need to go up to 2 V.

This shows that more turns of thinner wire makes the problem easier since you can use a higher voltage. A car battery can produce 50 A for well more than 20 ms. You say you only have room for 4 turns of wire, but with thinner wire you can fit more turns. At only 20 ms duration, you don't have to worry about the wire vaporizing.

Try maybe some #22 magnet wire to see how many turns you can pack in there. That will also have more resistance, but require less current. Both those effects will help in getting a voltage source that can power it for the bursts you need.

Answer 2

I've got a cheap Mastech power supply that does 10A.. 20 turns of wire.

A car battery will do 50A. For the current control perhaps an opamp driving a FET with a 10-20 milli ohm current shunt that feeds back to the opamp. (Kelvin connections on the shunt).
I'm assuming that your load can float.

Answer 3

Thanks for the suggestions. I actually dug around a little more after posting this question and came up with this solution:

The capacitor takes close to 10 seconds to charge, but is able to produce a 50 A spike for close to 1 millisecond. This should be enough to cause some magnetization in the current sensor (a LEM current transducer; if you are interested).

24V was chosen because it was a power supply that I had readily available. Because of the 24mA current used to charge the capacitor (i'll make it a capacitor bank in the final design) a regular mechanical switch will be used. Due to the high 50 surge current upon discharging the cap, I will most likely use a MOSFET as a switch and tie the end of the resistor network to the drain and the source to ground. 15V pulses will be sent to the gate to turn on the "switch". Depending on readily available MOSFETs here in the lab, I may have to attach a MOSFET to each resistor and then tie all the gates together.

Let me know if you guys have any suggestions or concerns with this design.