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I have an inductor and seller claims that inductor supports a maximum of 5A current and he asked me to test it myself. Can someone help how to test that how much this inductor can support/ draw maximum current?

I have the following components.

  1. 24VDC power supply which supports 5A current.
  2. I have multimeter
  3. Breadboard.
  4. 120ohm resistor.

Please tell me the procedure for testing the maximum inductor current and what all the supporting components required?

P.S. I am not an electrical student. Please support me in the practical experiment.

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    \$\begingroup\$ you need to borrow an oscilloscope, then we can talk. \$\endgroup\$
    – Neil_UK
    Commented May 22, 2017 at 6:09
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    \$\begingroup\$ there's a Dilbert cartoon of a few years ago that has Wally banging a nail into a piece of wood, that's the prototype for the new smartphone. That's essenatially your situation. \$\endgroup\$
    – Neil_UK
    Commented May 22, 2017 at 6:15
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    \$\begingroup\$ I'm sure the inductor will cconduct 5A, what you need to know is whether it still has its specified inductance at that current. It may do, it may not. A multimeter and 120ohm resistor will tell you diddly squat. \$\endgroup\$
    – Neil_UK
    Commented May 22, 2017 at 6:16
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    \$\begingroup\$ There are few dumb questions. I'm just pointing out that the equipment you have to hand cannot be configured to do the job you want it to, you need more stuff. \$\endgroup\$
    – Neil_UK
    Commented May 22, 2017 at 6:21
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    \$\begingroup\$ Some sanity checks can be made, perhaps. What's the inductance and what is the cross-sectional area of your inductor, around which the windings occur? \$\endgroup\$
    – jonk
    Commented May 22, 2017 at 7:15

3 Answers 3

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I usually measure inductor saturation current using the following circuit. You may need to adjust/replace components to your needs but the priciple is as follows.

x

Once the FET is turned ON the current through the Inductor starts to ramp up. If the inductor is not yet saturated this slope is constant. You can measure the current by measuring the voltage over the current sense resitor using an oscilloscope. You start with a very low duty cycle to make sure the inductor is not saturating at the beginning of the measurement. Then you start increasing the durty cycle while watching the current slope on the oscilloscope. At some point the slope of the inductor current will get more steep, thats where saturation starts.

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    \$\begingroup\$ I use basically the same method, but instead of the 555+MOSFET, I use a simple mechanical switch, and the scope is in single shot mode. I record the i=f(t) curve when I close the switch. \$\endgroup\$ Commented Nov 13, 2018 at 12:38
  • \$\begingroup\$ I wonder why this is done with a voltage as high as 12V? A lower voltage leads to a slower rise of the current, and therefore less critical timing and precise results presumably? \$\endgroup\$
    – mabartibin
    Commented Feb 14, 2022 at 15:44
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    \$\begingroup\$ Think you are right, but only if the series resistance of the inductor is low enough to allow a steep rise of current at saturation \$\endgroup\$
    – Jogitech
    Commented Feb 23, 2022 at 9:44
  • \$\begingroup\$ I figured out a reason for a higher voltage: most of the current through the inductor will be supplied by the capacitor bank. The drop of voltage during the discharge is proportional to the charge provided, and that charge is the integral of the current sent to the inductor: 1/2 * t_rise * I_max. So doing it quicker means a more stable voltage. \$\endgroup\$
    – mabartibin
    Commented Mar 3, 2022 at 19:01
  • \$\begingroup\$ Also note that the pins on the left and right sides of the 555 are switched :-) \$\endgroup\$
    – mabartibin
    Commented Mar 3, 2022 at 19:38
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You will need two types of equipment:

  • Low-value resistance
  • Oscilloscope

Place resistor before the inductor & measure the voltage over this resistance using oscilloscope then calculate the current by Ohm's law (V=IR)

Oscilloscope will give you the accurate & every second result while multimeters calculate average values.

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  • \$\begingroup\$ Welcome to EE.SE! Ohm's law won't help here. \$\endgroup\$
    – winny
    Commented Jan 16, 2019 at 10:25
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You need to add variable resistor and connect it in series before inductor to control the current from source. Now you will have two options.

  1. using ampere meter with 5A or more rang selection, If you can arrange it than place it in series and vary the current and see value on ampere meter.

  2. Arrange voltmeter and low resistance (high wattage) resistor. Put it into series with inductor and using voltmeter you can measure voltage across this resistor and apply Ohms law to find your current.

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  • \$\begingroup\$ Can you provide me a sample circuit for the 2nd option like how the connections need to be made? \$\endgroup\$
    – kaviarasan
    Commented May 22, 2017 at 6:30
  • \$\begingroup\$ Its simple series circuit, All components will be in series in below sequence. +Ve of supply-->Variable Resistor-->Inductor-->Low value resistor-->load-resistor-->-Ve of supply. Than place voltmeter across "Low value resistor (should be mili Ohm or lower).\ \$\endgroup\$
    – Deep
    Commented May 22, 2017 at 6:34
  • \$\begingroup\$ But how will he see if the core saturates from that? You're only passing 5 A though the wire, saturated core or not. \$\endgroup\$
    – winny
    Commented May 22, 2017 at 6:52
  • \$\begingroup\$ It doesn't occur with air coils. here, I assume iductor with air coils. Visually, if you apply a constant voltage across an inductor, the current will rise linear. At saturation, the (differential) inductance drops, and the current starts to rise much faster. Hope Its clear. \$\endgroup\$
    – Deep
    Commented May 22, 2017 at 7:32
  • \$\begingroup\$ This will test whether the inductor has caught fire at 5 A. But it won't tell you whether it has saturated at 5 A, which is what's usually meant by an inductor's current rating. \$\endgroup\$
    – The Photon
    Commented Nov 3, 2017 at 16:12

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