Measuring back EMF induced current with multimeter

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I have built a solenoid with copper.

It is 19 cm long, it has 64 turns and the radius is 1.5 cm.

To trigger back EMF, I use a permanent magnet that has the same surface as the cross sectional area of the solenoid.

I put the magnet on one end of the solenoid and then rapidly pull it away.

The voltmeter shows indeed 1.2 mV. This value is pretty the same each time I driftly pull the magnet. Now when I do the same experiment but by using an ammmeter in series, I see 0.12 mA.

What I can’t explain is this:

Given that the multimeter shows a total resistance of 0.1 ohms for the solenoid, I am expecting the ampmeter to show 12 mA instead of 0.12 mA. I checked the meter ranges and accuracy many times but can’t figure it out. My hypothesis is that the 12 mA is the maximum current. But the ampmeter is measuring the current before it reaches the max value. Does that make sense? Given that I have no battery in my circuit?currentinductormultimeterinductanceemfshareedit  follow  flagedited 3 hours agoJRE36.4k88 gold badges6363 silver badges110110 bronze badgesasked 3 hours agoRabih Sarieddine3133 bronze badges New contributor

• 51. I don’t think you can make any kind of meaningful measurements of short events while using a multimeter. Multimeters react slowly. – JRE 3 hours ago
• 1Ah, the back EMF spike is too fast for the multi-meter to react. I remember the digital multi-meter response time is of the order of 100mS. I once used a scope to display the back EMF voltage spike. Let me see if I can find find the picture. I will be back. – tlfong01 3 hours ago   
• 1is the load the same in the two measurements? – jsotola 3 hours ago 
• 1Rabih Sarieddine have you watched Legend of Korra or just the last Airbender? – Helena Wells 3 hours ago
• 2What range is the ammeter set to? DMMs usually read up to 0.2V across a shunt resistor, so a 1.2mV source and a 10 ohm shunt (20mA range) would give 1.2 mV corresponding to 0.12mA, whatever current the solenoid could provide. – Brian Drummond 3 hours ago
• 1@tlfong01 S for siemens. s for seconds. – winny 3 hours ago
• 1I am back. Let me tell you how I display the back EMF voltage spike. I used a buzzer to do the test. The buzzer is using a BJT switching on/off the current passing a coil. I know when the current is switching off the back EMF thing happens. I am not using any fly back diode, and I worry the back EMF might fry my Rpi. I checked out the voltages at the C. B, E of the BJT. I surprisingly found the back EMF is +15 to -1V, when Vcc is 3~5V. Now there you are the back EMF selfies. / to continue, … – tlfong01 3 hours ago   
• 1The back EMF selfies: (1) raspberrypi.org/forums/…, (2) raspberrypi.org/forums/…, (3) raspberrypi.org/forums/…, (4) raspberrypi.org/forums/…. – tlfong01 3 hours ago   
• 1Thanks for you comments and ideas. The range is 20 mA. Regarding the load, my circuit is only made of the solenoid and the multimeter. – Rabih Sarieddine 3 hours ago 
• 1Do a measurement storing the measurand in a large capacitor, this will remove the uncertain meter response from the equation. – Neil_UK 3 hours ago
• @winny, Many thanks for pointing out my careless spelling mistake. I have been wrongly spelling milliSeconds as mS and kiloHertz as kHz. I need to go back to check if half of my units are wrongly spelled for years. Cheers. – tlfong01 3 hours ago    
• 1@tlfong01 kHz is correct 🙂 – winny 46 mins ago
• @Rabih Sarieddine, I think we need to clarifying a couple of things. I am just brainstorming some ideas. Please fee free to attack my imaginations. (1) The scary Back EMF event only occurs during the very short split second when you try to break the path that a continuous current I is (steadily or steadily) flowing in an inductor L (2) If I remember correctly, at the point in time just before you try to break the circuit, the energy stored in the inductor is E = L * I ^ 2. / to continue, … – tlfong01 1 min ago   Edit   
• (3) Never mind the scary formulae, it is just as “simple” as the Einstein’s “simple” equation E = M * C ^ 2 (which I think only less than 0.00001% of the humans in our planet thoroughly understand the significance of what the three letters E, M, and C stand for). Just remember that the inductor guy has some energy stored in his pocket, … Ah bed time. I call it another locking down day. See you tomorrow. Cheers. – tlfong01 just now   Edit   

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The voltmeter shows indeed 1.2 mV. This value is pretty the same each time I driftly pull the magnet. Now when I do the same experiment but by using an ammmeter in series, I see 0.12 mA.

So, if your imperfect ammeter has an input resistance of 10 ohms (it’s never zero BTW) then you will see 0.12 mA. Try putting a 1 ohm resistor across the ends of the coil and seeing what the induced voltage is.

_{Voltage is induced, current is not; current flows because of the induced voltage and load}shareedit  follow  flag answered 3 hours agoAndy aka294k1717 gold badges229229 silver badges517517 bronze badges

• ‘Voltage is induced, current is not; current flows because of the induced voltage and load’ . – Helena Wells 2 hours ago
• @HelenaWells did you mean to add something else to your comment? – Andy aka 2 hours ago
• No it just remembered me something funny.There was a classmate in high school I was 12 by that time who used to say that current can exist without a voltage ( which is wrong) and one day we had a test and he wrote that and he got B while he thought he should have got A- and started crying.Lol. – Helena Wells 2 hours ago 

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1. I don’t think you can make any kind of meaningful measurements of short events while using a multimeter. Multimeters react slowly.
2. The resistance of the coil is of less importance than the impedance. The impedance depends on the rate of change and the inductance of the coil. The rate of change will vary as the magnet moves.
3. Measuring the current as you describe, the current from the coil will be shorted by the ammeter. You will be measuring the current through the ammeter shunt. At that moment, the back EMF voltage will be much lower because of the short circuit. Your calculated voltage based on the resistance of the coil would be wrong because you have to account for the resistance of the ammeter shunt in parallel.

What you can do is to attach a resistor to your coil, then measure the voltage across the resistor while moving the magnet. The resistor goes to both ends of the coil – it is in parallel to the voltmeter.

From the measured voltage and the known resistance, you can calculate the power dissipated in the resistor and from that you can calculate a current value.

You should use an oscilloscope to measure the voltage. You can then pick out the peak voltage and determine the peak power, or make an accurate average of the voltage and determine an average power dissipation and average current.

You will find that resistors with different values will give different results. With some experimentation, you will find a value that will result in the most power dissipation. That will be at approximately the impedance of your coil.shareedit  follow  flag answered 3 hours agoJRE36.4k88 gold badges6363 silver badges110110 bronze badgesadd a comment

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