I have an electromagnet that I want to switch on or off quickly. After making the prototype I found that the magnetic force remains for about 1 sec after I turn the current off . This is a problem as I want the magnetic force as quickly on or off as possible. Right now the electromagnet is made of a 2000 turn solenoid, and current in the solenoid stabilizes at 0.6 amps (after the annoying slow ramp up).

If I were to bundle 4 strands of magnet wire and coil the bundle 500 turns (500 x4 =2000 turns) and then connect each wire in parallel, then adjust the voltage so that the current in each wire is still 0.6, would that improve the response time of the magnet (by lowering the impedance and therefore reactance of each individual coil) ??


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Your inductance values seem too small \$\endgroup\$
    – Autistic
    Jan 9, 2017 at 11:17
  • \$\begingroup\$ Please ignore the induct and volt values, inreality the induct of the 1 schematic is 10 to 17 H and supply is 48 V \$\endgroup\$ Jan 9, 2017 at 12:24

2 Answers 2


If you are driving the coil with (say) a transistor you will have a diode across the coil to prevent back-emfs destroying the transistor however, this makes the turn off time quite long so you might consider, as a simple option, using a zener diode in series with the protection diode. This burns the stored energy much more quickly.

If I were to bundle 4 strands of magnet wire and coil the bundle 500 turns (500 x4 =2000 turns) and then connect each wire in parallel, then adjust the voltage so that the current in each wire is still 0.6, would that improve the response time of the magnet (by lowering the impedance and therefore reactance of each individual coil) ??

You have to keep ampere turns the same for producing the same mechanical force so, 500 turns x 2.4 amps = 2000 turns at 0.6 amps. In other words, it's a bit simpler than what you propose but you can do it your way to re-use the wire.

Force = \$(N\cdot I)^2\cdot 4\pi 10^{-7}\cdot \dfrac{A}{2g^2}\$

  • F = Force
  • I = Current
  • N = Number of turns
  • g = Length of the gap between the solenoid and the magnetizable metal
  • A = Area of solenoid/electromagnet

Reducing turns by a factor of 4 (irrespective of number of parallel windings) reduces the inductance by 16 (ratio of turns squared) so your response speed improves by a factor of 16. This assumes the same core and geometry of windings on that core (which I'm sure in this example will be true). So if you only wished to quarter the inductance (1/4 the response time) then halving the turns to 1000 would work and only requires 1.2 amps.

  • 1
    \$\begingroup\$ thanks for the instructive comment. Could you explain why the zenner in series with the protect diod would help ? \$\endgroup\$ Jan 9, 2017 at 12:11
  • \$\begingroup\$ Because the stored energy in the coil gets depleted more quickly - the down side is that a bigger back-emf is created but you choose the transistor to suit that higher back-emf. \$\endgroup\$
    – Andy aka
    Jan 9, 2017 at 12:34
  • \$\begingroup\$ what in the zenner makes it deplete more quickly, is the voltage drop different or something ? \$\endgroup\$ Jan 9, 2017 at 14:51
  • 2
    \$\begingroup\$ If you didn't have any reverse protection voltage limiter, all the stored energy would produce a spark that would dissipate all the energy very quickly. On the other hand, with a simple diode, voltage limits to about 0.7 volts and so the current that is flowing when the switch opens (0.6 amps) can only initially dissipate 0.42 watts. Therefore with a finite amount of energy to get rid of, a diode will take longer than a zener diode. Say it is a 15 volt zener - initial current is 0.6 amps and hence it is initially burning that energy at 15*0.6 watts (joules per second). Hence it is quicker. \$\endgroup\$
    – Andy aka
    Jan 9, 2017 at 14:56
  • \$\begingroup\$ I get it now, so a resistor would do the same job right ? \$\endgroup\$ Jan 9, 2017 at 15:50

Take a couple of steps back to energy, this will allow you to clear your head of the details of turns, volts, inductance etc.

When the magnet is on, and operating at its design flux, there will be a certain amount of energy stored in the magnetic field.

If you are happy to change the field slowly, so change the amount of stored energy slowly, then you only need a small amount of power. To change it quickly, you need a large amount of power. Power is rate of change of energy.

Power is volts times current. Now unfortunately, the copper wire has resistance, and this confuses the situation a bit, because we also need power to push the current through the resistance of the copper wire. However, let's simplify for a moment and imagine that we have a superconducting solenoid, like an MRI magnet. Just like a spherical cow in a vacuum beloved of physicists, this allows us to concentrate on the important bits.

To change the field quickly requires a lot of power, so more volts at any given current. If you alter the windings so you need more current to produce the field (reduce the inductance), then you will need fewer volts to slew the field at the same speed, and vice versa.

For any given electromagnet, you slew it as fast as possible by using the maximum drive voltage you can achieve. This might be limited by the breakdown voltage of your windings, or switching devices, or power supply. To slew the field up, you need to supply energy. To slew the field down, you are withdrawing energy, so you can generate the voltage by passing the current through a resistor, you do not need a power supply.

If the voltage range you can handle is limited, and you are not yet at your current limit, then you can improve the slew speed if you rewind your magnet to use more current, obviously limited by the maximum current you can tolerate. This will reduce the voltage needed to get the same slew power.

Finally, what about that pesky copper resistance? It means you have to supply power to keep the magnet on, and that your slew up takes longer, and your slew down is quicker. It does not alter the general conclusions that more slew power means faster field changes.

  • \$\begingroup\$ if you see the schematics above in all inductances the current is 0.6, however a the power supply level : in the 1st case U=u and I=i and the inductor has O=o Ohms, power is P=u x i in the 2nd case U=u/4 and I=ix4(because each induct has o/4, and I/4) . P = u/4 x i x 4 We see that power doesnt change but the response time does, right ? \$\endgroup\$ Jan 9, 2017 at 12:08
  • \$\begingroup\$ If you increase the current by 4x to 2.4A and keep the slewing voltage the same at 48v, you have increased the slewing power by 4, so you can change the field 4x as quickly. Once you get to operating field level you need to drop the applied voltage back to 12v, to keep the same resistive dissipation power. If OTOH you apply 12v, so that the current leisurely ramps up to its final value, then your time constant is just the same, as your slewing power is just the same. Another way to look at it is both R and L go as turns squared, so your L/R time constant stays the same \$\endgroup\$
    – Neil_UK
    Jan 9, 2017 at 14:34

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