# Slow Down Inductive KickBack

Thread Starter

#### Vinnie90

Joined Jul 7, 2016
86
Hello fellas,

I have a question on inductive kickback. To my understanding whenever you have an inductive load, you need to place in parallel a snubber diode, to provide a discharging path to the inductor's current one the supply is switched of. The time constant of this process will be given by the inductance of the inductor and the total resistance it sees (diode + inductor resistance). For almost all the applications usually one wants to reduce the discharge time to achieve fast OFF state. The question is, how could I slow down the de-magnetization process? Adding a resistance in series to the diode would increase the time constant but also will require to dissipate o lot of power because of the built in voltage of the inductor.

Thanks a lot for the hints

#### #12

Joined Nov 30, 2010
18,217
Adding a resistance in series to the diode would increase the time constant but also will require to dissipate o lot of power
The power you need to dissipate will be exactly the same, with or without the resistor, because a resistor can not amplify the power stored in the inductor. I think it is time to throw some numbers into this to experiment about how much power is in play. By working with some real numbers, the behavior will be revealed. How much inductance? How much current? Voltage? Time?

#### crutschow

Joined Mar 14, 2008
27,006
This is related to the TS's other post.
Better to keep all discussion in that one.

Thread Starter

#### Vinnie90

Joined Jul 7, 2016
86
The power you need to dissipate will be exactly the same, with or without the resistor, because a resistor can not amplify the power stored in the inductor. I think it is time to throw some numbers into this to experiment about how much power is in play. By working with some real numbers, the behavior will be revealed. How much inductance? How much current? Voltage? Time?
Thanks a lot for the reply! So the inductor is an electromagnet with a Inductance of 1.5H and resistance of 50 ohms. I'm driving it with a current source that supply 0.1 A.

#### tcmtech

Joined Nov 4, 2013
2,867
I work with large scrap yard magnet systems which are little more than huge inductors that run on 240 VDC 40 - 80+ amps and their load dropping characteristics are determined by how fast or slow their magnetic fields collapse.

The only way I know of to slow down field collapse time is to reduce the input current being anything else is going to be limited by the natural magnetic and resistive properties of the unit itself.

The problem is if the field coils are shorted out through a flyback diode that's the slowest natural collapse rate they will make on their own. Just adding resistance in series with the diode quickens things because the circulating current generated by the magnetic field collapse will drive the voltage higher trading off where the magnetic field energy gets transferred to which is the resistor and not the natural decay limits of the shorted out field coils as they fight the magnetic field collapse effects.

For example a large electromagnet may have a resistance of ~ 3 ohms giving it a ~60 amp input at 240 VDC.
With nothing but a flyback diode the field collapse time (load drop time ) can be upto 8- 10 seconds.
However with a 3 ohm resistor in series with the diode that drop time is cut in half and again proportionally cut with a 6 or more ohm series resistor.
The down side is the peak voltage the resistor may see is its proportional share of the total series resistance of the circuit which for a 3 ohm is ~ 120 VDC and so on.

For the units I work on I tend to run a ~10 - 12 ohm resistor in series with the diodes which give a pretty impressive 1 or so second load drop time but the trade off the resistor has to be an open wire wound 500+ watt unit that can take a few hundred volt DC peak hit for a few tenths of a second every time the magnet shuts off.

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