Answer 5—When the control signal is turned off, the current that was flowing through the transistor continues to flow through the diode, and this maintains the magnetic field until it dies away.

The current is reduced by the application of a voltage to oppose it. There are several ways to produce this voltage. One way is to rely on the distributed resistance of the circuit (mainly the solenoid and diode), which is normally on the order of a few tens of ohms.

In this case, the current collapses at an exponential rate, with a time constant of L/R, where L is the inductance of the solenoid and R is the total resistance. Increasing the resistance can reduce the time constant. This is accomplished by putting a resistor in series with the diode, where it has no effect when the transistor is on. The resistor must be sized so that the voltage across it (produced by the initial value of the coil current) doesn’t exceed the voltage rating of the transistor. In the example shown, a fivefold improvement in time constant (1.6 ms versus 8.3 ms) is achieved.

Another way to produce an opposing voltage is to put a Zener diode back-to-back with the existing diode. Again, the Zener should have a value that doesn't exceed the rating of the transistor. With this approach, the Zener voltage directly opposes the current, which ramps down linearly: dI/dt = V/L. In this example, the current decays from 200 mA to zero in approximately 2 ms.

In either case, the energy stored in the coil at switch-off (10 mJ) gets dumped into the resistor or Zener diode. If this happens often, it also will be necessary to make sure the device can handle the ongoing power dissipation.

Contributor: David Tweed