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[Ideas For Design]

Driver Kick-Starts Sluggish Solenoids



Contributing Author  |   ED Online ID #4996  |   December 4, 2000

Article Rating: Not Rated

A fundamental problem associated with driving solenoids is that the initial force is low when the moving part is in its farthest position. To overcome such an obstacle, this circuit briefly applies additional drive at the onset of activation (Fig. 1). It uses the solenoid’s flyback current to steer the discharge of the timing capacitor after the solenoid’s release.

In this circuit, a solenoid is driven with twice its rated voltage for a fraction of a second, just enough to let it complete its stroke, before letting the drive return to its nominal value. By doing so, a small part can be made to do the work of a much larger unit without exceeding its power rating. Figure 1 demonstrates the basic principle of this circuit, which was developed for a valve-drive circuit in an electrically controlled pipe organ.

The key contact turns on the solenoid. Assuming that the charge on C is zero, the key also turns on the p-channel MOSFET Q1. Consequently, Q1 switches the drive voltage to twice its nominal value. After a time determined by (R1 + R2) × C, the FET turns off, restoring the drive to normal. So far so good, but what happens when the key is released? Capacitor C is left with no possibility of discharge. And, since a fast repetition capability is essential, the circuit clearly has to be adjusted. Figure 2 shows a modified drive circuit, which makes use of the solenoid’s discharge current through flyback diodes D2 and D3.

Upon the release of the key contact, the solenoid sets up a discharge current through the flyback diodes. (The capacitor discharge circuit requires an additional diode for protection against reverse voltages. Therefore, two diodes in series are used.) Transistor Q2 mirrors the discharge current so that capacitor C is quickly discharged through its collector. The exponential decay of the solenoid current matches the exponential V/I characteristic of the transistor VBE, resulting in an almostconstant voltage across the emitter resistor R3 during the decay. This, in turn, results in a linear decay of the capacitor voltage, at a rate dependent on R3.

For a solenoid drawing approximately 300 mA of current at 24 V, suitable component values for a 100 ms turn-on kick are:

Diodes: 1N4004
FET: BUZ271
NPN: BC337
R1: 33k
R2: 6.8k
R3: 22 Ω




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