Relay and solenoid driver circuit doubles supply voltage to conserve sustaining power

Relay and solenoid driver circuit doubles supply voltage to conserve sustaining power when driving the device into an actuation state. The post Relay and solenoid driver circuit doubles supply voltage to conserve sustaining power appeared first on EDN.

Relay and solenoid driver circuit doubles supply voltage to conserve sustaining power

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A generally accepted fact about relays and solenoids is that after they’re driven into the actuated state, only half as much coil voltage and therefore only one fourth as much coil power, are required to reliably sustain it. Consequently, any solenoid or relay driver that continuously applies the full initial actuation voltage to merely sustain is wastefully squandering four times as much power as the job requires.

The simplest and cheapest (partial) solution to this problem is shown in Figure 1.

 Figure 1 Basic driver circuit where C1 actuates, current-halving R1 sustains, then C1 discharges through R1 during Toff.

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But as is often true of “simple and cheap,” Figure 1’s solution suffers from some costs and complications.

  1. While R1 successfully cuts sustaining current by half, it dissipates just as much power as the coil as it does so. Consequently, total sustaining power is ½ rather than ¼ of actuating power, so only half of the theoretical power savings are actually realized.
  2. When the driver is turned off, a long recovery delay must be imposed prior to the next actuation pulse to allow C1 enough time to discharge through R1. Otherwise, the next actuation pulse will have inadequate amplitude and may fail. This effect is aggravated by the fact that, during actuation, C1 charges through the parallel combination of R1 and Rm, but during Toff it discharges through R1 alone. This makes recovery take twice as long as actuation.

Figure 2 presents a better performing, albeit less simple and cheap, solution that’s the subject of this Design Idea.

Figure 2 Q1 and Q2 cooperate with C to double VL for actuation, Q2 and D2 sustain, then Q3 rapidly discharges C through R to quickly recover for the next cycle.

Actuation begins with a positive pulse at the input, turning Q1 on which drives the bottom end of the coil to -VL and turns on Q2 which pulls the top end of the coil to +VL. Thus, 2VL appears across the coil, insuring reliable actuation. As C charging completes, Schottky diode D2 takes over conduction from Q1. This cuts the sustaining voltage to ½ the actuation value, and therefore drops sustaining power to ¼.

At the end of the cycle when the incoming signal returns to V0, Q3 turns on, initiating a rapid discharge of C through D2 and R. In fact, recovery can easily be arranged to complete in less time than the relay or solenoid needs to drop out. Then no explicit inter-cycle delay is necessary and recovery time is therefore effectively zero!

Moral: You get what you pay for!

But what happens if even doubling the VL logic rail still doesn’t make enough voltage to drive the coil and a higher supply rail is needed? 

Figure 3 addresses that issue with some trickery described in an earlier Design Idea: Driving CMOS totem poles with logic signals, AC coupling, and grounded gates.

 

Figure 3 Level shifting Q4, R1, and R2 are added to accommodate ++V > VL.

 Stephen Woodward’s relationship with EDN’s DI column goes back quite a long way. Over 100 submissions have been accepted since his first contribution back in 1974.

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The post Relay and solenoid driver circuit doubles supply voltage to conserve sustaining power appeared first on EDN.

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