Circuit efficiently drives inductive loads
Carlisle Dolland, Honeywell Engines and Systems, Torrance, CA -- EDN, 5/25/2000
In the driver circuit in
Figure 1, the system controller provides the V
COMMAND signal. V
COMMAND equals the desired load current multiplied by R
8. When the controller applies this voltage to R
1, the output of IC
1 goes high, applying voltage to the gates of Q
1 and Q
2. These transistors turn on, allowing load current to flow to ground through Q
1 and R
8. The current in the load ramps up, and a voltage proportional to the load current, sensed by R
8, feeds back to the inverting input of the comparator IC
1. When this voltage exceeds the voltage at the noninverting input, the output of IC
1 goes to ground. Q
1 and Q
2 then switch off. The load current now circulates around the loop comprising D
1 and L
1. During this time, the slope of the load current becomes negative because of the dissipation in D
1 and the load resistance. The duration of this phase of the circuit's operation is a function of the hysteresis (set by R
1, R
2, and R
4) and the decay of the voltage across C
2 (essentially a function of R
9). C
2 and R
9 determine the ripple current in the load. The circuit cannot use a power MOSFET for Q
2, because of the intrinsic drain-to-source diode. You must use a device without the intrinsic diode, such as a 3N71. (DI #2535)
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