Circuit efficiently drives inductive loads
Carlisle Dolland, Honeywell Engines and Systems, Torrance, CA -- EDN, May 25, 2000
In the driver circuit in Figure 1, the system controller provides the VCOMMAND signal. VCOMMAND equals the desired load current multiplied by R8. When the controller applies this voltage to R1, the output of IC1 goes high, applying voltage to the gates of Q1 and Q2. These transistors turn on, allowing load current to flow to ground through Q1 and R8. The current in the load ramps up, and a voltage proportional to the load current, sensed by R8, feeds back to the inverting input of the comparator IC1. When this voltage exceeds the voltage at the noninverting input, the output of IC1 goes to ground. Q1 and Q2 then switch off. The load current now circulates around the loop comprising D1 and L1. During this time, the slope of the load current becomes negative because of the dissipation in D1 and the load resistance. The duration of this phase of the circuit's operation is a function of the hysteresis (set by R1, R2, and R4) and the decay of the voltage across C2 (essentially a function of R9). C2 and R9 determine the ripple current in the load. The circuit cannot use a power MOSFET for Q2, because of the intrinsic drain-to-source diode. You must use a device without the intrinsic diode, such as a 3N71. (DI #2535)
