Load-Disconnect Switch Halts Battery Discharge to Protect Rechargeable Batteries
Choosing the upper and lower voltage thresholds VU and VL lets you set values for R1, R2, and R3:
R1 = R2 × [(VL/1.15) - 1]To start the circuit, battery voltage (V+) must exceed VU. The micropower voltage detector IC1 then powers IC2, but only while V+ remains above VL. Otherwise, the loss of power to IC2 removes gate drive from Q1, turning it off. As shown, the circuit disconnects a 3-cell nickel-cadmium battery from its load when V+ reaches a VL of 3.1V. An approximate 0.5V hysteresis prevents the switch from turning on immediately when the load is removed; V+ must first return to VU (3.6V).
R3 = 1.15 × R1/(VU - VL)
IC2 is a dual charge-pump inverter that normally converts 5V to 10V. The capacitors C2, C3, and two diodes on the chip's positive-voltage side form a voltage tripler that generates an approximate 2(V+) gate drive for the high-side, floating-source MOSFET switch Q1.
Gate drive declines with battery voltage, causing the on-resistance of Q1 to reach a maximum of ≈0.1Ω just before V+ reaches its 3.1V threshold. A 300mA load current at that time will cause a 30mV drop at the disconnect switch; the drop will be 2mV to 3mV less for higher battery voltages. Resistor R4 assures turn-off for Q1 by providing a discharge path for C3.
Figure 1. When battery voltage drops below a threshold set by R1 and R2, the voltage-detector chip (IC1) removes power from the charge pump IC2, which turns off the high-side switch Q1 by removing its gate drive.