アプリケーションノート 968

# Load-Disconnect Switch Halts Battery Discharge to Protect Rechargeable Batteries

Deep discharge can damage a rechargeable battery. The Figure 1 circuit, by disconnecting the battery from its load, halts battery discharge at a predetermined level of declining terminal voltage. Transistor Q1 acts as the switch. The overall circuit draws about 500µA when the switch is closed and about 8µA when the switch is open.

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]
R3 = 1.15 × R1/(VU - VL)
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).

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.