EDN Access -- 09.01.94 Solid-state relay saves battery's lif

-September 01, 1994

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Design Ideas:September 1, 1994

Solid-state relay saves battery's life

Tom Gay,
Siemens, Regensburg, Germany

The circuit in Fig 1 uses a low-resistance p-channel FET, Q4, to disconnect a load from the battery when the battery's voltage drops below a certain value. After you disconnect the circuit, it remains off until you remove and replace the battery.

Q4's minimum gate-source threshold determines the minimum operating battery voltage of the circuit. For most FETs, this minimum battery voltage is 5V.

The circuit's operating current is low; using the resistor values in Fig 1, the circuit consumes less than 6 µA from a 9V battery. The circuit also features a wide input-voltage range, which depends on the maximum collector-emitter voltage rating of the transistors Q1 through Q3 and the FET Q4.

In operation, Q1 and Q2 form a Schmitt trigger that senses the circuit's output voltage. In normal operation, Q1 is on while Q2 is off. R5 and R6 pull the base of Q6 high, such that Q4's gate-source voltage across R8 equals -VBATT. For battery voltage higher than 10V, select R9 according to

R9=R8(VBATT/VGS max -1).

However, the circuit does not need R9 if the battery's voltage is 10V or less. D1 generates a reference voltage of approximately 1.4V. Because the current through D1 is low, use the common low-current trick of substituting an LED for a Zener diode. An LED has lower dynamic resistance than a Zener diode does at low currents.

The voltage divider R1, R2, and R3 trim the minimum output voltage at which the Schmitt trigger toggles. When the Schmitt trigger toggles, it turns Q3 off and Q2 on. To turn 3V on, Q3's base voltage must decrease to a value below the transistor's VBE min:

VBE(Q3)=R7/(R6+R7)(VLED+VCE sat(Q2)).

Q3 turns off, and Q4's gate-source voltage goes to zero. Q4 disconnects the load. Depleted batteries sometimes recover some of their voltage (but none of their capacity) after disconnection. To avoid reconnection of a depleted battery, the load remains disconnected, even if the input voltage slowly rises again.

Only inserting a new battery resets the circuit. After you remove the old battery from the circuit, leakage currents discharge capacitors C1 and C1 in a couple of seconds. When you insert a new battery, the gate-source capacitance of Q4 charges with the current flowing into C1, connecting the load. For safe turn-on, C1 m ust now be less than the gate-source capacitance of Q4. Keeping C1 below three times Q4's gate-source capacitance ensures a reasonable reset time for the circuit.

Replacing potentiometer R2 with an NTC resistor protects the battery from excessive temperatures. Diode D2 should be in good thermal contact with transistors Q1 and Q2 to reduce the influence of junction temperatures on the turn-off voltage. You could also consider paralleling FETs to increase output-current capability and decrease resistance. The optional switch S1 resets the circuit to an on state and proves useful if the load experiences a temporary short circuit. (DI # 1584)

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