Circuit extends battery life
Using a relay, the circuit completely disconnects the batter during to eliminate power consumption.
José M Espí, Rafael García-Gil, and Jaime Castelló, University of Valencia, Valencia, Spain; Edited by Martin Rowe and Fran Granville -- EDN, July 29, 2010
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Two previous Design Ideas describe simple ways to automatically disconnect a battery from its load after a preset on period, which extends battery life (references 1 and 2). These circuits have little loss in standby operation, but they do draw some current. The circuit in this Design Idea presents a simpler way to perform the same function with fewer components and with no power consumption during standby operation (Figure 1). Moreover, the network comprising R2, D2, and C2 activates and deactivates the circuit. An additional control signal, control on/off, becomes slower than the battery’s on/off cycle.
Switching S1 to Position 1, the on position, the 24V battery quickly charges capacitor C1 through diode D1. That voltage drives transistor Q1 into saturation. Q1’s saturation magnetizes and activates relay coil L1, connecting the battery to the main power and control board. Meanwhile, capacitor C2 charges more slowly through 100-kΩ resistor R2, thus generating the control on/off signal with some delay relative to the relay coil’s closing. That scenario occurs after the proper power supply to the power stage and control circuits.
Switching S1 to Position 2, the off position, causes capacitor C1 to slowly discharge through resistor R1 when diode D1 is off. That action delays Q1’s turn-off. Before Q1 turns off, C2 quickly discharges through D2, indicating that the control should shut down the power. The relay switches off with minimum current. Once Q1 is off, the relay coil demagnetizes through R4 and D3. The relay switches off, disconnecting the main power and control board from the battery. During this off state, current flows neither in the on/off circuit of the management board nor to the main board.
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Talkback
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Q1 should be a FET; otherwise C1 will discharge through R3 and the base-emitter rather than R1.
Ron Bauerle - 2010-13-9 11:27:25 PDT -
Especially the diode D1 should deserve to connect a "small-value" resistor serially with it, so that resistor R1 would be paralelled with series connection of the D1 and the added resistor.
The reason is an excessive surge current flowing through the D1 at changing position 2-->1 at S1.
If you estimate a total resistance of the S1, leads and ESR of C1 as 1 Ohm; then 24A peak current can be expected to flow through D1. The corresponding time constant is 1 microsec.
The datasheet of the 1N4148 gives 2.0 A maximum surge current with a duration of 1 microsec.
So the mentioned "small" resistor could have a value of about 10 Ohm and its use might contribute to increasing the reliability of the circuit presented.
Marian Stofka - 2010-29-7 08:56:44 PDT
