Scheme provides automatic power-off for batteries
Edited by Bill Travis
Miguel Gimenez, Altair-Equipos Europeos Electronicos, Madrid, Spain -- EDN, May 13, 2004
The circuit in Figure 1 provides a simple and inexpensive way to protect one of the most valuable components in portable applications: the battery. Applications include all portable equipment that requires a limited time of operation, such as test instruments, guitar tuners, and electronic toys. Pressing the on/off momentary switch starts the cycle, and the circuit provides power to the application circuit. If you again press the switch at any instant, the circuit switches off and "sleeps" until the next cycle. In case you forget to switch off the circuit, the circuit incorporates an auto-power-off function with a time period that is a function of preprogrammed time constants.
IC1 and related parts provide a bistable toggle function and also ensure protection against switch-contact bounce. IC1C buffers the toggled signal and isolates the R1-C1 charging current. This signal feeds the IC2 timer, configured as a monostable multivibrator that remains activated until it times out, according to the expression t=1.1×RTCT. This figure is the auto-power-off time. In the example, this interval is approximately six minutes. The timer's output feeds the Q1 inverter that activates medium-power, pass-through Q2 transistor. This circuit is configured as a pnp block to ensure low losses to the load. The loss comes only from VCE(SAT)—approximately 0.2V at 100 mA, or 20 mW. For applications demanding more current, you can choose a more suitable transistor. A MOSFET can be an efficient approach when you need either lower standby losses or a lower voltage drop between the battery and the application circuit.
Standby losses in the switched-off state are negligible, because the circuit draws power only from the CMOS gate in the inactive off-state. LED D1 indicates the on-off status of the circuit. No extra power comes from the battery to drive this LED, because it is connected in the current-source leg of the driver transistor. The output transition to 0V during time-out ensures the timed power-off by means of the C5 feedback loop that toggles the bistable circuit to the off state, performing the same role you might have forgotten with the on/off switch. This simple circuit is useful when the application doesn't require a microcontroller.
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I an certain the schematic of fig.1 is wrong. While I haven't built this circuit my intuition says it can't work as shown.
The line from the junction of R2, R3 and IC1b pin 6 to the line shown as connecting to the Vcc supply pin on IC1c and C6 etc should not be there. The correct arrangement should be that the 9V battery +ve should instead connect to IC1 Vcc supply pin, C6 and Q2 emitter.
Ross Herbert - 2005-6-4 21:49:00 PDT -
Sorry Mr Steve Winder, you forgot two main problems:
-Switch-off of the Mosfet driven with a simple RC (6 minutes) is not a good and abrupt step and it can damage the application or the mosfet in the linear transition.
-You can not switch-off the circuit at any time.
the author - 2004-24-5 15:13:00 PDT -
I suggest using a P-channel MOSFET in place of Q2, with a parallel R & C between source and gate, and a push-putton connecting the gate to 0V. Throw the rest of the circuit away. When the button is pressed C charges and Q2 turns on. When R discharges C over time and Q2 turns off.
Steve Winder - 2004-13-5 10:58:00 PDT
