Voltage cut off circuit

I have a 2 way radio that will run with wide range of input voltages(5V-16V). I mostly run it on batteries. Since in stand by mode it draws very low current and will work with as little as 5V, it often runs the NiCd/NIMh batteries down to too low of a voltage. Since I put together my own battery packs, I was wondering if there was a way for me to add a simple circuit that would just cut off the power when the battery voltage dropped below some set point. The radio draws about 2 amps on high power transmit. Of course I don't want this circuit to draw much additional current and preferable zero current once the shut off point is reached. When in standby the radio only draws about 15 ma. The set point would vary with the battery pack, some are 6 cell some

8 and some 10. The set point will probably be 1.1V/cell
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Chris W
KE5GIX

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Reply to
Chris W
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try this (view in fixed font)

p-channel mosfet s d bat------+--------- -----------+------load pos | |___| | | _ | 10K | | | | | | | ___/ zener '---------+ /_\\ diode | | \\c | npn |-----R-----' /e | gnd

R can probably be a few K, haven't taken the time to figure ought what it ought to be really.

When the voltage at the load drops below Vz+Vbe, the bipolar transistor will turn off, releasing the mosfet gate, which will turn the mosfet off (the gate-source resistor makes sure of that). Load on your battery will then be zip. Not even a microamp, my guess. You must also put a momentary contact switch across the mosfet. Put in a fresh battery and push the button, turning on power to the load. It will then stay turned on until the voltage falls, either because the battery runs down or the load presents a short. In other words, this circuit can also act like a circuit breaker tripped by voltage drop at the load.

A resistor from the zener anode to ground might not be a bad idea, but I'm not sure that's necessary. Another option instead of the zener is a voltage divider from load to ground, tie the transistor base to the divider midpoint and dispense with the base resistor R.

Reply to
kell

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What\'s going to turn on the MOSFET?
Reply to
John Fields

You have to turn the circuit on, because as you see it doesn't turn on by itself. A momentary pushbutton parallel to the mosfet ought to accomplish this.

Reply to
kell

On 23 Jun 2006 07:14:38 -0700, "kell" wrote:

--- Here's a better way:

IRF7404 VBAT>--+----------+-----------+-------S D---> LOAD | | | G | [82.5K] | | | | +---|---[1M]--+ [412K] | | | | | +-[1K]--+--|+\\U1A | | A | | >-------+ +---O [37.4K] | +---|-/ | C | | +---O | | | K|2.5V | [150K] [LM385] | | | | GND>---+----------+-----------+---------------> GND

With a Vbat of 13.2V, U1A- is connected to "C", with a Vbat of 9.6V to "B", and with a Vbat of 7.2V to "A." That will cause the circuit to shut down power to the load when battery voltages fall to 11V,

8.8V, and 6.6V, respectively.

Total current should be less than about 200µA, but simulating it with:

Version 4 SHEET 1 944 680 WIRE -624 256 -624 0 WIRE -624 576 -624 336 WIRE -624 640 -624 576 WIRE -400 0 -624 0 WIRE -400 32 -400 0 WIRE -400 128 -400 112 WIRE -400 144 -400 128 WIRE -400 256 -400 224 WIRE -400 304 -400 256 WIRE -400 400 -400 384 WIRE -400 432 -400 400 WIRE -400 576 -624 576 WIRE -400 576 -400 512 WIRE -368 128 -400 128 WIRE -368 256 -400 256 WIRE -272 400 -400 400 WIRE -272 400 -272 272 WIRE -224 432 -224 304 WIRE -224 576 -400 576 WIRE -224 576 -224 512 WIRE -176 304 -224 304 WIRE -64 304 -96 304 WIRE -64 368 -64 304 WIRE -48 368 -64 368 WIRE 32 272 -272 272 WIRE 32 304 -64 304 WIRE 64 0 -400 0 WIRE 64 96 64 0 WIRE 64 256 64 176 WIRE 64 576 -224 576 WIRE 64 576 64 320 WIRE 128 288 96 288 WIRE 128 368 32 368 WIRE 128 368 128 288 WIRE 240 0 64 0 WIRE 256 288 128 288 WIRE 256 288 256 48 WIRE 256 384 256 288 WIRE 256 576 64 576 WIRE 256 576 256 464 WIRE 432 0 336 0 WIRE 432 256 432 0 WIRE 432 576 256 576 WIRE 432 576 432 336 FLAG -624 640 0 SYMBOL pmos 336 48 M270 SYMATTR InstName M2 SYMATTR Value IRF7404 SYMBOL voltage -624 240 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR Value PULSE(0 13.2 0 1e-3 5s 1s) SYMATTR InstName V1 SYMBOL Comparators\\\\LT1017 64 288 R0 WINDOW 0 -76 -59 Left 0 WINDOW 3 -96 -91 Left 0 SYMATTR InstName U1 SYMBOL res -416 128 R0 SYMATTR InstName R2 SYMATTR Value 61.9K SYMBOL res -416 416 R0 SYMATTR InstName R3 SYMATTR Value 150K SYMBOL voltage -224 416 R0 WINDOW 0 -51 101 Left 0 WINDOW 3 -60 132 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 2.5 SYMBOL res 416 240 R0 SYMATTR InstName R4 SYMATTR Value 6.6 SYMBOL res -80 288 R90 WINDOW 0 61 59 VBottom 0 WINDOW 3 68 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 1000 SYMBOL res 48 352 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R6 SYMATTR Value 1E6 SYMBOL res -416 16 R0 SYMATTR InstName R5 SYMATTR Value 412K SYMBOL res -416 288 R0 SYMATTR InstName R7 SYMATTR Value 37.4K SYMBOL res 240 368 R0 SYMATTR InstName R8 SYMATTR Value 1e6 SYMBOL res 48 80 R0 SYMATTR InstName R9 SYMATTR Value 1 TEXT -594 602 Left 0 !.tran 30 TEXT -520 400 Left 0 ;V1 = 13.2V TEXT -512 256 Left 0 ;V1 = 9.6V TEXT -512 128 Left 0 ;V1 = 7.2V

Shows that a single comparator is drawing roughly Vbat/1000R, when the data sheet, at:

formatting link

shows it as 100µA worst case with A 40V supply and the chip ay 125C.

Dunno what's up with that, but unless I made a mistake, there's probably something wrong with the model. If Helmut's listening, maybe he'd care to comment?

-- John Fields Professional Circuit Designer

Reply to
John Fields

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