Simple Low Voltage Shut Down

. . --+-----E C---- . | B . 6.2k | ZTX718 . | 470 . | | . | K . +-----R TL431 . | A $0.40 . 10k | . | | . --+-------+------

What does the emitter follower do? If it's a linear circuit and the emitter follower has a passive pull-down, that might waste a lot of power.

Got a schematic?

First, there's a lot of useful energy you can get from an alkaline cell between 1V an 0.9V (NiMH, too, but not as much).

Second, you want some hysteresis, or at a certain level of charge the thing will start madly turning on and off. Without hysteresis the current will get shut off, the battery voltage will rebound, the circuit will turn on, the battery voltage will get pulled low, and the cycle will repeat.

I'd design the circuit to work well down to 0.9V/cell, and I'd design the turn-off circuit to latch off -- maybe design the thing with an "on" pushbutton and an "off" pushbutton, with "off" coming either from the user or from the low-voltage cutout.

A little hysteresis would be good.

. 4.0-volt low-V shutoff. . . ZTX718 . --+-----E C-----+---- . | B | . 6.8k | 120k . | 470 5% hyst . | | | . +------ | -|

True, but a 1-volt battery with no load doesn't rebound very much. If the hysteresis is, say, 0.25V, it should stay off.

An Internet acquaintance of mine who used to be a battery engineer for GE claimed that nearly-fully-discharged NiCd cells would rebound to 1.2V if you gave them enough time, even if they couldn't deliver any significant charge.

But -- I'd have to test that theory on alkalines.

Alkalines are rather different, in a useful way. Their open-circuit voltage drops from 1.55V to 1.0 or 0.9V, etc, as they're depleted, providing a convenient way to evaluate their condition.

Is there a simple explanation for why this happens? It's a chemical reaction with voltages determined by chemistry/physics. I can understand why the series resistance would increase as the reagents are "used up". But, why does the open-circuit voltage change?

I don't mean to hijack the thread, just a few words in layman's terms or a link will suffice. Thanks,

That's a neat little circuit, but I was hoping for something passive.

Maybe a dumb question, but, given the low current, would it be possible to power the amp through a zener diode?

Robert Miller

What would that accomplish other than losing a lot of power and voltage to the amp?

The amp would be powered through the zener until the supply voltage is too low to cause breakdown. Then the zener would stop conducting.

The effect of dynamic resistance may need to be factored in.

I am willing to accept the standard diode voltage drop.

Robert Miller

The Zener would drop a constant voltage as the supply voltage drops leaving very little for the amp to operate from. As the voltage drops to the Zener breakdown voltage the current will tend to zero, but largely because the voltage on the amp will tend toward zero.

What standard diode voltage drop? Try drawing the circuit and calculate the voltages.

In engineering terms, passive means no transistors. Really? How could that work? My simple circuit is passive in the sense that no other power is required. But I do suggest the hysteresis version, with more hysteresis to force shutoff after the battery dies.

Since this is a battery-powered circuit, maybe a BSS209 PFET ($0.07 @

Cheers

Phil Hobbs