Re: 12v battery protector circuit
From: andy (news4_at_earthsong.free-online.co.uk)
Date: 08/27/04
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Date: Fri, 27 Aug 2004 02:29:06 +0100
On Thu, 26 Aug 2004 19:16:16 +0000, Jonathan Kirwan wrote:
> On Thu, 26 Aug 2004 17:06:12 +0100, andy <news4@earthsong.free-online.co.uk>
> wrote:
>
>>Could do, but I'm still looking for a way where there's no current draw at
>>all apart from leakage once the circuit has turned off.
>
> Andy, your current circuit draws current because the BJT applies some gain to
> the small leakage current through the zener (perhaps 100X or so) and this is
> then allowed to flow through the collector. The zener leakage is probably
> something like a microamp or so, so this suggests something like 100uA of
> consumption for your circuit.
I hadn't thought of that, I admit.
> Frankly, I'm not even sure what your circuit is supposed to be doing. Those
> 100k resistors seem very big.
I was trying to keep the current down as much as possible - the main
circuit only uses 0.8 mA, so I'd like this part to use maybe up to 0.2 or
0.3mA when it's on, and under 10 uA when it's off.
The idea of it is -
- when Vin is below Vzener plus 0.7 volts, there's no current except for
leakage, because both devices are off (but I forgot about this being
amplified, as you said).
- when Vin goes above Vz+0.7, then current will start to flow in the
zener. This will create a collector current, which will start to increase
the voltage across R2 and R3. R3 is meant to give a negative feedback - as
the collector current increases, this will push the base voltage on the
transistor closer to Vin again, tending to switch it off.
- so Vg will rise until the transistor saturates. The maths I worked out
has this happening when Vin=(Vz+0.7)*(R2+R3)/R2
- then the voltage across R1 starts to rise, which stops the base current
going too high after the transistor has saturated.
> For one thing, zeners are spec'd to operate at
> much higher currents than you'll get with your 100k resistor. For another, your
> zener will leak current to the base of the BJT and the BJT will apply gain to it
> to yield collector currents that are some 100X higher.
>
> In the case where the zener hasn't yet reached its zener voltage, it will just
> leak at say, 1uA. Thus, your Vg will be:
>
> Izener ~= 1uA (rough guess)
> V(R1) = Izener * R1 ~= 0.1V (negligible)
> so, Ib(BJT) ~= Izener, Ic(BJT) = beta * Ib(BJT) ~= 100uA (estimate, for now)
>
> With an Ic of 100uA and an (R2+R3) of slightly over 100k, you'd expect to see
> about 10V across them. Since we already know as a rough guess that the voltage
> supply is below the zener voltage and thus below 10V, we can be pretty sure that
> V(CE) is small and that the transistor is probably saturated (reducing the beta
> so that the actual Ic matches up with (R2+R3) to yield about the supply voltage.
>
> So, in effect Vg would basically just track your supply voltage before the zener
> voltage kicks in, placing the supply voltage directly on your MOSFET gate.
> Mostly because of that huge R2 value, though.
>
> But what happens after the zener has kicked in? Well, Ib will be even higher of
> course. And V(CE) will just be that much less (not much less, because it was
> already saturated to begin with.) So Vg would be about the same then, too.
>
> So you might as well just hook up the supply to Vg and be done with it.
It didn't do that, honest. I'll try building again to be sure.
> This all changes some if you make R2 a smaller value. I'm not sure if you are
> telling us correctly about the 100k value on R2, or not. But if R2 is a lot
> smaller than 100k, then the case before breakdown and the case after breakdown
> look different, which is probably what you are after. Then Vg is held down to a
> low value by the much smaller R2 and R3 against the relatively smaller looking
> Ic=beta*(Ib=leakage) until the zener breakdown when the Ib rises (although
> again, I suspect your R1 should be something less than 100k as even these tiny
> Ib currents present a real voltage drop across something that big and your zener
> really needs more substantial currents after breakdown, anyway) and then more
> substantial Ic values can begin to work against R2+R3 and cause Vg to rise up
> towards the supply voltage.
>
> But any reasonable way you work it, your Ic is going to be at least beta times
> the zener leakage and that's going to set the total leakage of your circuit,
> when off.
>
> I also think Gene's point about having to have some kind of comparison circuit
> operating, even when the MOSFET is disabled, is true enough. You'll need
> something running.
The idea of that circuit was that you don't, because it's relying on the
transistor and zener shutting off when Vin goes below their combined
switch off voltage. But I hadn't thought about the transistor amplifying
the leakage, as you've said.
> But I believe you can get this down to about a microamp or
> two with predictable performance, with some careful design. (I'm not a good or
> professional designer, but I do have an idea how this might be achieved.)
>
> Jon
I did build it as in the diagram, and it worked as I said with those
resistor values - Vg stuck at 0V for Vin = 0 to ~10V, and swung up to near
Vin as Vin went from ~10 to ~10.5V. I probably am using the devices
outside the specs though, so maybe I was just lucky.
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