Need to drop 0.2 V with 2-ternimal series device

cell.

so

much

to

invert

Yes. But with a .15V reverse bias I don't see it working.

oops that was 0.2V and not .15. Hard trying to work and play. It'd seem the original purpose of the diode isn't there ... because you're forcing a 0.2 voltage onto it. I'd opt for a non inverting amplifier circuit.

Looks interesting... but won't it need a large-value resistor to the anode of the LT1389 to get it to start up? Looks to me like it would come up with the output = 0 and just stay there. Or could we rely on noise to give it enough of a "push" to get it started?

Excuse me, you're considering an opamp "like a 741" in a circuit with a 1.5V total supply voltage? Darlington stages? * Cough * Have you studied low-voltage RRIO opamps? Ahem, internally these are about as far from a 741 configuration, etc., as you can get.

Another consideration for these modern LV amps is that the amplifier gain approaches 0 at rail saturation voltages- so that Vin,diff=Vout/Aol is untenable, the differential cannot exist with enough magnitude to support that locked up state.

No it's not- look at the battery polarity again.

off

Who said I was considering a 741 in that circuit? darlington? Didn't you say that the opamp shouldn't necessarily be RRIO type? My post was to show that an opamp with the same internal circuitry like a 741 in the circuit above with other slight modifications wouldn't have start up problems and even if it did they could easily be solved.

BTW The circuit would have a worse problem. you've got a saturation conditions at the opamp inputs. V(-)= 1.434V and V(+)= 0.115V.

I'm not sure where you get those numbers, or what point you're trying to make, but I'll say this, sub-100mV collector-saturation voltages for sub-100uA currents (i.e., Rsat under 1k) is entirely reasonable for some low-power opamp output stages.

Notice I haven't suggested a specific low-voltage opamp - that's too much like work (and it's the OPs task). There is a useful industry term, sub-1V, which yields 1170 hits on Google. Good reading.

For example According to the zener specs it should drop 1.25V. That would mean .3 and 1.35 between the output and the V(+) node , across the 68K resistor. Meaning current through resistor is 1.05/60.8K which doesn't add up to whats written there.

I tried a Microcap simulation and i got V(-) almost equal to the supply voltage, which is more or less what I'd expect from Positive feedback increasing to infinite the output impedance which somewhat agrees with my earlier calculation.

Why not ground both the zener anode and the 1M node connected to the supply. Less fidgeting and circuit is more predictable. Vout is then easily determined and controlled by the expression Vzener*(1 + R2/R1)? R2, R1 now are the 1M and 80.2K resistors

You seem rather badly confused, there's 100mV across the resistor when the circuit is at equilibrium. The circuit is an output-voltage regulator, not a fixed-voltage-dropper, as the OP originally proposed, perhaps that explains your confusion?

The overall feedback is negative, not positive at equilibrium. That's because the zener dynamic impedance is much much lower than 68k. The output impedance is certainly NOT infinite. I'm going to sign off.

proposed,

The circuit I proposed is a fixed output voltage regulator so its all clear on my side. I see confusion somewhere else. Unless we are looking at 2 different circuits. Stating that there's 100mV across the resistance doesn't help the argument much. Writing the loop equation you get -1.55 +1.25+ IR*VR + Vout =0; Now you say Vout is 1.35Volt. Show me how that gets you 100mV for IR*VR?

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This problem with the camera battery has been beaten to death. A number of solutions within the parts count and size limitations have been proposed.