I am working on a project that has a single 1.5 volt battery in series with it. What I need to do when the battery is depleted or below .8 volts I need it to be permanently bypassed with a small value resistor in the order of 5 -50 ohms. I need very little current drawn < 50 ua with the monitoring circuit.

I have looked at depletion mode FETs as they are "ON" with 0 Vgs( battery dead). Ive found some with very low Vgs and all I can find is N channel. I could connect the positive of the battery through a 5 - 50 ohm resistor to the drain and connect the negative of the battery to the source, But the gate then has to be negative, how can I possible drive the gate without making up a inverter to give me a - v.

Thanks
Geoff

 

Do you have a schematic so we can see better what might work?

 

I can't go into exact details of exactly what this is as I had to sign a NDA. But basically BT1 can be approx .1 to .5 volts. BT2 is a 1.5 v dry cell. R2 varies but the current through it is only a few milliamps. R1 is in the range of 5 -50 ohms and its only real purpose is to limit current when the FET is turned on.

What I want it to do is when BT2 has depleted to its useful life around .8 volts or so I want to use Q1 to bypass it and basically have R1 across BT2 so current can still flow from BT1. Q1 is a depletion mode FET so it is on when there is no Vgs but somehow I have to get a negative Vgs to turn it off until BT1 reaches 0.8 volts. I cannot figure out how to get a negative voltage on the gate whichout making some kind of inverter cct. I have found some N channel depletion mode FETs with a Vgs of -.4 to -.8




Thanks
Geoff

 

Hi GRG,

My solution requires a small button battery to supply biasing either to keep the Enh-mode mosfet on, or keep a depletion mode off to do as you require. It should last a long time as not much current is needed.

A small high gain transistor is also needed to invert the signal needed. You will likely have to tweak R3 and R5 to get the activation point you need of 0.8V on BT2.

You can manipulate the circuit to use what ever polarity of components you thing will work best. The circuit presented here is only an example, but works in LTspice simulation to prove the concept.

Good Luck,
Ifixit

 

Interesting I didn't think about a small battery, I'll play around a bit with the same concept with a depletion mode as when the 3.6 V battery goes dead BT2 won't be bypassed anymore.

I was heading in the direction of a small oscillator that would run on down to 0.8 volts or so then make a doubler or tripler with the - v. the - v would hold the depletion mode fet open till the voltage went down enough for it to stop. I tried making a tripler with some schoctky diodes and 1 volt p-p signal but I only got 1.5 volts. I guess the drop across the diode must be coming into play. I was trying to get around 3 volts so I could use a more common depletion mode fet.

Thanks