Baffling simple circuit

Is the ground, measured at each ground symbol, really 0V? It "smells" like something is floating.

Ed

It looks to me look like that diode in the positive feedback path stops it from monitoring the battery until the relay turns on.

Sorry, (I'm confused) the circuit works with the DMM monitor in place? If not a ground issue (Re Ed.) then maybe it's oscillating?? Try some capacitance??

George H.

Thanks - that's just the kind of tip that I was looking for.

I should have said: the relay powers a timer and does not affect the monitor circuit.

Also, the diode is there to pull down the input when the output goes negative due to falling battery voltage, and keep the op amp from going high when the battery recovers. That was my intent, anyhow.

Bob

Yes, you should be confused: connecting the DMM "causes" the circuit to work.

Right after I check the grounds I'm going to get out the scope and look at things.

I am guesing you may have created a RF detector at the input using that diode..

also take note that you did say before you are using a +/- 12 supply or close to that? That would mean the output is going to be -10/+10 or there abouts. Looking at that diode in the loop back makes me think the output when -10 is going to bleed back to the input divider and thus current will be present and since small diodes like the generic 914 etc love to act as a detector, you could have noise etc..

Try placing a cap around the diode. Jamie

OK ... I didn't do full disclosure - there was circuitry that I didn't show 'cause I thought that it was irrelevant but wasn't. Not the 1st time - I should know by now. The whole circuit:

There was oscillation & ringing on the MOSFET gate. 130kHz base + 10x that ringing. -2v to +5v.

I grabbed a TLAR* capacitor and put it on the gate, to ground. Stabilized it perfectly & the whole thing works. A 10v 6.8uF tantalum electro.

I don't know anything about capacitor choice (this is S.E.Basics) and I've used electros in power supplies to smooth full wave rectified. But I think that I've read that they shouldn't be used on AC (+&- excursions) - they are polarized.

What would be a proper cap to use here?

Thanks, Bob

• - TLAR - That Looks About Right

i woupd put a .1uf at least around the diode.

Right (That circuit's an old friend of mine), put ~10 k ohm in the feedback path from the top of the sense resistor to the inverting input. Then roll off the opamp gain with an integrating cap ~100pF is a good start/gyess. Cap from output to inverting input.

George H.

Thank you! It's the perfect reply: it's the answer to the question that I should have asked.

Thanks - I'll do that.

Thanks again - it's done and working perfectly.

For my edification, if you don't mind: what do the resistor and cap do?

Bob

The gain of the circuit is set by the ratio of the feedback impedance to the impedance to ground. The resistor is added to reduce the gain across the board to something reasonable rather than the huge full gain of the opamp and the capacitor reduces it further for higher frequencies since it only needs to respond to lower frequencies.

Thanks