Need Op Amp for design

I am puzzled why you are puzzled, or why any part of this is a mystery.  I'll try another tack:

I have a battery alarm monitor which attaches to a 6S LiPo battery to insure the voltage levels do not fall to dangerously low values while in operation, damaging the battery.  The battery monitor attaches to each cell of the 6S battery, and will automatically turn on and draw current from the battery if any one or all of the leads have a voltage on it greater than 1V or so.  This situation is perfectly fine during operation, but when the device - a flashlight - is shut off, at least 6 of the 7 leads coming from the battery must be disconnected from the battery monitor.

Currently I am using 4 DPST relays between the battery and the monitor to shut off all 7 leads whenever power is shut off.  The battery has an output voltage ranging from about 21V to 25V, so I employ a 12V supply to provide power to the relays and the two cooling fans.

I am looking for a solution to replace the relays with something smaller and if possible less expensive.  Whatever device I use, relays, SSRs, Op Amps, etc, will be shut down by removing voltage from the device or the switch trigger.  In the case of the Op amps, it is Vcc that is removed.

I hope I have made it clear.  It is really not particularly complicated.

Oh, by the way, you are referring to the outputs of the Op Amps.  They are not relevant to the discussion, because we know they will perform as required, which is to say they will fall below 1V WRT ground whenever their power is removed.  The only issue here is the non-inverting inputs, at least 2 of which will be subjected to potentials higher than 15V when Vcc is removed.

tabbypurr wrote in message 
news:5bddf5d7-7296-4ab9-b920-dded0567911f@googlegroups.com...

> You're not giving us the relevant info here. What signal a powered opamp 
> sends depends entirely on your circuit. Why your schematic shows power 
> switched to the opamps is another mystery. I don't see how can we help if 
> you don't explain why you've made the slightly puzzling choices you have.

I tried a simulation using an OP-07, and with power disconnected and a 1k 
resistor to the input, it draws 8 mA from a 12V battery. With 24V power 
connected, the op-amp draws 2.25 mA, while the + input draws femtoamps.

http://enginuitysystems.com/pix/electronics/BMS_OpAmp.png

I think some 2V gate P-MOSFETS would do the job inexpensively. You could 
pull all the gates low through an NMOS that has its gate tied to the op-amp 
power pin. When you switch power off, the PMOS devices will turn off.

Paul 

That doesn't work, at all.  It's entirely backwards.  During operation, the battery can be delivering in excess of 4.5 amperes, so a few milliamps is not really significant.  The flashlight has a variable output, so the output current may be as low as 100 milliamps, so I don't want the current draw to be too high, of course, but 10 or even 20 mA should be fine.

What I can't have is any significant current draw when the device is off. An 8 mA draw to the input for 6 inputs is way, way too much.  The battery would drain completely in less than 3 days and would be destroyed in less than a week sitting on the shelf.

You guys are making this far more complicated than it is or needs to be.  All that is needed is a switch ( x 6 ) of some sort between the battery and the monitor that doesn't consume very much current when on and no more than 100 microamps or so when off.  Reed relays do the job pretty well, but are too large.  SSRs would do the job extremely well, but are too large and too expensive.  Low power Op Amps should work very well as long as they are not destroyed by having more than 25V on their non-inverting input with Vcc disconnected.

On Monday, November 20, 2017 at 2:50:48 AM UTC-5, Leslie Rhorer wrote:
> This will not be in any significant quantity.  Probably 20 at most.
> 
> A diode to the power rail won't work.  That will cause the Op Amp to power up and send signal to the battery monitor, powering it up.  The battery monitor is powered from its input leads.
> 
> The 324 can survive an input 32V (actually, around 25V in this case) higher than its V+?  I thought not.  Looking at the design, and the specs, it sure enough looks like it can.  That may be my solution.
> 
> A schematic?  Really?  OK.  I don't see how to attach a file to the thread, but I will put one up on my web site.
> 
> http://fletchergeek.com/images/Battery%20Monitor.png

I'm not sure and it may depend on the opamp, but the opamp input 
may draw current,even when the power is not applied.  (It may draw more 
current.)  Since it's all one polarity can you use some fet's to turn the 
inputs on and off?  (You show some other power supply or is the 12V the battery?)  

George H. 
> 
> Yes, I think the battery leakage is on the order of a microamp or so.  That's why I specified a few nanoamps as the off current.  Let's see, a typical worst case shelf discharge for a LiPo battery is about 5% per month.  Given this is a 20AH battery, that amounts to over a milliamp.  Good heavens!  So you are correct, I don't really need much better than a few microamps.  The battery monitor pulls several milliamps, though, so a switch is definitely needed.
> 
> The FTR-B3GA is not all that cheap, nor all that small.  They run about $3.75 each, and I would need 4 per light.  Compare that to a quad LM324 for under $.10, and I only need 2 per light.  The FTR-B3GA is 10.6 mm x 7.2 mm, for a total of .305 square cm.  That's not too bad, but the LM324 is only 8.65 mm x 6.5 mm for a total of .112 square cm.  It's true the relay solution offers virtually zero off-current, but the tiny leakage into the op amp inputs is not significant, as one can see above.  I think the op-amps win hands down in this case.

On 2017-11-20 04:28, Leslie Rhorer wrote:
> That doesn't work, at all.  It's entirely backwards.  During
> operation, the battery can be delivering in excess of 4.5 amperes, so
> a few milliamps is not really significant.  The flashlight has a
> variable output, so the output current may be as low as 100
> milliamps, so I don't want the current draw to be too high, of
> course, but 10 or even 20 mA should be fine.
>
> What I can't have is any significant current draw when the device is
> off. An 8 mA draw to the input for 6 inputs is way, way too much.
> The battery would drain completely in less than 3 days and would be
> destroyed in less than a week sitting on the shelf.
>
> You guys are making this far more complicated than it is or needs to
> be.  All that is needed is a switch ( x 6 ) of some sort between the
> battery and the monitor that doesn't consume very much current when
> on and no more than 100 microamps or so when off.  Reed relays do the
> job pretty well, but are too large.  SSRs would do the job extremely
> well, but are too large and too expensive.  Low power Op Amps should
> work very well as long as they are not destroyed by having more than
> 25V on their non-inverting input with Vcc disconnected.
>

Paul isn't making it complicated, he is on the right track. You can use 
FETs to replace the Reed switches you have now. If your stack remains 
below 25V you could pick some that have a Vgs breakdown rating of 30V or 
higher. If using N-channels (more available and usually cheaper) the top 
FET would have to be bootstrap-driven but that's not difficult.

This way you can use any old opamp and it just gets turned off along 
with all the FETs when the system is shut down. The gate drives drop 
away -> all FETs go high resistance. If you want to use cheap FETs with 
only 20V max for Vgs the top ones need the usual resistor plus zener 
limiter. Some even have a built-in zener.

What I don't understand is where you mention that cost matters while you 
only plan on producung 20 units.

-- 
Regards, Joerg

http://www.analogconsultants.com/

On Monday, 20 November 2017 11:48:18 UTC, Leslie Rhorer  wrote:

> I am puzzled why you are puzzled, or why any part of this is a mystery.  I'll try another tack:
> 
> I have a battery alarm monitor which attaches to a 6S LiPo battery to insure the voltage levels do not fall to dangerously low values while in operation, damaging the battery.  The battery monitor attaches to each cell of the 6S battery, and will automatically turn on and draw current from the battery if any one or all of the leads have a voltage on it greater than 1V or so.  This situation is perfectly fine during operation, but when the device - a flashlight - is shut off, at least 6 of the 7 leads coming from the battery must be disconnected from the battery monitor.
> 
> Currently I am using 4 DPST relays between the battery and the monitor to shut off all 7 leads whenever power is shut off.  The battery has an output voltage ranging from about 21V to 25V, so I employ a 12V supply to provide power to the relays and the two cooling fans.
> 
> I am looking for a solution to replace the relays with something smaller and if possible less expensive.  Whatever device I use, relays, SSRs, Op Amps, etc, will be shut down by removing voltage from the device or the switch trigger.  In the case of the Op amps, it is Vcc that is removed.
> 
> I hope I have made it clear.  It is really not particularly complicated.

I can see where the problem is, it's your approach to the situation. Good luck.


NT

> I'm not sure and it may depend on the opamp,

Yes, it does.

> but the opamp input 
> may draw current,even when the power is not applied. (It may draw more 
> current.)  Since it's all one polarity can you use some fet's to turn the 
> inputs on and off?

Indeed some Op Amps will.  The LM324 will not because the inputs are the bases of PNP transistors, unless the input voltage exceeds the reverse breakdown voltage of the base-emitter junction.  If a negative voltage is applied, then the inputs can potentially pull a significant current, or if one exceeds the breakdown voltage.  Similarly, an FET displays a very large impedance between the gate and both the source and drain, so again unless one exceeds the G-S or D-S breakdown voltage, the current is going to be even smaller.  Yes, FETs should be able to be used to shut off the path between the battery and the monitor, but it doesn't save me anything to do so.  It is neither less expensive nor more conservative of space to use any of the FETs I found.  If someone has a specific recommendation for a quad or larger FET switch chip that doesn't cost much, I'm surely willing to take a look.


  (You show some other power supply or is the 12V the battery?)  
> 
> George H. 
> > 
> > Yes, I think the battery leakage is on the order of a microamp or so.  That's why I specified a few nanoamps as the off current.  Let's see, a typical worst case shelf discharge for a LiPo battery is about 5% per month.  Given this is a 20AH battery, that amounts to over a milliamp.  Good heavens!  So you are correct, I don't really need much better than a few microamps.  The battery monitor pulls several milliamps, though, so a switch is definitely needed.
> > 
> > The FTR-B3GA is not all that cheap, nor all that small.  They run about $3.75 each, and I would need 4 per light.  Compare that to a quad LM324 for under $.10, and I only need 2 per light.  The FTR-B3GA is 10.6 mm x 7.2 mm, for a total of .305 square cm.  That's not too bad, but the LM324 is only 8.65 mm x 6.5 mm for a total of .112 square cm.  It's true the relay solution offers virtually zero off-current, but the tiny leakage into the op amp inputs is not significant, as one can see above.  I think the op-amps win hands down in this case.

> Paul isn't making it complicated, he is on the right track. You can use 
> FETs to replace the Reed switches you have now. If your stack remains 
> below 25V you could pick some that have a Vgs breakdown rating of 30V or 
> higher. If using N-channels (more available and usually cheaper) the top 
> FET would have to be bootstrap-driven but that's not difficult.
> 
> This way you can use any old opamp and it just gets turned off along 
> with all the FETs when the system is shut down.


 The gate drives drop 
> away -> all FETs go high resistance. If you want to use cheap FETs with 
> only 20V max for Vgs the top ones need the usual resistor plus zener 
> limiter. Some even have a built-in zener.
> 
> What I don't understand is where you mention that cost matters while you 
> only plan on producung 20 units.
> 
> -- 
> Regards, Joerg
> 
> http://www.analogconsultants.com/

> Paul isn't making it complicated, he is on the right track. You can use 
> FETs to replace the Reed switches you have now. If your stack remains 
> below 25V you could pick some that have a Vgs breakdown rating of 30V or 
> higher. If using N-channels (more available and usually cheaper) the top 
> FET would have to be bootstrap-driven but that's not difficult.
> 
> This way you can use any old opamp and it just gets turned off along 
> with all the FETs when the system is shut down.

Yes, but what is that saving me?  Not cost.  The FETs run about $.50 each, and I will need at least 6, so that is $3.10 per light, vs. less than $.20.  They certainly would not save me space, as 6 discrete FETs plus one Op Amp will take more space than 2 quad Op Amps.


> What I don't understand is where you mention that cost matters while you 
> only plan on producung 20 units.

I don't know about you, but I am not made of money, and since I will not be selling these units, cost matters very much.  Yes, I surely could afford $3, but why spend the money if it does not provide a superior solution?  In what way would discrete FETs provide a superior solution?  (In terms of operational superiority - I realize their current draw is a tiny fraction of the Op Amp solution.)