Hello all,

I remember, there was some problem when using opamps with their inputs near
the supplyvoltage.

I now have a cirquit, I hope I can post that too, where the inputs are at +supply - 100mV.

Could I use any OP, or is there something to consider ?
The OPs should not be too expensive, bcs I need a lot.

Kind regards,

Edwin

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Hi,
Why dont you adjust the values of the components in order to bring the inputs within the operating range of standard OPA's.?

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Eric " Good enough is Perfect "
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Hi Eric, thanks for the very quick reply.

It is used to charge a second battery in a car, and when I go lower, it will not get charged.
Now I should get the 2nd battery up to some 200...300 mV below the main battery.

Edwin

 

hi,
Most OPA's as may already know dont make 'good' comparators.

A comparator like the LM393 etc, will not work reliably at Vin > Vsup -1V.

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Well, I might have to think of something else then :-(

Thanks anyway Eric,

regards,

Edwin

 

There are opamps that work "rail-to-rail" rather than within two diode drops of V+ and V-. If you can find an online "selection guide" they will probably show these types.

 

hi willbe,
IIRC most of the rail/rail amps are about 5V supplies.

I would use a power schottky to isolate the two batteries, a 400mV drop should be acceptable.

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Eric " Good enough is Perfect "
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If you are looking for RR in and out OPA's the TLV2371 is not that pricey.

National has a good selection guide here.

Possibly for your application you could use a TLV431 low voltage,or 431 2.5V ref as a comparator. Mind you I'm not 100% sure what it is your trying to do.But if your looking for cheap you aren't going to do much better then an LM431SCCMFX 20 cents small quantity.

 

Thanks for the replys.

The cirquit should prevent `short-cirquits` when the 2nd battery is completely flat.
Any relay would burn away, with, say, 200 Amps, when you connect a flat battery
to a charged one.
This cirquit is meant to keep the current to a max of 5 or 6 Amp.

Regards,

Edwin

 

Hi Edwin,

There is a much simpler solution if you can avoid ground referencing the device
you intend to power. If you rearrange the circuit you will be connecting the
resistor in series with the ground leg of the device to be powered rather than
in series with its plus power supply pin. Doing this eliminates the rail to rail
input op amp requirement, and a standard LM358 will work in many cases.

So the million dollar question is...
Can your device to be powered stand to have a small resistor in its ground
lead rather than its supply lead? Measurements can still be made with
a multimeter across the device, it's just that one leg will no longer be
connected to ground.

 

You have the drain and source of the P-channel Mosfet connected upside-down.
The source is supposed to connect to the resistor and the drain is the output.

Many FET-input opamps like the TL071 work fine when their input is at the positive supply voltage because their FETs are P-channel.

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Uncle $crooge

 

I know, re D/S, in real the FET will be connected the right way.
It is only the wrong symbol I used in the diagram. Sorry, should have told in the first post.

No, it is not possible to lift the 2nd battery from ground, I already thought about it.
I think I etch 1 PCB, and try the circuit with different opamps, starting with the mentioned TL 071.

There is a much easier circuit with only 2 transistors, but that has a drop of at least 1,5 Volt.
That is why I thought of this one, with the opamp.

Thanks !

Regards,

Edwin

 

hi Edwin,
Dont forget the main car battery could fall to 9V when you pull the starter.!

Anything you have connected in series with the two batteries will be reversed biassed for a while..
Regards,

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Eric " Good enough is Perfect "
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Yes, very good to mention, but the chargingcircuit is switched on at least 1 minute after
the motor is running.
The 2nd battery might be even 200 Ah, so, the starterbattery has to get recharged from starting,
before one can start charging the 2nd battery.

With a series of these circuits in parallel, the chargingcurrent is limited to 25 or 30 Amps.
To keep a bit power left for the rest of the car.

Rgds,

Edwin

 

The + and - pins are the wrong way round, there's no compensation so it will oscillate and there needs to be a capacitor on the output to improve the transient response.

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