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Which opamp would be best for this ?

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You can stabilize the loop like this. You might have to play with the values of C1, R2, and R3, although they worked well in simulation with IRF7404, IRF7204, and several others that I tried.
 

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Thank you all very much for all the effort you are making.

Roff: Looks nice, but what about the mostly needed 10 V for the FET ?
The opamp is hanging on 4V7 below the supply, so having a maximum output of 4,7V.

Mr Al: Your Idea went also through my mind before, to lower each signal in the same way,
to get in a more comfortable area. I didn`t do it, bcs the voltages are already very `small`.
But it might be the better way, indeed.

I think I can have a go now, will implement the advices and let you know, abt next week I think,
how it works.

Kind regards,

Edwin
 
Thank you all very much for all the effort you are making.

Roff: Looks nice, but what about the mostly needed 10 V for the FET ?
The opamp is hanging on 4V7 below the supply, so having a maximum output of 4,7V.
Kind regards,

Edwin
Most power MOSFETs will conduct 5 amps with much less than 4.7V gate-to-source voltage. I ran the simulation with 6 different MOSFETs, and only one of them didn't work. It was an FQB11P06, which has relatively high Rds(on) of 175mΩ. I admit that, if you wanted to build a lot of these, you would probably want a higher voltage op amp, but as a one-off, a 5V op amp should work fine. I searched Linear Technology's site and found LT1677, which is an RRIO op amp which will work over the supply range of 3V to 36V, so you can eliminate the zener.
 

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  • 5A current source.PNG
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Thank you all very much for all the effort you are making.

Roff: Looks nice, but what about the mostly needed 10 V for the FET ?
The opamp is hanging on 4V7 below the supply, so having a maximum output of 4,7V.

Mr Al: Your Idea went also through my mind before, to lower each signal in the same way,
to get in a more comfortable area. I didn`t do it, bcs the voltages are already very `small`.
But it might be the better way, indeed.

I think I can have a go now, will implement the advices and let you know, abt next week I think,
how it works.

Kind regards,

Edwin

Hi Edwin,

Well, dont worry that the signal is small to begin with, the op amp
doesnt care :)
If you are that worried, then you can change both lower 10k resistors to
20k or something like that. This will still keep the common mode
range within that which the op amp can handle.

Some frequency compensation might also be good, but with the
cap connected directly between the output of the op amp and
the inverting input of the op amp. Test with your required
waveform.
 
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Hi Al, Thanks.

What I meant was, that in the beginning there is a 100 mV-range, and with those
10k resistors it is halved into 50 mV.
OK, a microphone sometimes gives only 50 uV !! HAHA

What about replacing the upper 10k-resistors by a zenerdiode of 4,7 V ?
That is just shifting the signal down, not halving the signal.
Problem will be temperaturedrift I think.

I will buiild the circuit up, with a 1 Ohm senseresistor, making the current to 100 mA.
A bit easier to test on a breadboard than 5 Amps.

Although, if it is 5 or 6 Amps, doesn`t even matter, as long as it is not going to 1 or 2 Amp :)

Oh, what I wanted to explain: The output shows "uit", many of you all would have asked what
that would be....
I forgot to change it, as I uploaded the diagram. It means OUT in dutch.... that`s all.

Nice weekend !

Edwin
 
Hi again Edwin,

I really think you are worrying too much about this. What if you wanted
to sense a 50mv signal? I understand your point however, in that the basic
op amp will not see the entire signal change, but that's pretty common
anyway with voltage regulators, as very often the output is divided wayyyyy
down to meet the reference voltage.
Oh yeah, i made a slight error in wording with my last post too, i meant
to say "change the lower two 10k resistors to 20k". That will help improve
the problem which you are worried about. We just have to make sure that
the highest voltage that reaches the op amp inputs is Vcc-2 volts or so.
If that requirement is met i think this will fly.
BTW the simulation looks just like any other regulator.

Yes, zeners might work but the differential temperature drift
might be a problem depending on what accuracy you are
looking for with this circuit.
 
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Hi Al,

yes you`re right, re 50 mV. That is what I meant with the 50 uV of a microphonesignal.

I already modified the pcb-layout for this.

I kept the current this low, that there will be no overheating of the FETs, even when the 2nd
battery is flat, let`s say 1 or 2 Volt. bcs then you have a difference of some 12 Volts, multiplied
by 5 Amp is `only` 60 Watts. Doing 30 a 35 Amps in one go, it is 420 Watts.....
ok even splitted over 5 or 6 supplies, you have the same dissipation, but no FET is burning out.
I might even include a temperaturesensor, that could switch on a fan.

Kind regards,

Edwin
 
There is a potential problem with Mr Al's circuit. The LM358 output may not go near enough to the positive rail if a large-geometry MOSFET is used, especially if it happens to have a low threshold voltage and/or it is running at relatively low current. I discovered this while running a simulation.

This can be remedied with a zener level shifter.
 

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Hi Ron, thanks for the reply.

I just switched on the other pc where I draw the pcb on, to see if it can be
implemented.

Although I think I remembered that the 358 was able to almost reach the supplyrail.
But even when it reaches +U - 1 Volt, the FET will be `off` far enough I think.

I might be able to make the pcb so, that I can choose with or without zener, by
placing the seriesresistor (gateresistor) in two different ways. So no need for
a bridge instead of the zener when it is not needed :)

10 mins later....
Yepp, implemented, two places for the resistor (1k) at the output of the opamp,
one of them makes it in series with a zener.

Thanks Ron!

Kind regards,

Edwin

OOPS ! I forgot the pull-up resistor! Doing now! ;-)
 
Hi again,


Roff, yes that is correct, i should have specified that it had to be a
'regular' MOSFET, not a logic level type. Good catch there.

Anyway, the output of the LM358 goes within 1.5v of the positive supply
rail, so with a logic level MOSFET that turns off with 1v that means even
1 diode drop should do it. The diode would have to be connected with
the cathode to the output of the LM358, and of course the bias resistor
would still be required. If you are REALLY squeamish, use two diodes
in series. Of course with the typical 4v threshold MOSFET no diodes
should be required.
It might also be possible to use two resistors as a voltage divider, say
1k on top and 2k on bottom. This would work for a logic level MOSFET.
The upper voltage swing would be about 0.5v down from Vcc, and the
lower voltage swing would be 5v down from Vcc, which should be enough
to turn this type of MOSFET fully on.

Another thing we havent talked about yet is the turn on transient
and if any slow start mechanism would be needed. I have no idea
what the end application is going to be.
 
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When it's turned on won't the MOSFET be off anyway?

It will only turn on when the gate capacitance becomes charged up enough.
 
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.

Hi, Mr. G!
I've had a few, so please ignore this post. . .
Regards,
Willbe
 
I was responding to MrAl's remark regarding a slow start mechanism. I think he was implying that it's possible that a high current could flow when the circuit is powered on. I don't think this will happen because the MOSFET will be in the off state when the circuit is first powered up.
 
I was responding to MrAl's remark regarding a slow start mechanism. I think he was implying that it's possible that a high current could flow when the circuit is powered on. I don't think this will happen because the MOSFET will be in the off state when the circuit is first powered up.


Hi Hero,


Yes, but the MOSFET gate will probably charge faster than the op amp can work into the linear mode when the feedback starts to make a difference. I think the output of the LM358 will start up at ground, and gradually ramp up during this time. As it's ramping, the MOSFET is turned on pretty hard. The LM358 is pretty slow during times when it's not in the linear mode.
Something to force the output high might help.
 
Hi Hero,


Yes, but the MOSFET gate will probably charge faster than the op amp can work into the linear mode when the feedback starts to make a difference. I think the output of the LM358 will start up at ground, and gradually ramp up during this time. As it's ramping, the MOSFET is turned on pretty hard. The LM358 is pretty slow during times when it's not in the linear mode.
Something to force the output high might help.
I ran some sims, and the op amp output did slew from zero to max, causing a high current spike in simulation. I thought that a cap from the + in to Vcc would help, but the problem still existed. The circuit below solves the problem in simulation, so long as the power supply takes 200us or more to slew from zero to max.
I emphasize that these are simulated results, and may not represent the real world.
 

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  • Mr Al's current source2 waves.PNG
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WOW !! You`re making a serious job out of it ! Beautifull !

Nice to have a lot of extra ideas about the circuit, so if something goes wrong,
well, wrong.... not doing as expected, I know where to look for the cause.

To clear some questions, although already explained: It is used for in a camper,
to charge the householdbattery from out of the starterbattery, while driving.

To keep the dissipation per FET low, I limited the current to some 5 or 6 Amps.
To get round 30 Amp as maximum current, I will use 5 or 6 of these circuits in parallel.
Maybe selectable, that when not a high current is needed, and in winter one need the
energy from the alternator more for driving, you can switch to only 1 currentsupply,
to get only 5 Amp chargingcurrent, just for buffering.

Besides that, these currentsupplys are (going to be) automatically switched off as soon
as the motor stops, and only allowed to switch on abt a minute after the motor is
running.
This also prevents returncurrent through the currentsupply.

To calculate the dissipation one should assume the output shorted !
Bcs it is POSSIBLE, that the 2nd battery is completely FLAT.
Won`t happen too often, but bcs it is possible, it should be taken in the calculations.
And, to make the heatsink not too big, there will be a fan, with thermoswitch mounted.

Oh, AND a heavy relay to switch both batteries parallel, when the differencevoltage
is < 0,5 Volt.

Kind regards,

Edwin

PS: I just got the idea, that, when the 2nd battery is fully charged and thus via the relay
is in parallel with the starterbattery, it will also get (a bit) discharged when the motor is
idling with all the lights on.... :-(

Yepp, a second alternator is always the best option !
aehhh.... when the relay is not yet activated.... there still will be returncurrent via the
currentsupply(s) :eek: :mad:

Back to the drawingboard ... :(
 
HAHA ! The idea of the total circuit, (including the not drawn automatical switching)
is better as I thought.

What happens, when the differencevoltage is < 0,5 Volt ?
Correct, the relay comes in.

So, if during idlingspeed of the motor the starterbattery becomes lower, and finally
lower than the HH-battery ?

The differencevoltage gets < 0,5 Volt => the relay comes in, and bridges both batteries !!
So, STILL NO returncurrent can flow through the FET..... ;)

Kind regards,

Edwin
 
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