> Hi All,
> I have an application to measure the load current of a 24V PSU.
> The current will range from 0 to about 5 Amps. I can put a 0.1R 5W
> resistor in series with the +24V line to measure the voltage drop.
> Does anyone have a simple circuit that can operate off a 5V supply
> to provide 0-5V to feed into the A2D input of a PIC?  Some sort of
> differential amp is required but how to measure the difference across
> the 0.1R resistor which is sitting up at 24V is a problem. I might be
> able to but the resistor in the 0V line which would make it all a lot
> easier to measure but that may muck up the common earth system.
> Any ideas appreciated.
> Regards...
>
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>I can get ya current sensing with a lower series R, and a plain old
>precision opamp (like OP07 or 177).  The only trick is that I need a
>supply voltage higher than the measure line.  This works well if there is
>a linear regulator involved, which has a higher line in front of the reg,
>or if there is a stragegic spot to rectify a higher line voltage in front
>of an inductor, such as forward or buck converters.  If you can get the
>higher line, we can avoid the complication of special application IC's.
>Lemme know if you have this voltage, and I will fax a sketch.
>
>Chris~
>
>David Duffy wrote:
>
> > Hi All,
> > I have an application to measure the load current of a 24V PSU.
> > The current will range from 0 to about 5 Amps. I can put a 0.1R 5W
> > resistor in series with the +24V line to measure the voltage drop.
> > Does anyone have a simple circuit that can operate off a 5V supply
> > to provide 0-5V to feed into the A2D input of a PIC?  Some sort of
> > differential amp is required but how to measure the difference across
> > the 0.1R resistor which is sitting up at 24V is a problem. I might be
> > able to but the resistor in the 0V line which would make it all a lot
> > easier to measure but that may muck up the common earth system.
> > Any ideas appreciated.
> > Regards...

> >No, the +24V supply is the highest point in the system, hence the problem.
> >I re-configured it short term for low side measurement with a simple op-amp.
> >The Maxim chip (MAX4172) looks perfect for the application but I need to
> >design the PCB now! Maybe I will upgrade to a better solution in the next
> >revision of the board. Thanks for your offer to help though. :-)
>
>Huh? All you need is seven resistors(six if you have a great collection)
>and an opamp to do high-side current measurement. If your opamp is a
>rail-to-rail output type, you can run it on the same supply as your PIC(5
>volts):
>
>Note: I give myself an "0x41" in ascii art!
>
>         0.1R
>24in-*--\/\/\/\--*---24out
>      |           |
>      \           \
>      / 1M    1M  /
>      \           \
>      |           |     10M
>      |           *----/\/\/\/----|
>      |           |    |\         |
>      |           *----|-\        |
>      |           |    |  >-------*----  5v=5A
>      *-------------*--|+/
>      |           | |  |/
>      \           \ \
>      / 100K  100K/ / 10M
>      \           \ \
>      |           | |
>   ---*-----------*-*-- ground
>
>The voltage dividers(1Meg,100K) lose 11 to 1, but the opamp gain is set at
>110, net gain is 10. The opamp's inputs are sitting at about 2.4 volts, so
>they are quite happy. Use one of them new-fangled Microchip-brand opamps
>and you're sitting pretty. Expect a little offset error using a circuit
>with this much gain.

>David Duffy wrote:
> >
> > Hi Chris,
> > No, the +24V supply is the highest point in the system, hence the problem.
> > I re-configured it short term for low side measurement with a simple
> op-amp.
> > The Maxim chip (MAX4172) looks perfect for the application but I need to
> > design the PCB now! Maybe I will upgrade to a better solution in the next
> > revision of the board. Thanks for your offer to help though. :-)
> > David...
>
>
>Here's a simple suggestion, run two simple resistor dividers from
>each side of the resistor to GND. These can feed 2 ADC inputs on
>the PIC. So instead of measureing 0.5v across the resistor you
>are measuring 24v and 23.5v both with reference to ground,
>this is very low parts count and will work fine. Just subtract
>them in software.
>
>Resolution will be a bit lower, and it uses two ADC inputs, not
>one. But it might do what you need. :o)

>
> > But I don't think you necessarily need a negative supply. The gate does sit
> > below the source (depletion mode). But the source is still positive wrt
> > ground. You just need to make sure R3 is large enough such that the lowest
> > source voltage still allows the op-amp to bring the gate voltage low enough
> > to turn the JFET on.
>
> Well, at zero current the source is at zero volts, and the gate needs to
> be more negative than that. Also depletion mode means it needs to be
> turned off, so at zero signal you need the most negative drive.
>
> Note that this circuit will work fine if you use a regular bipolar NPN
> transistor instead, no negative supply needed, however the base current is
> not negligible so there will be small error, on the order of 1 percent.
>
> Cheers,
>
> Bob
>
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> I've used a bipolar transistor in a feedback loop to effectively provide
> logarithmic gain but I've never used a FET like in this circuit.  Why did
> you use it here?  What are the advantages of using it and when do you or
> don't you?
> Thanks.
>
> Mark P
>
>
>         0.1R
> 24in-*--\/\/\/\--*---24out
>      |           |
>      |           |
>      |           |
>      |           |
>      |           |
>      \           |
>      /R1         |    |\
>      \           +----|-\
>      |                |  >-------*----  Vout = 0.1I(load)R3/R2
>      *-------------*--|+/        |
>      |                |/         |
>      +-|                         |
>        | 2N3684 (JFET)           |
>      +-|<------------------------+
>      |
>      |
>      \
>      /R3
>      \
>      |
>   ---*---------------- ground
>
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>
>
>

> Bob, it seems to me that you could get the best of both worlds by using
> a MOSFET with a low turn-on threshold voltage. Am I right?
>
> Bob Blick wrote:
> > > But I don't think you necessarily need a negative supply. The gate does sit
> > > below the source (depletion mode). But the source is still positive wrt
> > > ground. You just need to make sure R3 is large enough such that the lowest
> > > source voltage still allows the op-amp to bring the gate voltage low enough
> > > to turn the JFET on.
> >
> > Well, at zero current the source is at zero volts, and the gate needs to
> > be more negative than that. Also depletion mode means it needs to be
> > turned off, so at zero signal you need the most negative drive.
> >
> > Note that this circuit will work fine if you use a regular bipolar NPN
> > transistor instead, no negative supply needed, however the base current is
> > not negligible so there will be small error, on the order of 1 percent.
> > Cheers,
> > Bob

> The opamp alone could balance out the voltage across the .1 ohm resistor,
> but by using the fet, that current comes comes from the lower resistor,
> nicely imposing a proportional voltage across it.
> >
> >         0.1R
> > 24in-*--\/\/\/\--*---24out
> >      |           |
> >      |           |
> >      |           |
> >      |           |
> >      |           |
> >      \           |
> >      /R1         |    |\
> >      \           +----|-\
> >      |                |  >-------+
> >      *----------------|+/        |
> >      |                |/         |
> >      +-|                         |
> >        | 2N3684 (JFET)           |
> >      +-|<------------------------+
> >      |
> >      *------ Vout
> >      \
> >      /R3
> >      \
> >      |
> >   ---*---------------- ground