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'[EE]: Current Measurement'
2000\12\11@214537 by David Duffy

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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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2000\12\11@220503 by rwyoung

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>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.

Maxim has several current sensing parts as to Linear Technologies.  Check
out the MAX4172 and its cousins.

Robert Young
YR Consulting  785-842-6211 v/f
.....rwyoungKILLspamspam@spam@ieee.org

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2000\12\11@231817 by Chris Eddy

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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:

{Quote hidden}

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2000\12\11@233039 by David Duffy

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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...

At 09:22 PM 11/12/00 -0500, you wrote:
{Quote hidden}

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2000\12\12@003528 by Bob Blick

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>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.

Cheerful regards,

Bob Blick

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2000\12\12@005342 by David Duffy

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At 09:07 PM 11/12/00 -0800, you wrote:
{Quote hidden}

Hi Bob,
This is exactly what I drew on paper when I started out this morning.
Maybe I made a mistake transferring to my breadboard layout.
I may have some LMV358 dual op-amps around here. (somewhere !)
The high resistor values might be affected (humidity ?) in this application
but I could always scale them down - current draw is not a major concern.
Regards...

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2000\12\12@031529 by Roman Black

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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)
-Roman

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2000\12\12@083906 by David Kott

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> 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...

Have you considered a GMR?  Measure the current through a trace or wire by
quantitizing the magnetic field around it.  Take a look at
http://www.nve.com

-d

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2000\12\12@085131 by Olin Lathrop

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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.

A higher line would be the easiest, but you could also use divide the
voltages on either side of the current sensing resistor 5% or 10% towards
ground and run the diff amp from the 24V.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, TakeThisOuTolinEraseMEspamspam_OUTembedinc.com, http://www.embedinc.com

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2000\12\12@105221 by Eisermann, Phil [Ridg/CO]

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I posted such a circuit a while back. It was directly off of NatSemi's app
note 32. It uses two resistors, plus the sensing resistor, a JFET, and an
op-amp. Here's the circuit.


       0.1R
24in-*--\/\/\/\--*---24out
    |           |
    |           |
    |           |
    |           |
    |           |
    \           |
    /R1         |    |\
    \           +----|-\
    |                |  >-------*----  Vout = 0.1I(load)R3/R2
    *-------------*--|+/        |
    |                |/         |
    +-|                         |
      | 2N3684 (JFET)           |
    +-|<------------------------+
    |
    |
    \
    /R3
    \
    |
 ---*---------------- ground


{Original Message removed}

2000\12\12@121532 by Eisermann, Phil [Ridg/CO]

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oops, i mislabeled the resistor connected to V+. That should be R2 (it's
labeled as R1). R1 would be the sense resistor.

{Original Message removed}

2000\12\12@123403 by Bob Blick

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Also the output is taken from the top of R3, not the opamp output.

That circuit requires an opamp with inputs valid at V+ and to have V+
connected to the +24 supply. You also need a V- supply, because the JFET
gate is below ground most of the time. An LM741 would work, but LM358 or
LM324 would absolutely not work. A V- supply of -5 volts would probably be
adequate.

Cheers,

Bob

On Tue, 12 Dec 2000, Eisermann, Phil [Ridg/CO] wrote:

> oops, i mislabeled the resistor connected to V+. That should be R2 (it's
> labeled as R1). R1 would be the sense resistor.
>
> {Original Message removed}

2000\12\12@132952 by Eisermann, Phil [Ridg/CO]

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ok, i'm not doing so well this morning! The output is indeed taken from the
top of R3. Thanks for pointing that out.

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.

Or am I messing something else up here? Can't seem to get the brain in gear
this morning.


{Original Message removed}

2000\12\12@142314 by Bob Blick

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> 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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2000\12\12@173754 by David Duffy

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Roman wrote:
{Quote hidden}

Already drew that one out too !  Resolution went too coarse for the
application.
I've decided to go with the dividers and an op-amp with gain to compensate.
Thanks to all than replied.
Regards...

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2000\12\13@082526 by Mark Peterson

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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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2000\12\13@113517 by Eisermann, Phil [Ridg/CO]

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well, aside from the things pointed out by Bob Blick...

If you were to use a bipolar transistor in this circuit, you would have some
error due to base current. A FET, on the other hand, effectively uses no
gate current. That reduces the errors introduced by a bipolar considerably.

The main limitation in this application, as Bob pointed out, is that in
order to turn the JFET off, you need to bring the gate voltage below the
source voltage. For example, if the current to be measured is zero, the gate
has to go negative. That means the op-amp now requires a negative supply.


{Original Message removed}

2000\12\13@115000 by rottosen

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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?

-- Rich


Bob Blick wrote:
{Quote hidden}

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2000\12\13@130317 by Bob Blick

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Hi Mark,
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.

Cheers,

Bob

On Wed, 13 Dec 2000, Mark Peterson wrote:

{Quote hidden}

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2000\12\13@130524 by Bob Blick

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Yes, that'd work fine. Also probably be a lot more stable than using a
bipolar transistor(unless you really did want an oscillator!).

Cheers,

Bob

On Wed, 13 Dec 2000, Richard Ottosen wrote:

{Quote hidden}

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2000\12\13@134851 by Bob Blick

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Note that this schematic shows the correct place to take the output:

{Quote hidden}

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2000\12\14@153450 by Chris Carr

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Instead of connecting the low side switches of the H Bridge directly to
earth, connect them through a low value resistor. Commutator pulses can be
picked off the resistor by AC coupling to an Amplifier.

Regards

Chris
----- Original Message -----
From: "Duane Brantley" <RemoveMEDuane.Brantleyspam_OUTspamKILLspamGENBAND.COM>
To: <RemoveMEPICLISTTakeThisOuTspamspamMITVMA.MIT.EDU>
Sent: Thursday, December 14, 2000 2:04 PM
Subject: Re: [EE]: Current Measurement


> Hello everyone!
>
> With the motor/wheel combo that I'm using, I haven't found a way to
> incorporate an optical encoder.  I had heard that there is some way, using
> and instrumentation amp, to isolate the current pulses of a PM motor on
the
> return side of an H-bridge setup.  Is this true, and if so, how would I go
> about implementing this?
>
> Cheers,
>
> Duane
>
> {Original Message removed}

2000\12\14@164547 by Chris Carr

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All this ASCII Art is doing my brain cell in.

Could I suggest

http://pdfserv.maxim-ic.com/arpdf/AppNotes/A0911.pdf

as a starting point for a solution.

My appologies if this has already been submitted as a solution by anyone
either directly or as ASCII Artwork.

Regards
Chris

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