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PICList Thread
'[PIC]: Measuring Current Consumption During Sleep'
2002\01\21@140234 by John Hansen

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I am currently doing my first project where current consumption is an
issue.  I've configured a keypad on portB of a 16F628 and have my program
set up so that I'm using the internal port B pullups.  I have the PIC sleep
except when a key is pressed.  The program works great.   However, I am a
little concerned by the current consumption.  In measuring the current
drawn by the circuit I'm getting 85 microamps.  I've checked my meter by
measuring the current consumed by a circuit that simply has a 1 megaohm
resistor in it, and I get the expected 5 microamps.

That's pretty good, but Microchip seems to indicate the current consumption
of this device during sleep should be much lower.  So I set up a simpler
circuit that has only a PIC, all pins configured as outputs and all pins
tied to ground.  I still get about 85 microamps.  I tried it with all pins
configured as inputs.  Still 85 microamps.

Then I read application note AN606, where it talks about a circuit that
measures current consumption of the PIC by putting 100 ohm resistors on the
input and output and alternately shorting them and measuring the current
differential.  This would seem to imply that there is something wrong with
my technique of simply using a multimeter to measure the current
consumption of the device.

As I said, I'm pretty new to very low power circuits.  Is my device
actually consuming 85 microamps, or is there something wrong with my
measurement technique?

Thanks,

John Hansen

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2002\01\21@150405 by Don Hyde

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You should be able to get it down to a couple of microamps.

It is very easy to consume a few extra microamps.  To get down to the
minumum, you need to turn off the brownout detector, make sure than you have
not left any peripherals running -- such as analog voltage reference, and
take great care with each and every I/O pin.

An input pin that's left to float will (sometimes, and not always repatable)
float into the linear region, which will cause both of its CMOS transistors
to turn on, wasting a few microamps.  Either connect it to ground or Vcc
with a resistor (1M is enough), or if it's not connected to anything at all,
make it an output.  It doesn't matter whether it's high or low, as long as
it's not somewhere in the middle.

An output pin that is driving any sort of load will consume power.  This
includes pullup/down resistors and devices which you have powered down.  You
need to look carefully at each one and make sure it's not sourcing current
to something.

I often find it helpful to separately measure the current through Vcc and
GND to the PIC.  This will give you a clue as to whether you have a pin
that's pulling up or down.  If you have an output that's pulling down
against a pullup resistor, you'll see more current through Vcc than Gnd.  If
you have a high output going to a powered-down chip, then you'll see more
current through Vcc.



> {Original Message removed}

2002\01\21@155903 by John Hansen

picon face
Thank you!   I had assumed it had something to do with the I/O pins.  It
seems I was barking up the wrong tree.  I turned the brownout detector off
and current consumption dropped to less than 1 microamp.  Thanks so much, I
should now be able to run this on a battery for the shelf life of the battery.

Thanks!

John Hansen


At 02:02 PM 1/21/2002 -0600, you wrote:
{Quote hidden}

> > {Original Message removed}

2002\01\22@003214 by Russell McMahon

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> You should be able to get it down to a couple of microamps.
>
> It is very easy to consume a few extra microamps.


Presumably you are using a crystal or resonator.
When using an RC clock with external R the oscillator tends to add about
Vcc/2R to the current drain.



       RM

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'[EE]: Measuring Current'
2002\02\18@140812 by Donovan Parks
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Hello,

Will placing an ammeter in series along the positive end of my power supply
tell me how much current my circuit requires?  Seems like it should, but my
thumbs are green when it comes to actually doing EE.  Thanks for taking the
time to enlighten me.

+5V ---> Ammeter ----> +5V terminal of circuit

Regards,
Donovan

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2002\02\18@141659 by Jim

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Yes.

Connect an Ammeter *in series* for current

and a Voltmeter *in parallel* for Voltage.

Jim


{Original Message removed}

2002\02\18@141702 by Dale Botkin

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On Mon, 18 Feb 2002, Donovan Parks wrote:

> +5V ---> Ammeter ----> +5V terminal of circuit

That's the way, yes.

Dale

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2002\02\18@150051 by Michael Vinson

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Donovan Parks wrote:
>Will placing an ammeter in series along the positive end of my power supply
>tell me how much current my circuit requires?  Seems like it should, but my
>thumbs are green when it comes to actually doing EE.  Thanks for taking the
>time to enlighten me.
>
>+5V ---> Ammeter ----> +5V terminal of circuit

Yes, unless your application is supplied from several different
voltage sources (e.g. 5V and 12V). In that case, you can connect
the ammeter between circuit ground and the (common) power supply
ground. That will allow you to measure TOTAL current drain.

Michael V

Thank you for reading my little posting.


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2002\02\18@150433 by Dal Wheeler

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Just make sure you have the meter set on the amperage setting and in the
expected range before you power up the circuit...  You could pop a fuse in
the meter if you aren't careful.  I seem to manage to do this every other
month or so...

----- Original Message -----
From: Dale Botkin <.....daleKILLspamspam@spam@BOTKIN.ORG>

>
> > +5V ---> Ammeter ----> +5V terminal of circuit
>
> That's the way, yes.
>
> Dale

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2002\02\18@152838 by Thomas C. Sefranek

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Michael Vinson wrote:

> Donovan Parks wrote:
>
>> Will placing an ammeter in series along the positive end of my power
>> supply
>> tell me how much current my circuit requires?  Seems like it should,
>> but my
>> thumbs are green when it comes to actually doing EE.  Thanks for
>> taking the
>> time to enlighten me.
>>
>> +5V ---> Ammeter ----> +5V terminal of circuit
>
>
> Yes, unless your application is supplied from several different
> voltage sources (e.g. 5V and 12V).

OF THE SAME POLARITY!!!

{Quote hidden}

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2002\02\18@153845 by Jim

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----- Original Message -----
From: "Michael Vinson" <.....mjvinsonKILLspamspam.....HOTMAIL.COM>
To: <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU>
Sent: Monday, February 18, 2002 1:59 PM
Subject: Re: [EE]: Measuring Current


> Donovan Parks wrote:
> >Will placing an ammeter in series along the positive end of my power
supply
> >tell me how much current my circuit requires?  Seems like it should, but
my
> >thumbs are green when it comes to actually doing EE.  Thanks for taking
the
{Quote hidden}

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2002\02\18@154750 by Olin Lathrop

face picon face
> Will placing an ammeter in series along the positive end of my power
supply
> tell me how much current my circuit requires?  Seems like it should, but
my
> thumbs are green when it comes to actually doing EE.  Thanks for taking
the
> time to enlighten me.
>
> +5V ---> Ammeter ----> +5V terminal of circuit

That should work fine most of the time.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olinspamspam_OUTembedinc.com, http://www.embedinc.com

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2002\02\18@181038 by Donovan Parks

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Hello,

Thanks for all the help.  Another question: how do I measure the current
from a 3-phase, 120Vac voltage source (i.e. a wall plug-in)?

Regards,
Donovan Parks

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2002\02\18@182918 by michael brown

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Donovan submits this for piclist perusal:
> Hello,
>
> Thanks for all the help.  Another question: how do I measure the current
> from a 3-phase, 120Vac voltage source (i.e. a wall plug-in)?

I don't know where you live, but here in the US, the 120VAC electricity that
you would find in a home is single-phase.  But, if that is the type of
current you wish to measure, I suggest you get a "clamp on" type device.
This will save you much trouble and danger.

On the lighter side:

The only guy I ever knew that had 3-phase 440VAC coming into his house was
putting out the REALY BIG signal on 11 meters.  The FCC subsequently removed
his equipment (44,000 watts PEP on AM) amidst much fanfare and cheering by
the neighbors.  This alleviated him of his need for 3-phase power in the
home.

michael brown

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2002\02\18@190214 by Donovan Parks

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Hello,

I'm up here in Canada and indeed we have single-phase.  Was thinking 3 phase
because of the 3 wires - brain fart.  Anyway, to measure the current can I
put the "clamp on" device in series with any wire?

Regards,
Donovan


> Donovan submits this for piclist perusal:
> > Hello,
> >
> > Thanks for all the help.  Another question: how do I measure the current
> > from a 3-phase, 120Vac voltage source (i.e. a wall plug-in)?
>
> I don't know where you live, but here in the US, the 120VAC electricity
that
> you would find in a home is single-phase.  But, if that is the type of
> current you wish to measure, I suggest you get a "clamp on" type device.
> This will save you much trouble and danger.
>

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2002\02\18@193209 by Dipperstein, Michael

face picon face
> From: Donovan Parks [@spam@Donovan.ParksKILLspamspamNRC.CA]
>
> Hello,
>
> I'm up here in Canada and indeed we have single-phase.  Was
> thinking 3 phase
> because of the 3 wires - brain fart.  Anyway, to measure the
> current can I
> put the "clamp on" device in series with any wire?
>
> Regards,
> Donovan

Current clamps go around the wire, not in series with anything.  Imagine
attaching a vise-grip to a wire, and then connecting the vice-grip to your
multi-meter.

-Mike

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2002\02\18@193213 by Sean H. Breheny

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Dear Donovan,

The great thing about clamp on devices is that you don't put them in series
with anything! You just clamp it around the fully insulated wire without
having to make any changes to the circuit. In the case of measuring the
current drawn from a regular AC outlet, you would want to clamp it on the
hot wire, I'd think, since it is possible for some of the current to flow
in both neutral and ground (although there is a problem if it is flowing in
ground), but the total current must flow in the hot wire.

Clamp on probes work by measuring the magnetic field around the wire due to
the current flowing in the wire; that's why they need no electrical
connection to the circuit. There is a law in electromagnetics that says
that the integral (think average) of the magnetic field around any closed
loop is equal to the current flowing through the hole in the loop times a
constant (which is a fundamental constant). That's why clamp on probes are
not position-critical: as long as the sensing coil in the pickup loop goes
around the wire, it will measure properly.

Sean

At 04:11 PM 2/18/02 -0800, you wrote:
{Quote hidden}

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2002\02\18@194040 by Herbert Graf

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No, clamp on meters are design to clamp AROUND he wire carrying the current.
Note though that this is an additive process, the meter will show the SUM of
current going through the cables, so you need to clamp only ONE wire,
clamping both would result in a reading of zero. TTYL


> {Original Message removed}

2002\02\18@194043 by michael brown

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Donovan says:
> Hello,
>
> I'm up here in Canada and indeed we have single-phase.  Was thinking 3
phase
> because of the 3 wires - brain fart.  Anyway, to measure the current can I
> put the "clamp on" device in series with any wire?

The "clamp on" types don't actually have "live" current going thru them.
They inductively measure the amount of current flowing thru the wire.  It
does this by measuring the magnetic field around the conductor.

http://www.appatech.com/a-30.htm  <--- A picture is worth a thousand words

michael brown

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2002\02\19@054747 by Vasile Surducan

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On Mon, 18 Feb 2002, Jim wrote:

> Yes.
>
> Connect an Ammeter *in series* for current
>
> and a Voltmeter *in parallel* for Voltage.
>

 connect an ohmmeter in parallel with your mains socket to measure the
 internal mains resistence....

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2002\02\19@082633 by M. Adam Davis

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Herbert Graf wrote:

>No, clamp on meters are design to clamp AROUND he wire carrying the current.
>Note though that this is an additive process, the meter will show the SUM of
>current going through the cables, so you need to clamp only ONE wire,
>clamping both would result in a reading of zero. TTYL
>
Unless you're making the reading in a bathtub... ;-)

Adam

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'[EE]: measuring current pulses to imply torque'
2003\02\10@181914 by Derek Cowburn
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Hello piclisters.
I'd like to use a PIC to measure the torque on an R/C servo.  For my
project, it would suffice to measure the pulse widths of current sent to the
motor by the h-bridge in the servo.

This site (http://www.web-hobbies.com/servo8t.html) has a controller with
the feature I'm looking for, but I want to add this feature to a small
PIC12C671 (or similar) and place it inside the servo (space permitting).

Has anyone made such a circuit?  I'm much better at software than
electronics.  Any help appreciated.

-Derek

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'[EE:] Measuring Current in a High Voltage Circuit'
2004\03\12@152912 by Roberts II, Charles K.
picon face
Hello All

I am working on a PIC based High Voltage module. The schematic of the
current meter portion of the module is attached. I am measuring the
voltage drop across a knowm resistor, but to do it I had to use two
voltage dividers to bring the levels to something I can read. I measure
about the same current with or without a load. What could be going on?
Loooking into the devider there are 50Meg Ohms to Ground, Is this what
is messing up my reading?

Thanks in Advance

Chuck Roberts

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2004\03\12@153536 by Roberts II, Charles K.

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part 1 1187 bytes content-type:text/plain; (decoded quoted-printable)



The Circuit I was talking about in my orginal post.

Charles K Roberts II



{Original Message removed}
part 2 8934 bytes content-type:application/octet-stream; (decode)

2004\03\12@154403 by Alexander Rice

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On Fri, 12 Mar 2004 15:29:05 -0500, Roberts II, Charles K.
<TakeThisOuTrobertsckEraseMEspamspam_OUTORNL.GOV> wrote:

> Hello All
>
> I am working on a PIC based High Voltage module. The schematic of the
> current meter portion of the module is attached. I am measuring the
> voltage drop across a knowm resistor, but to do it I had to use two
> voltage dividers to bring the levels to something I can read. I measure
> about the same current with or without a load. What could be going on?
> Loooking into the devider there are 50Meg Ohms to Ground, Is this what
> is messing up my reading?

Quite possibly, i don't think that the source impedence for the A2D
modules is reccomended to be greater then 10K Ohms, try putting a 0.1uF
capacitor across it to lower the impedence during A2D.

Regards

Alex Rice

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2004\03\12@155709 by Mike Hawkshaw

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>----- Original Message -----
>From: "Roberts II, Charles K." <RemoveMErobertsckspamTakeThisOuTORNL.GOV>
>To: <PICLISTEraseMEspam.....MITVMA.MIT.EDU>
>Sent: Friday, March 12, 2004 8:29 PM
>Subject: [EE:] Measuring Current in a High Voltage Circuit
>

>Hello All
>
>I am working on a PIC based High Voltage module....

Chuck,

If at all possible, put the current sensing resistor in the negative return
of the power supply. That way you will have a much simpler amplifier / level
shifter circuit, and the whole thing should be safer to work on.

Failing that, I would use an optical isolater to do the 3kV isolation.

All the best...Mike.

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2004\03\12@160125 by Madhu Annapragada

picon face
but, but...your A2D converter on the PIC is isolated from the divider. You
have a unity gain follower after the divider and then a difference amp which
has a very low output impedance. So why should the divider have any affect
on the A2D..as far as the PIC RA1 input is concerned there is a voltage
source with a very low impedance it is looking at. Am I missing something
obvious here?
Madhu


>{Original Message removed}

2004\03\12@160745 by David Minkler

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Chuck,

Wrong approach.  Do a quick tolerance analysis of your dividers and you
can see that even a small variation between the two dividers will easily
swamp any signal that you get across your metering resistor.  Measure
the output current on the low side.  Do you have access to the inside of
the "HV MODULE"?  These little supplies typically use a flyback or
cockroft-walton bridge in their output stage.  You need to insert your
metering resistor between the low side of the output stage and its local
ground.

Dave

Roberts II, Charles K. wrote:

{Quote hidden}

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2004\03\12@161159 by Spehro Pefhany

picon face
At 03:29 PM 3/12/2004 -0500, you wrote:
>Hello All
>
>I am working on a PIC based High Voltage module. The schematic of the
>current meter portion of the module is attached. I am measuring the
>voltage drop across a knowm resistor, but to do it I had to use two
>voltage dividers to bring the levels to something I can read. I measure
>about the same current with or without a load. What could be going on?
>Loooking into the devider there are 50Meg Ohms to Ground, Is this what
>is messing up my reading?

Are you trying to measure the voltage across the 1K resistor? That's
REALLY not feasible, your whole signal will be lost by a 100ppm mismatch
between the 100M resistors, and if you wanted 10% accuracy, you'd need
them to match and track within +/-10ppm. That's just not going to happen
at that resistance level.

Howzabout you put the 10K in the TP10 to ground net and measure the
voltage across that? Add some protection in case the output gets
shorted you don't lose the op-amp etc. every time.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2004\03\12@162236 by Dwayne Reid

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At 01:33 PM 3/12/2004, Roberts II, Charles K. wrote:

>I am working on a PIC based High Voltage module. The schematic of the
>current meter portion of the module is attached. I am measuring the
>voltage drop across a knowm resistor, but to do it I had to use two
>voltage dividers to bring the levels to something I can read. I measure
>about the same current with or without a load. What could be going on?
>Loooking into the devider there are 50Meg Ohms to Ground, Is this what
>is messing up my reading?

I don't think that you have a hope of making this work well.  The problem
is matching the divider resistors accurately.  I see two easy alternatives:
monitor the ground current instead or do it magnetically with a hall effect
current sensor or flux gate current sensor.

Monitoring the ground current is the easiest.  Hall effect current sensors
(LEM modules, etc) aren't very sensitive - they are normally intended for
measuring current in Amperes, not mA or uA.  Flux gate current sensors are
much more sensitive but need to be shielded or have some other mechanism
for ignoring the effects of external magnetic fields.  This makes them
expensive.

dwayne

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2004\03\12@162650 by Roberts II, Charles K.

picon face
To give some back ground I am a technician working on the code for this
beast and I didn't design the circuit. So my questions may seem simple.


Spehro Pefhany Wrote:

>Are you trying to measure the voltage across the 1K resistor? That's
>REALLY not feasible, your whole signal will be lost by a 100ppm
mismatch
>between the 100M resistors, and if you wanted 10% accuracy, you'd need
>them to match and track within +/-10ppm. That's just not going to
happen
>at that resistance level.


So the mismatch in resistance is drowning out my measurement, and it
can't be accounted for as an offset?
So what type of tolerance would that be to get it to work?

>Howzabout you put the 10K in the TP10 to ground net and measure the
>voltage across that? Add some protection in case the output gets
>shorted you don't lose the op-amp etc. every time.

So it will not work as it is done now? My boss designed this baby, if
his design won't work how should I tell him and convince him it won't
work?

P.S This is my first job out of college. And thanks for all the help.

Chuck

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2004\03\12@163314 by Tom

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Chuck,

As pointed out by several others, try to arrange your current sense
resistor in the ground side of the load.  Many problems will go away if you
do that.

If you cannot do that for some good reason, you will have to solve each of
the various problems.  Large resistors tend to not be as stable as low
value ones unless you pay *lots* for them. So you will need to calculate
drift effects over the intended operating range.

Your circuit shows no offset adjustment; it will almost certainly need one
just to dial out initial tolerance differences.

Something not shown is protection circuitry.  Your 100Meg resistors should
either be big physically and potted -or- comprised of several discrete
parts to reduce the voltage across each one.  Resistors have a voltage
coefficient - you must find out what it is for the parts you plan on using.
Also, multiple resistors offer more protection against arcing.

A common approach at the bottom of the divider is 2 series resistors to
your amp inputs with clamp diodes in between.  Then if the top 100meg
resistor arcs, the first series resistor limits the current, the diodes
clamp it, the second resistor in series makes sure the diodes clamp and not
the internal amplifier diodes.

Finally, highvoltage modules often exhibit highvoltage, high frequency
noise.  With 100meg resistors and lets say 100khz converter noise, you get
a corner with 16 femtofarads - easily seen across such resistors.  So keep
in mind that you may need filter caps across the bottoms of the dividers.

If possible, really, really try to move your sense resistor! ;)
HTH,
Tom

Who used to work in highvoltage power supplies a long time ago in a galaxy
far far away.

At 03:29 PM 3/12/04 -0500, you wrote:
{Quote hidden}

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2004\03\12@170504 by Paul Hutchinson

picon face
> -----Original Message-----
> [RemoveMEPICLISTspam_OUTspamKILLspamMITVMA.MIT.EDU]On Behalf Of Roberts II, Charles K.
> Sent: Friday, March 12, 2004 4:26 PM
>
> To give some back ground I am a technician working on the code for this
> beast and I didn't design the circuit. So my questions may seem simple.
<snip>
> So it will not work as it is done now? My boss designed this baby, if
> his design won't work how should I tell him and convince him it won't
> work?

Don't _tell_ him about it, _show_ your boss that it doesn't work. He
designed the circuit so, either he should be able to make it work or, he
should admit he doesn't know how to make it work. At that point you can
mention the good suggestions other here have made.

Paul

P.S. Your boss may actually know that the circuit doesn't work and wants to
see how you react.

> P.S This is my first job out of college. And thanks for all the help.
>
> Chuck
>
>

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2004\03\12@175035 by Hopkins

flavicon
face
What is the voltage you expect out of the high voltage unit and how much
current?

As has been suggested a current transformer is the way to go as they come in
different current ratings and will out put a voltage proportional to the
current.

A wire is passed through the toroidal section of the transformer and the
windings are designed to give a current or voltage out at the designed
value - usually 5A or 5v depending on its design. You can also get 4 to 20ma
loop transformers if that is what you want - let me know what you are
measuring and I will see if I can find a suitable supplier for you.

Are you using a PIC i.e. you want 5v maximum level?

*************************************************
Roy Hopkins   :-)

Tauranga
New Zealand
*************************************************

----- Original Message -----
From: "Roberts II, Charles K." <RemoveMErobertsckTakeThisOuTspamspamORNL.GOV>
To: <EraseMEPICLISTspamspamspamBeGoneMITVMA.MIT.EDU>
Sent: Saturday, March 13, 2004 9:33 AM
Subject: Re: ] Measuring Current in a High Voltage Circuit




The Circuit I was talking about in my orginal post.

Charles K Roberts II



{Original Message removed}

2004\03\12@182740 by Spehro Pefhany

picon face
At 04:25 PM 3/12/2004 -0500, you wrote:
>To give some back ground I am a technician working on the code for this
>beast and I didn't design the circuit. So my questions may seem simple.

Okay, fair enough.

{Quote hidden}

Yes. The offset is large and unstable.

>So what type of tolerance would that be to get it to work?

Forget it. You can't get stable enough resistors at that resistance
level and with that much DC voltage across them. You'll be lucky
if they don't shift a percent or more in time and with temperature.
To put it bluntly, it's a really silly circuit.

Work it out in detail if you'd like, it's only Ohm's law and you
should be able to do it easily.

Roughly, you should be able to see that the voltage change
across the 1K due to the load current is Ix*1K. That results in
a change in current through the 100K load resistor of

-Ix * 1K/100M

IOW, a +1mA current change results in a -10nA change

Now suppose you want 5% accuracy at 1mA. The changes due to the
resistors should be less than 5% of this, or +/- 0.5nA

A small change in the 100M resistor results in a change in current of
roughly -a * Vhv /100M, where a is the fractional change in the 100M
resistor. IOW, a +1% change, if Vhv is 2kV, results in a -200nA change.

(if you don't see this, then try calculating dI/dR (differentiating
the resistor value) and make approximations for small changes)

I think it's obvious that |200nA| >> |0.5nA|

> >Howzabout you put the 10K in the TP10 to ground net and measure the
> >voltage across that? Add some protection in case the output gets
> >shorted you don't lose the op-amp etc. every time.
>
>So it will not work as it is done now? My boss designed this baby, if
>his design won't work how should I tell him and convince him it won't
>work?

No, it won't work, if the goal is to measure the voltage across the
1k. If your boss is an engineer, ask him what happens when the
resistors are 1% off (keep in mind that they WON'T stay put, they
will vary with time, temperature, the voltage across them, the length
of time the voltage has been across them, physical stress, humidity
and probably more things than that). High meg resistors are unstable
beasts at the best of times, and all the more so under your conditions.

>P.S This is my first job out of college. And thanks for all the help.

No problem. Time to chuck that circuit and stop wasting time on it.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
RemoveMEspeffKILLspamspaminterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

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2004\03\12@190347 by David Minkler

flavicon
face
Roberts II, Charles K. wrote:
>> and if you wanted 10% accuracy, you'd need them to match and track
within +/-10ppm.
>  So the mismatch in resistance is drowning out my measurement, and it
can't be accounted for as an offset?
>  So what type of tolerance would that be to get it to work?

That works out to a combined tolerance of 0.001%.  This includes initial
tolerance (probably 1% at best), tempco (typically 100ppm/degC for
"good" dividers), voltage coefficient, phase of the moon, light in the
side office over there ... you get the idea.  As has been pointed out,
it's not going to work this way.

>  So it will not work as it is done now? My boss designed this baby,
if his design won't work how should I tell him and
>  convince him it won't work?

So, your real problem isn't technical but political.  Presenting a
problem is always easier if you can simultaneously present the
solution.  Any attempt at showing that the circuit doesn't work will
inevitably force you to conduct the tolerance analysis so, you may as
well do it ahead of time.  Start with one of the dividers perfect at
1000:1 and show what happens when you alter the other divider by its
nameplate tolerance.  The argument will then be made that you can
calibrate the error out.  Then show what happens if that tolerance is
reduced by a factor of ten (say from 1% to 0.1%) but that the value
changes (use real numbers and work through the math, assume ideal op
amps) from before the offset adjustment to after the offset adjustment.
This is comparable to the effect of tempco over a normal operating range
unless you have a fairly good ratio tempco (great ones can be 10ppm/degC
or better but note, this is per degree C).  Then show that all the
problems go away if you can put the sense resistor in the ground leg of
the HV output section (then it becomes principally an initial tolerance
problem).  If your boss is reasonable, he'll appreciate what you've done
and work with you to implement the solution.  If he's not, he'll spend a
lot of time, effort and money trying to get the wrong approach to work
and eventually, natural selection will take over.

Good luck,

Dave

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2004\03\12@191347 by David Minkler

flavicon
face
Hopkins wrote:

>As has been suggested a current transformer is the way to go as they come in
>different current ratings and will out put a voltage proportional to the
>current.
>
A current transformer is NOT the way to go.  The OP wants to measure DC
current from the output.  A current transformer won't tell him
anything.  The correct/traditional approach is to move the current sense
resistor into the ground leg of the HV output (or the return path of the
load).  Hall effect and other high side approaches were mentioned as
alternatives but none of them is as simple or straightforward as moving
the sense resistor to the correct location.

Dave

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2004\03\15@084138 by Alan B. Pearce

face picon face
>A current transformer is NOT the way to go.  The OP wants
>to measure DC current from the output.  A current
>transformer won't tell him anything.

Well it will if he gets one of the ones that incorporate hall effect sensors
such as those from LEM. However I am not sure what the temperature
co-efficient is, and as others have said they are really designed for
currents in the amp range, rather than tiny currents.

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2004\03\15@105614 by Roberts II, Charles K.

picon face
Thanks to everyone who helped me with this problem. And a special thanks
to David and Spehro for there detailed response. You guys have been very
helpful and supportive.
Chuck Roberts

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2004\03\15@135150 by Richard.Prosser

flavicon
face
Is that Chuck Roberts Hero & super fix-it-up-man
or Chuck Roberts, now unemployed ???

RP





Thanks to everyone who helped me with this problem. And a special thanks
to David and Spehro for there detailed response. You guys have been very
helpful and supportive.

Chuck Roberts

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2004\03\15@141851 by Roberts II, Charles K.

picon face
>Is that Chuck Roberts Hero & super fix-it-up-man
>or Chuck Roberts, now unemployed ???

>RP

I am still employed, my boss is out of town. He will not get the news
until next week. But I now I know how to give him the news.

Charles K Roberts II

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'[EE] uTutorial: Measuring current with elcheapo ('
2005\07\20@052853 by Russell McMahon
face
flavicon
face
Excerpted and expanded from another post as this is a trap that's
easily fallen into.

When measuring current on ANY cheapo meter (and some not so cheap
ones) measure the voltage drop across the meter when operating to get
an idea of the meter's operating resistance . this may vary somewhat
with current. The result may surprise you. Resistances of 10 to 20
ohms on eg 200 mA range are not unknown and can cause major problems
in some cases if not accounted for. Ten ohms at 100 mA causes a 1 volt
drop. More current and higher resistance can more than double this.
Such a drop may give an entirely inaccurate indication of what is
really happening in a circuit or cause problems directly. eg a 10 ohm
undecoupled resistance in a circuit may cause oscillations or drop
voltages well below their expected value. A circuit that runs properly
on batteries may malfunction when a current meter is added to the
circuit. (Ask me how I know :-) ). (Still easy to get caught out after
many years experience).

When current measurement at low voltage drops is required this may be
achieved using a selected series resistor and a meter set to a low
voltage range. For example, imagine that a circuit draws up to 100 mA
and that a drop of up to 0.1 volt is acceptable. This is the
equivalent of R = V/I = 0.1/0.1= 1 ohm. Placing a 1 ohm resistor in
circuit, decoupling it's downstream side with a suitable capacitor*
and measuring the voltage drop across it will yield an acceptable
ammeter. Set to the 2 volt range a multimeter will read 0.100 at 100
mA. ie volts = amps. Substituting a 0.1 ohm resistor and switching to
the 200 mV range will give a reading of 10.0 at 100 mA.

If you want to keep your brain young you can use resistor values which
are not multiples of 10 :-)

* If this is measuring current from eg a power supply rail then
downstream decoupling is desirable. In some cases the circuit may be
better not decoupled. In such cases the effect of the series resistor
must be allowed for.

Use of an opamp based solution allows far smaller sense resistor
values but in most cases the above will suffice.




               RM


'[PIC] measuring current in a buck dc-dc converter'
2006\05\04@200602 by rlistas
picon face
Hi,


I need some suggestion on how to measure the output
current in a buck dc-dc converter used to charge
a battery.

I´m using a PIC 16F88 as PWM generator and charge
controller (MPPT).
The problem of  measuring the current is the ground referencial
that the microcontroller uses. I´m using a N channel
mosfet as the switch (so the negative is switched and the positive is
always conected) and therefore I´m getting trouble using the AD of the PIC.

I want to get the value of the current by means of a low value resistor
connected to the battery.

Any suggestion will be welcome.

thanks

Rubens

2006\05\04@203835 by PicDude

flavicon
face
Op-amp across the sense resistor?

-Neil.


On Thursday 04 May 2006 19:05, rlistas wrote:
{Quote hidden}

2006\05\05@063945 by istopher Holmes

flavicon
face
Hi Rubens,

 The INA138 from TI is a nice part for measuring current. Keep in mind
though that the minimum common mode input is 2.7V, so you couldn't use
it for charging a low voltage battery.

 You might also use an instrumentation amplifier. It's simpler than
building a differential amplifier with an op amp, plus you can easily
set the output's reference voltage to whatever you like. I did a circuit
like yours once using an AD623 to control the output of a switching
current source at very low voltage (i.e. low resistance load). Worked
pretty well.

Chris

rlistas wrote:
{Quote hidden}

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2006\05\05@071523 by Gerhard Fiedler

picon face
rlistas wrote:

> I need some suggestion on how to measure the output current in a buck
> dc-dc converter used to charge a battery.
>
> I´m using a PIC 16F88 as PWM generator and charge controller (MPPT). The
> problem of  measuring the current is the ground referencial that the
> microcontroller uses. I´m using a N channel mosfet as the switch (so the
> negative is switched and the positive is always conected) and therefore
> I´m getting trouble using the AD of the PIC.
>
> I want to get the value of the current by means of a low value resistor
> connected to the battery.

If I understand you correctly, you can do that with an opamp or a high-side
current sensor amp (which is a specialized opamp). Look at the LT1637 or
similar. (For a one-off, Maxim also has such parts :)

Gerhard


'[PIC] Measuring current'
2007\04\30@171201 by Rikard Bosnjakovic
picon face
My next PIC-project is a small - very basic - component tester. It
will measure (together with shorts) the polarity of a diode, if a BJT
is NPN/PNP and cables. I have written down all necessary logic
required, and so far it looks like a doable project.

When thinking of the BJT-test, I thought that it might actually be
nice to get an approximate value of hFE/beta as well. Do be able to do
this, I need to feed the base and collector with a (known) current,
and measure / calculate current/beta at the emitter, and this is also
where I'm stuck; I haven't found any PICs that can measure current.

So I went to piclist.org and searched the archives in case someone
else already did this project. I only found one thread (from 1998)
where the OP wanted to measure around 2-3 amps with a uC, and got the
reply that he should take a peek at Maxim's MAX4172-IC which is a
current-to-voltage-converter and which would do what the OP wanted.
There is no problem sampling the IC from Maxim, but it's in SO-package
and I'm only able to do DIP)

Is there some other way to accomplish what I want to do? Construct a
CVC using an opamp, or is there some better way?


--
- Rikard - http://bos.hack.org/cv/

2007\04\30@173657 by Harold Hallikainen

face
flavicon
face
Well, for your collector current sampler, you could do something like this:

1. Put a resistor from the output of an op amp to the inverting input, say
1k.
2. Drive the non-inverting input with the desired collector voltage, say
+10V.
3. Connect the collector of the transistor to the non-inverting input of
the op amp.
4. Ideally, the inverting input is at the same voltage as the
non-inverting voltage, so it's at 10V. Ideally no current flows into
either input of the op-amp, so all the collector current must flow through
the resistor. The output voltage will then be 10V + Ic*1k. You can measure
that output voltage and calculate the collector current.

Harold



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2007\04\30@205214 by Richard Prosser

picon face
Rikard,

For a (very) rough check,
Tie a 10mA source to the collector of the bipolar. Increase the the
base current until the collector voltage drops to less than 0.6 (or
even just registers "low" on a pic input.) The ratio of 10mA to the
base current will give you the approximate hfe. at 10mA. (or 1mA or
10A etc.) Since the base current will be low, and the base voltage
will be in the 0.5 - 0.7 V area, a pwm generated voltage feeding a 10k
resistor to the base may be enough.

(50% pwm = 2.5V,
2.5V - 0.5V = 2V,
2/10e3 = 200uA Base Current)

hfe = 10mA/200e-6 = 50.



RP
On 01/05/07, Harold Hallikainen <@spam@harold@spam@spamspam_OUThallikainen.org> wrote:
{Quote hidden}

> -


'[PIC] Measuring current'
2007\05\01@120254 by Brooke Clarke
flavicon
face
Hi Rikard:

You might want to check out the Semiconductor Analyzer at:
http://www.m3electronix.com/sa.html

It's a very capable device and reasonable priced kit.
But you do need a 40 watt or so iron to solder to the think copper traces and a
good flux cleaner to get rid of the flux residue.

Have Fun,

Brooke Clarke
--
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www.pacificsites.com/~brooke/PRC68COM.shtml
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'[EE] measuring current through an inductor'
2008\10\03@233515 by Jesse Lackey
flavicon
face
Hello all, as time goes by I find myself doing more and more dc/dc work,
including things like switcher battery chargers and high power LED
drivers, mostly successfully but sometimes not.

In doing the design of such things, the inductor is often the biggest
question.  Datasheets for whatever IC is running the show of course give
guidance, but sometimes this is less than complete, and sometimes
they'll give typical but not corner-case information, and sometimes
they'll give info needed to run whatever it is at max power, and I'm
designing for 1/5 of that, and don't want to use a way overkill inductor.

Ok, all this leads up to the question: how do I measure current through
an inductor as it is operating under normal loads in such designs?  If I
put a 0.1R resistor in series, I can get some kind of measurement (with
a scope, 2 channel, in difference mode), but for some inductors adding
0.1R to its resistance changes the overall operation considerably.

The goal of all this is twofold - firstly to check that the inductor
isn't saturating at corner cases, and secondly to see how predicted
operation compares with actual and maybe optimize the design to use
smaller/cheaper inductors.

Suggestions?

Thanks everyone!
J

2008\10\03@235332 by Xiaofan Chen

face picon face
On Sat, Oct 4, 2008 at 11:34 AM, Jesse Lackey <spamBeGonejsl-mlspamKILLspamcelestialaudio.com> wrote:
> Ok, all this leads up to the question: how do I measure current through
> an inductor as it is operating under normal loads in such designs?  If I
> put a 0.1R resistor in series, I can get some kind of measurement (with
> a scope, 2 channel, in difference mode), but for some inductors adding
> 0.1R to its resistance changes the overall operation considerably.

Typically we use a current probe and the associated amplifier and
a oscilloscope. Tektronix is popular.
http://www.tek.com/products/accessories/oscilloscope_probes/current.html

Xiaofan

2008\10\04@025041 by Sean Breheny

face picon face
Note that some excellent Tek current probes can be purchased via eBay.

If your switching frequency isn't too high (less than 25kHz), or you
only care about measuring the DC current through the inductor, then
you could get away with a hall-effect clamp on current probe. AEMC
makes some of these. At work I have one (SL261) which covers 0-10A and
0-100A (on a less sensitive scale), 100kHz bandwidth. It was about
$450.

I also have a Tek A6303 along with the AM503A amplifier which it needs
and a TM502 mainframe to power it. I bought each of these separately
and in total I paid about $800 on eBay. However, this one is vastly
more capable than the SL261. 0-100A with the ability to accurately
measure 10s of mA. 15MHz bandwidth.

Depending on exactly what you want to do, you might get away with a
regular clamp on AC/DC ammeter, or a true RMS model, or a current
transformer (would not show DC but could show AC).

Sean


On Fri, Oct 3, 2008 at 11:53 PM, Xiaofan Chen <.....xiaofancspam_OUTspamgmail.com> wrote:
{Quote hidden}

> -

2008\10\04@044034 by Vasile Surducan

face picon face
If you're using AC, the simplest current probe is one wire to n wire
transformer. You can do it yourself with the price of a litle work.
Then you must measure the probe and find the exact ratio. That's all.

Vasile

On 10/3/08, Jesse Lackey <TakeThisOuTjsl-mlKILLspamspamspamcelestialaudio.com> wrote:
{Quote hidden}

> -

2008\10\06@121126 by alan smith

picon face
Another choice is to simulate using SwitcherCAD


--- On Fri, 10/3/08, Jesse Lackey <.....jsl-mlspamRemoveMEcelestialaudio.com> wrote:

{Quote hidden}

> --

2008\10\06@135102 by 'Grif'

flavicon
face
For development or in a production environment?  Anyway, from the old days, I'm still found of my Tectronics current probe and a scope.

{Original Message removed}


'[EE] Measuring current via a shunt, can it go belo'
2018\01\26@100842 by Mario
picon face

Hello,
I'm playing with my SMPS step up (boost) design, where the controller is a PIC.

I thus have +12V power input, an inductor, a MOSFET, and a diode that charges a
high voltage capacitor, and a voltage divider before the diode, to monitor voltage
(via a first PIC ADC input) of the cap without actually discharging it continuously.

The gate of the MOSFET is connected to a +5V rail through a 1k resistor and I use
a PIC output in open drain configuration to ~quickly stop the MOSFET from conducting
(the opposite doesn't need to be ultrafast, as the SMPS will be always used in
discontinuos mode, thus current is zero whenever the MOSFET starts to conduct).
Switching frequency is pretty low.

While it's certainly not the most performant boost circuit, I like its simplicity
and low components count.

Now, I also need to monitor current (for which I have devoted a second ADC input),
and thus I have placed a shunt resistor between the source of the MOSFET and the
0V rail. I can be 100% sure that the voltage out of it will never exceed the PIC
ADC max input voltage, also due to insufficient input current from the power supply.
The shunt has also pretty low resistance so the voltage never reachs 0.5V (not much
resolution from the ADC, but enough to not want to add an OpAmp, as space on board
is very very limited).

Question is: should I put a series resistor between the PIC ADC input and the
shunt? Theoretically the voltage will never go below 0V and will never go above
0.5V (as explained above), but I fear some insidious aspects that maybe I haven't
been able to consider, maybe for some hard to grasp parasitic capacitance, etc..
could the shunt voltage go below 0V and thus fry my PIC ADC input pin unless I put
a series resistor? Is this series resistor necessary?

If it's totally useless, I don't want to put it, it's a matter of principle.. :D

Thank you very much for augmenting my design knowledge and experience.

Kind regards,
Mario

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2018\01\26@182209 by Brent Brown

flavicon
face
Hi Mario,

I think my reply on PIC input protection will be relevant.

I'll just add that a simple diode across the shunt resistor can add a degree of protection. A 1A or more silicon diode is relatively robust and cheap, and is a good fit if the knee of the conduction curve is above your analog input range. If AC is necessary, add another diode the other way. For higher voltages you might use multiple diodes in series, but pretty soon a zener or TVS diode becomes a better choice, then add series R to analog input and...

On 26 Jan 2018 at 16:08, Mario wrote:

{Quote hidden}

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2018\01\27@114142 by Mario

picon face

My question really wasn't about overloading a PIC input (I even specified it
won't happen, at least from the shunt normal operation) but if there could be
some "parasitic" or anyhow difficult to understand phenomena that will make
the shunt go negative. I'm thinking about overshoot, etc.. although theoretically
this should not happen.

Anybody that can elaborate on it please?



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2018\01\29@063250 by Mario

picon face

Oh well, I found myself one case when the voltage at the shunt can go negative:
after the MOSFET is turned OFF and the zener stops conducting as well, the body
diode of the MOSFET (or the zener itself, depends if it's connected to the source
of the MOSFET instead of to the ground, and I prefer the former solution as it
lets me monitor current also during the decay phase) then there won't be anything
to absorb the (small) remaining energy of the coil, but the parasitic capacitance,
resistance, etc.. and the circuit will thus exhibit ringing, which can indeed go
negative, if the Q is high.

So, in answer to my own question, a resistor is really needed to protect the PIC
ADC input.

Cheers,
Mario


At 17:41 2018-01-27, Mario wrote:
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