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PICList Thread
'[OT]:Current Sensor'
2001\04\04@203751 by Thomas N

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Hi everyone,

I am building a robot and I want to monitor the current drawn by the motor.
The motor can draw from 250mA up to 2.5Amp!  What do I have to do to measure
this current?

I was thinking of puting a 1-ohm resistor probe in series with the motor,
and feed the voltage drop across the resistor into a ADC, but at 2.5A, the
voltage drops across the resistor is too high and it leaves nothing for the
motor!

Is there any current sensor circuit out there that have very low loading
resistance?  Is there any IC doing this job?

Thanks

Thomas
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2001\04\05@043146 by Vasile Surducan

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Use a small resistor and a curent to voltage converter.
example: 0.1 ohm , 25mV to 250mV drop-out,
convert this potential with a good operational amplifier to 0-5V
Vasile

On Thu, 5 Apr 2001, Thomas N wrote:

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2001\04\05@091240 by t F. Touchton

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part 1 2648 bytes content-type:text/plain; charset=us-asciiHall effect sensors can also be used in this application to detect the flux
lines radiating from the motor supply leads.  Don't know a part number off the
top of my head, but a quick search on the web should turn up some hits.

Scott F. Touchton
1550 Engineering Manager
JDS Uniphase


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Use a small resistor and a curent to voltage converter.
example: 0.1 ohm , 25mV to 250mV drop-out,
convert this potential with a good operational amplifier to 0-5V
Vasile

On Thu, 5 Apr 2001, Thomas N wrote:

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part 3 136 bytes
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2001\04\05@122859 by Thomas N

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Do you know the part number for the current to Voltage converter IC?
THomas


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2001\04\05@140725 by Eisermann, Phil [Ridg/CO]

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       The resistor *is* your current to voltage converter. Not sure how
you're controlling the motor (PWM? FET/BJT? simple on/off?), but if one side
of the resistor is tied to ground, you can use an op-amp (non-inverting
configuration) to scale the small voltage drop across the resistor into 0-5V
for A/D conversion. Beware, though. The motor current will have a lot of
noise. Some  low-pass filtering before the A/D wouldn't hurt.

       I like the SenseFET for this application. International Rectifier
IRCZ44 comes to mind, although you might not get much accuracy because
that's a 50A device. It's a 5 pin MOSFET. The extra two pins are an
integrated 'current tap' This current tap reflects a fraction of the
drain-source current. You convert this scaled down current with a resistor.
The advantage is that you don't have the current-sensing resistor wasting
power. You only have the on-state resistance of the FET to worry about. For
the IRZC44, that's 0.028 Ohm. Only drawback is that since the motor is 2.5A
and the FET is 50A, you might not get much accuracy because the load is such
a small fraction. You'd need a large gain on the op-amp, and possibly
correlate some actual current readings with output to get good results. Or
you could use a different version of this FET (one with less range). DigiKey
now carries these types of FET's, so you might start there to see what you
can easily get.


{Original Message removed}

2001\04\05@141949 by Thomas N

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I will use PWM to control the speed of the Motor.  If I put the sensing
resistor in series with it and feed the resistor's voltage drop to the ADC,
then the input to the ADC will be a square wave! (has the frequency and duty
cycle of the PWM) which is bad.  I think I will need some circuit to average
it out...

Thomas


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>{Original Message removed}

2001\04\05@150814 by Eisermann, Phil [Ridg/CO]
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No, it won't :) well, it depends on the off time, i suppose. The motor
current can't decay instantaneously to zero because the motor is inductive.
The current will decay somewhat during the off period, but it's not going to
be zero unless you have a *very* slow PWM frequency. I don't suppose you
have access to a scope, so you? If you do, look at the current waveform of
the dropping resistor during PWM. But then, if you had a scope, you probably
would have done that already.

You should add a low pass filter anyway. Obviously, you want to set the
break frequency lower than your PWM frequency. That will go a long way
towards removing noise. Or you could add a peak-detect circuit if you want
the peak current instead of the average. But I don't think that is necessary
if your PWM frequency is reasonable (depends on the motor). How fast are you
running (or planning on running) the PWM?

{Original Message removed}

2001\04\05@152455 by Thomas N

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I am a bit confused.  Let's say I have:

+5V---0/ 0--------(0)-------/\/\/\------------|

+5V   switch    motor    Sense Resistor   Ground


Let's say the switch is closed and the motor is running, then suddenly I
open the switch, should the voltage drop across the sense resitor be at 0
mA? If not, where is the current comming from?

I am planing to use PWM (open and close the switch) at around 10Khz.  How to
I measure the average current flowing thru the motor while it is running
accurately?

Thomas



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>{Original Message removed}

2001\04\05@171535 by Eisermann, Phil [Ridg/CO]

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       The trick is that the motor is inductive. I don't know what your
background is, mathematical or otherwise, so i'll try to offer a simple
explanation without any math. I hope this makes sense. Its been a while :)
If I omitted something, or goofed it up, I hope someone will point it out
and correct me.

       An ideal inductor behaves according to v=L(di/dt). In non-math
speak, that means the voltage (v) is proportional (via the inductance, L) to
the time rate of change of the current through the inductor. In other words,
the faster the current is changing, the higher the voltage across the
inductor. Similarly, if the current isn't changing (e.g a constant current
through the inductor), there is no voltage across the inductor. Of course,
that's the ideal. A real inductor has to be made of wire, and wire has
resistance. But you get the idea.

       Now, consider what would happen if the current were to somehow
change from 2A to 0. Never mind why. What's the voltage across the inductor?
It depends on how fast the current changes, as stated above. The faster it
goes to zero, the higher the voltage that develops across the inductor has
to be. So what would happen if it were to change instantaneously? The
voltage across it would have to be infinite! Can this happen? Only in a
textbook. (This is the principle behind spark coils, btw).

       In the motor cicuit, once you open the switch, the current has got
to go somewhere. It cannot instantaneously dissappear (because that would
require the voltage to be infinite). Where does it go? back to ground,
through your resistor. It decays exponentially, not linearly. The faster
your PWM frequency, the less the motor current decays in the off period.

       So, does that make any sense at all? Also, your 10kHz sounds like a
good place to start.

{Original Message removed}

2001\04\05@191609 by Thomas N

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Thank you for your reply.  It clears up the cloud a little bit for me.
thanks!

Thomas


{Quote hidden}

>{Original Message removed}

2001\04\05@192019 by Bob Ammerman

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>         In the motor cicuit, once you open the switch, the current has got
> to go somewhere. It cannot instantaneously dissappear (because that would
> require the voltage to be infinite). Where does it go? back to ground,
> through your resistor. It decays exponentially, not linearly. The faster
> your PWM frequency, the less the motor current decays in the off period.

And the circuit is probably completed by the nice bright arc on your switch
contacts.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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2001\04\06@011635 by Dwayne Reid

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At 01:04 PM 4/5/01 -0500, Eisermann, Phil [Ridg/CO] wrote:
{Quote hidden}

Exactly right!  So use the IRCZ24 (much lower current rating).  We use a
lot of them . . .

The exact on resistance is not critical since it is a current ratio
arrangement (assuming you use IR's recommended amplifier circuit which,
unfortunately, requires a negative supply).

dwayne



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2001\04\06@040257 by Matt Pobursky

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Go to Maxim's web site and check out the MAX471 and MAX472.

http://dbserv.maxim-ic.com/quick_view2.cfm?pdf_num=1108

It's a single supply "high side current sense" IC that outputs a voltage
proportional to the current through a very low value (0.035 Ohm) sense
resistor. The sense resistor itself is built into the IC and can handle up to 3
Amps I believe. It comes in 8 pin DIP and SO packages, so it doesn't take up
much space and it runs on microamps of supply current.

Maxim also provides some good applications notes with the datasheet. Follow
them and you should have very few problems.

Matt Pobursky
Maximum Performance Systems

On Thu, 5 Apr 2001 00:37:33 -0000, Thomas N wrote:
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2001\04\06@045808 by Roman Black

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Thomas N wrote:
>
> I am a bit confused.  Let's say I have:
>
> +5V---0/ 0--------(0)-------/\/\/\------------|
              |                         |
               ---------|<--------------

> +5V   switch    motor    Sense Resistor   Ground
                       diode

> Let's say the switch is closed and the motor is running, then suddenly I
> open the switch, should the voltage drop across the sense resitor be at 0
> mA? If not, where is the current comming from?
>
> I am planing to use PWM (open and close the switch) at around 10Khz.  How to
> I measure the average current flowing thru the motor while it is running
> accurately?


First add the diode if you are using PWM on a
motor that has inductance (that's all of them!).

Thomas, search the net for "Jones on steppers"
he has a very nice tutorial that covers motor
controlling, and the "constant current" chapter
applies to all motors, not just stepper motors.
Well worth reading if you plan of getting good
performance from any motor.
-Roman

PS. Start your PWM experiment at a SLOW speed
first. This will be easier to control and
measure. Once you speed it up to 10kHz you start
to get quirky things like effects from switching
speed limits of the transistor and diode, and
switching surge losses get much higher. It might
work great at 500Hz but cook parts at 10kHz. :o)

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2001\04\06@144757 by Harold M Hallikainen

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On Thu, 5 Apr 2001 16:10:10 -0500 "Eisermann, Phil [Ridg/CO]"
<@spam@peiserma@spam@spamspam_OUTRIDGID.COM> writes:
>
>         In the motor cicuit, once you open the switch, the current
> has got
> to go somewhere. It cannot instantaneously dissappear (because that
> would
> require the voltage to be infinite). Where does it go? back to
> ground,
> through your resistor. It decays exponentially, not linearly. The
> faster
> your PWM frequency, the less the motor current decays in the off
> period.


       It would with a catch diode! Without a catch diode, the energy in the
inductor (LI^2/2) is dissipated in the switch as an arc. The voltage will
increase until it is high enough to maintain the current at the instant
the switch opened. It will then ramp down as that energy is dissipated.
       So... is a catch diode typically included in motor pwm circuits?

Harold

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2001\04\06@145003 by Harold M Hallikainen

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       Is it common to put a "catch diode" to ground after the switch in these
applications? If the motor appears as an inductive load, the current will
ramp up when the switch is closed and ramp down (through the diode) when
the switch is open. The voltages present, the inductance, and the PWM
frequency would determine the ripple in the current.

Harold

On Thu, 5 Apr 2001 19:24:30 -0000 Thomas N <.....thomasn101spam_OUTspamHOTMAIL.COM>
writes:
{Quote hidden}

> >{Original Message removed}

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