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Discussion Starter #1
It's interesting to know that the shunt can go anywhere. I ain't saying you=
're wrong but that surprised me. An ammeter being a millivoltmeter, I would=
think that in order to measure the -all- the millivolts in the pack, it wo=
uld have to be at one end so that it all flows through the shunt. Otherwise=
, you're cutting down on the millivolts. This might be flawed understanding=
on my part.
=

At any rate, the shut is in fact, at the end. The positive end. Roland's go=
t me all interested in buying another shunt for motor amps. It was recommen=
ded that I just use one meter with a selector switch to toggle between batt=
ery and motor amps.
=

Since we're talking millivolts here, does the milli-ohm resistance of the s=
elector switch throw off the ammeter by very much?
=

Rich A.
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Discussion Starter #2
Not critical. The battery amp meter is going to swing from 0 to about 200
amps while the motor amps will read 0 to 600 amps in a movement of
fractional of a seconds. Trying to read that little tiny of bit of ampere
difference with non-recording type of meters is about impossible to do.

I like to use separate meters for every system I have, so I can try to read
all with a glance.

It may not be a good ideal to have the motor and battery shunts on a
selector switch. If you do, fuse each sensor wire with a clip on fuse that
connected to the shunt sensor screw.

Some selector switches contacts may not have the clearance and some are
design to make before they break the other circuit.

If you selector switch shorts, the main battery voltage by passes the
controller circuit and it will try to run the motor and you could have the
selector switch melt and the sensor wires melt and burn.

Roland


----- Original Message -----
From: "Richard Acuti" <[email protected]>
To: <[email protected]>
Sent: Thursday, November 01, 2007 6:52 PM
Subject: [EVDL] Shunt placement issue solved



It's interesting to know that the shunt can go anywhere. I ain't saying
you're wrong but that surprised me. An ammeter being a millivoltmeter, I
would think that in order to measure the -all- the millivolts in the pack,
it would have to be at one end so that it all flows through the shunt.
Otherwise, you're cutting down on the millivolts. This might be flawed
understanding on my part.

At any rate, the shut is in fact, at the end. The positive end. Roland's got
me all interested in buying another shunt for motor amps. It was recommended
that I just use one meter with a selector switch to toggle between battery
and motor amps.

Since we're talking millivolts here, does the milli-ohm resistance of the
selector switch throw off the ammeter by very much?

Rich A.
_________________________________________________________________
Climb to the top of the charts! Play Star Shuffle: the word scramble
challenge with star power.
http://club.live.com/star_shuffle.aspx?icid=starshuffle_wlmailtextlink_oct
_______________________________________________
For subscription options, see
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Discussion Starter #3
The milivolts are created by the current flowing through the low resistace
of the shunt.
It's not measure the pack voltage AT ALL, it's measuring the current.

The voltage different it is measuring is from one side of the shunt to the
other side.
Again, assuming an analog meter, the only conection is to these two points
(opposite sides of the shunt).

The amp meter doesn't know, and doesn't care what the pack voltage is,
only the millivoltage across the shunt.

Note: Ohms law Voltage =3D Current * resistance

In this case
Measured voltage =3D current * shunt resistance

You'll notice that pack voltage does NOT appear anywhere in the equation.

So until we tie one side of the shunt to something other than a simple
meter movement, the location of the shunt is unimportant.
>
> It's interesting to know that the shunt can go anywhere. I ain't saying
> you're wrong but that surprised me. An ammeter being a millivoltmeter, I
> would think that in order to measure the -all- the millivolts in the pack,
> it would have to be at one end so that it all flows through the shunt.
> Otherwise, you're cutting down on the millivolts. This might be flawed
> understanding on my part.
>
> At any rate, the shut is in fact, at the end. The positive end. Roland's
> got me all interested in buying another shunt for motor amps. It was
> recommended that I just use one meter with a selector switch to toggle
> between battery and motor amps.
>
> Since we're talking millivolts here, does the milli-ohm resistance of the
> selector switch throw off the ammeter by very much?
>
> Rich A.
> _________________________________________________________________
> Climb to the top of the charts! Play Star Shuffle: the word scramble
> challenge with star power.
> http://club.live.com/star_shuffle.aspx?icid=3Dstarshuffle_wlmailtextlink_=
oct
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev
>


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Discussion Starter #5
My understanding is that the shunt is of a known, very precisely
calibrated resistance. A resistive load creates a voltage drop - by
measuring the voltage difference from a point immediately before and
after the shunt, you get a millivolt reading on that voltage drop
across the shunt.

Ben
Not an EE, but I read too much

Joseph T. <[email protected]> wrote:
> "The milivolts are created by the current flowing through the low resistance
> of the shunt."
>
> Does this mean that the current flowing through the wire "creates"
> milivolts, and that an amp meter is then measuring milivolts? So,
> basically, the amp meter indirectly reads the amperage by reading the
> milivolts?
> And by the way, how do amps, flowing through a wire of course,
> "create" milivolts?
>
>
> On 11/2/07, Peter VanDerWal <[email protected]> wrote:
> > The milivolts are created by the current flowing through the low resistace
> > of the shunt.
> > It's not measure the pack voltage AT ALL, it's measuring the current.
> >
> > The voltage different it is measuring is from one side of the shunt to the
> > other side.
> > Again, assuming an analog meter, the only conection is to these two points
> > (opposite sides of the shunt).
> >
> > The amp meter doesn't know, and doesn't care what the pack voltage is,
> > only the millivoltage across the shunt.
> >
> > Note: Ohms law Voltage = Current * resistance
> >
> > In this case
> > Measured voltage = current * shunt resistance
> >
> > You'll notice that pack voltage does NOT appear anywhere in the equation.
> >
> > So until we tie one side of the shunt to something other than a simple
> > meter movement, the location of the shunt is unimportant.
> > >
> > > It's interesting to know that the shunt can go anywhere. I ain't saying
> > > you're wrong but that surprised me. An ammeter being a millivoltmeter, I
> > > would think that in order to measure the -all- the millivolts in the pack,
> > > it would have to be at one end so that it all flows through the shunt.
> > > Otherwise, you're cutting down on the millivolts. This might be flawed
> > > understanding on my part.
> > >
> > > At any rate, the shut is in fact, at the end. The positive end. Roland's
> > > got me all interested in buying another shunt for motor amps. It was
> > > recommended that I just use one meter with a selector switch to toggle
> > > between battery and motor amps.
> > >
> > > Since we're talking millivolts here, does the milli-ohm resistance of the
> > > selector switch throw off the ammeter by very much?
> > >
> > > Rich A.
> > > _________________________________________________________________
> > > Climb to the top of the charts! Play Star Shuffle: the word scramble
> > > challenge with star power.
> > > http://club.live.com/star_shuffle.aspx?icid=starshuffle_wlmailtextlink_oct
> > > _______________________________________________
> > > For subscription options, see
> > > http://lists.sjsu.edu/mailman/listinfo/ev
> > >
> >
> >
> > --
> > If you send email to me, or the EVDL, that has > 4 lines of legalistic
> > junk at the end; then you are specifically authorizing me to do whatever I
> > wish with the message. By posting the message you agree that your long
> > legalistic signature is void.
> >
> > _______________________________________________
> > For subscription options, see
> > http://lists.sjsu.edu/mailman/listinfo/ev
> >
>
> _______________________________________________
> For subscription options, see
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Discussion Starter #6
> Does this mean that the current flowing through the wire "creates"
> milivolts, and that an amp meter is then measuring milivolts? So,
> basically, the amp meter indirectly reads the amperage by reading the
> milivolts?
> And by the way, how do amps, flowing through a wire of course,
> "create" milivolts?

Yes, that's exactly what it means. It's just Ohm's Law, V=I/R

If it sounds like free energy ("creating" millivolts), it's not. It's
creating a "negative" voltage that fights the battery and wastes power
(but a very small amount of power as long as you use thick wires).

And yes, the "amp meter" is really just a millivolt meters. If you
have a 500 amp shunt, it has a resistance of 100 uOhms (microOhms).
When you put 500A through it, V=I/R=500A/100uOhm=50mV; then the meter
takes that 50mV and displays the maximum output.

The amps flowing through the wire create a voltage the same way that
any resistor creates a voltage when current flows through it. When you
have current through a resistor, you get a voltage drop opposing the
current.

I hope that helps!

-Morgan LaMoore

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Discussion Starter #8
hehe - try V = IR ;)

john

Morgan LaMoore wrote:
> Yes, that's exactly what it means. It's just Ohm's Law, V=I/R

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