J
John Fields
Guest
On Thu, 02 Mar 2006 12:50:09 -0500, Ken C <cprstn54@nospam.att.net>
wrote:
Not true. Let's say the shunt drops 300 millivolts with 30 amps
across it.
With 30 amps going through the shunt, your circuit will look like
this:
30A--->
+V---------+
|
[RL] 300mV
| /
+--------+
| |
[SHUNT] [METER]
| |
GND--------+--------+
and if your meter impedance is 10 megohms, the circuit will look
like this:
30A--->
+V---------+
|
[RL] 300mV
| /
+--------+
| |
[SHUNT] [10MR]
| |
GND--------+--------+
Now, 300 millivolts across 10 megohms will cause:
E 3.0E-1V
I = --- = --------- = 3.0E-8A = 300nA
R 1.0E7R
to flow through the wires connecting the meter to the shunt.
Now, just for grins, lets say that you've got the meter located 1000
feet away from the shunt and you've got them connected up with #24
AWG copper wire. That's a total length of wire of 2000 feet, and a
total resistance of 51.4 ohms, so the voltage that'll be dropped
across the wire will be:
E = IR = 3.0E-8A * 51.4R ~ 1.54E-6V = 1.54ľV,
which is:
1.54E-6V
---------- * 100 ~ 0.00051% of full scale.
3.0e-1V
the meter will have, best case, a resolution of 1 part in 2999,
which is:
100
------ = 0.0333%
2999
so you can see that the effect the length of the wire is going to
have on the system will be miniscule.
The readings are going to stay pretty much the same no matter what
you do, DC wise, but as the length of wire between the shunt and the
meter increases you may wind up with an antenna... :-(
--
John Fields
Professional Circuit Designer
wrote:
---On Thu, 02 Mar 2006 11:29:39 -0600, John Fields
jfields@austininstruments.com> wrote:
This is really a digital millivoltmeter reading the drop across the
shunt and, typically, the input impedance to these voltmeters is
about 10 megohms, so the current in the leads (and, thus, the drop
across them) will be negligible compared to what's going through the
shunt.
Thank you. I still don't think I have enough information to avoid
trial & error.
The voltage drop along the meter leads will indeed be very small in
absolute amount, but not as a fraction of the voltage drop across the
shunt -- which will also be very small.
Not true. Let's say the shunt drops 300 millivolts with 30 amps
across it.
With 30 amps going through the shunt, your circuit will look like
this:
30A--->
+V---------+
|
[RL] 300mV
| /
+--------+
| |
[SHUNT] [METER]
| |
GND--------+--------+
and if your meter impedance is 10 megohms, the circuit will look
like this:
30A--->
+V---------+
|
[RL] 300mV
| /
+--------+
| |
[SHUNT] [10MR]
| |
GND--------+--------+
Now, 300 millivolts across 10 megohms will cause:
E 3.0E-1V
I = --- = --------- = 3.0E-8A = 300nA
R 1.0E7R
to flow through the wires connecting the meter to the shunt.
Now, just for grins, lets say that you've got the meter located 1000
feet away from the shunt and you've got them connected up with #24
AWG copper wire. That's a total length of wire of 2000 feet, and a
total resistance of 51.4 ohms, so the voltage that'll be dropped
across the wire will be:
E = IR = 3.0E-8A * 51.4R ~ 1.54E-6V = 1.54ľV,
which is:
1.54E-6V
---------- * 100 ~ 0.00051% of full scale.
3.0e-1V
the meter will have, best case, a resolution of 1 part in 2999,
which is:
100
------ = 0.0333%
2999
so you can see that the effect the length of the wire is going to
have on the system will be miniscule.
---I think I need to know either the typical current through the meter
leads or the typical voltage drop across this kind of shunt.
I don't have instruments accurate enough to measure such small
variables. My only recourse is to try different lead lengths and to
see what happens to the readings.
The readings are going to stay pretty much the same no matter what
you do, DC wise, but as the length of wire between the shunt and the
meter increases you may wind up with an antenna... :-(
--
John Fields
Professional Circuit Designer