# Ammeter internal current draw

Discussion in 'Electronic Basics' started by Ken C, Mar 2, 2006.

2. ### John FieldsGuest

---
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.

3. ### Ken CGuest

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.

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.

Ken C

4. ### John FieldsGuest

---
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 = 1.54µV,

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.

6. ### Ken CGuest

Thank you very much. You are a gentleman and a scholar (no kidding!).

Ken C

7. ### Ken CGuest

Will I minimize the antenna effect if I use coax for the signal leads,
like RG-174? There will be RF radiation (1.8-432 mHz) in the
immediate vicinity. RF currents in conductors is a mystery to me and

Ken C

8. ### ChrisGuest

Hi, Ken. The input impedance of your ICL7107-based digital panel meter
is almost certainly 10 megohms. Mr. Fields is correct -- a longer line
might act like an antenna, picking up electrical noise. It sounds like
he's been there before. I have, too.

The ICL7107 is very good at cancelling out 60Hz noise, but less good at
other frequencies. If you've got a longer wiring length from the shunt
resistor to the DPM (say, over 6 feet or so), you've got a close source
of RF energy, or if your wiring is going through wiring duct with high
current lines (could cause inductive pickup), you'll want to provide
yourself some protection for your DPM input.

Note here that you will need a separate, floating power supply for the
DPM -- they work that way. Unless special circuitry is included on
board, the 7107 can't measure its own power supply.

Omitting the power supply, here's what your basic circuit should look
like (view in fixed font or M\$ Notepad):

|
| ___
| .---o-o---|___|---o-o----.
| | |1 milliohm | |
| | | | |
| | | _ | |
| +| | + / \ | .-.
| --- '---(DPM)---' | |
| - \_/ | |1 ohm
| 12VDC| '-'
| | |
| '------------------------'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Your "30 amp" shunt is almost certainly a 1 milliohm resistor, set so
the DPM will read 29.9 for 29.9 amps. One of the easiest ways to
reduce the effect of electrical noise is to provide a load resistor
across the DPM to directly provide a lower input impedance:

|
| ___
| .---o-o---|___|---o-o----.
| | |1 milliohm | |
| | | | |
| | | | |
| | | | |
| | | ___ | |
| | o---|___|---o |
| | | 220 ohm | |
| | | _ | |
| +| | + / \ | .-.
| --- '---(DPM)---' | |
| - \_/ | |1 ohm
| 12VDC| '-'
| | |
| '------------------------'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

This simple step helps most problems. But for higher energy
interference, you can either use protection diodes or a load capacitor
here.

| ___
| .------o-o---|___|---o-o----------.
| | |1 milliohm | |
| | | | |
| | 100 ohm| | |
| | .-. | |
| | | | | |
| | | | | |
| | '-' | |
| | | 2 X | |
| | | 1N4001 | |
| | | | |
| | o-----|<----o |
| | | | |
| | o----->|----o |
| | | _ | |
| +| | / \ | .-.
| --- '---(m V)---' | |
| - \_/ | |1 ohm
| 12VDC| '-'
| | |
| | |
| '---------------------------------'
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

The diodes here are valuable, because they limit any voltage excursion
to 600mV or so, which is overrange for the DPM, but won't cause damage.
This setup is good for inductively coupled noise, which could be
higher energy.

Here's another setup, which is probably better for "antenna-type"
noise:

| ___
| .------o-o---|___|---o-o----------.
| | |1 milliohm | |
| | .-. | |
| |220 ohm| | | |
| | | | | |
| | '-' 0.1uF | |
| | | || | |
| | o----||-----o |
| | | || | |
| | | _ | |
| +| | / \ | .-.
| --- '---(m V)---' | |
| - \_/ | |1 ohm
| 12VDC| '-'
| | |
| '---------------------------------'
|
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Make sure you use a good ceramic cap which is made to be low impedance
at your frequencies of interest. And attach the resistor right at the
DPM.

Or use a combination of these. Component selection kind of depends on
potential sources of noise, &c.

By the way, coax won't hurt, but you should be able to use plain
twisted pair for just about any application.
don't have the equipment to track down problems here. It might be best
just to build in protection, test it well, and then test it better. If
it doesn't cause problems, it's probably good.

Many times, people forget that there's another solution to this type of
problem. If you have problems with other sensitive electronics in this
installation, it might be better to go with a time-tested solution
that's impervious to RF energy and doesn't require an external power
supply -- namely, an analog meter with external shunt. Your total
installed cost will be competitive, even if you didn't have any
problems with your DPM. If you have to spend hours of fiddle-factor
time on your DPM, the analog meter comes out way ahead. Assuming
everyone who's using the meter can actually read analog meters, you can
just install it, enjoy a quick win, and move on to the next problem.

Since your DPM is on the low end of its range (you're only going to be
using 299 counts of the 2000 count full scale range), a good analog
meter would probably be just about as accurate as your DPM, too.

Good luck
Chris

9. ### Rich GriseGuest

I wouldn't use coax cable for something like this - especially since
the meter is isolated (i.e., floating) I'd personally be more comfortable
with twisted pair. But, if you have a pile of RG-174 just lying around,
then by all means, give it a try!

Good Luck!
Rich

10. ### Ralph MoweryGuest

Unless you can ground one side of the coax, it may be worse than other
methods. It will unbalance the wires as far as RF is concerned. In that
case the twisted pair would be beter if needed.

11. ### John FieldsGuest

---
If your circuit looks like this:

Vin>-------+
|
[RL]
|
+--------+
| |+
[SHUNT] [METER]
| |
GND>-------+--------+

which it should,

Then what I would do would be to decouple the meter from the coax
line like this:

Vin>-------+
|
[RL]
| R1
+----------------+--[1kR]--+---------+
| | | |+
[SHUNT] [0.1µF] [0.1µF] [METER]
| |C1 |C2 |
GND>-------+----------------+---------+---------+

Where the 1000 ohm resistor and the 0.1µF caps are mounted at the
meter end of the cable, as close to the meter as possible.

At 1.8MHz the reactance of the caps will look like:

1 1
Xc = --------- = -------------------------- ~ 1 ohm
2pi f C 6.28 * 1.8E6Hz * 1.0E-7F

So even if you wind up with something horrible like, say, 100
millivolts of RF across C1, your circuit will look like this:

0.1V
|
+------+--E1
| |
| [1000R]
| |R1
[1R] +-----+--E2
|X1 | |
| [1R] [10M]
| |X2 |R2
+------+-----+

Where X1 and X2 are the reactances of C1 and C2, and R2 is the

The [AC] voltage at E2 can be determined from:

E1 X2
E2 = ---------------
R1 + (X2||R2)

Since 1 megohm in parallel with 1 ohm is essentially 1 ohm, we have:

0.1V * 1R
E2 = ------------ = 100µV
1000R + 1R

Now, assuming that the sensitivity of your meter is 200mV full scale
and is a 2000 count meter means that its resolution will be:

200mV
------------- = 100µV
2000 counts

Voila! we have reached the point where 100mV of RF at 1.8MHz on the
center conductor of the cable could cause a reading error of 1 LSD
on the meter.

As frequency increases, the reactance of the caps will decrease,
decreasing the amount of RF presemt at the meter, further decreasing

One thing: Be sure that the caps you get have a self-resonant
frequency (SRF) higher than 432MHz, asince if its lower than that
they'll start looking inductive and could cause trouble.

Finally, what about that 1000 ohm resistor and the DC drop across
it? OK, Here's that circuit:

Vin E1
|
[1000R]
|
+-----E2
|
[10MR]
|
GND

Since:

E1 R2
E2 = ---------,
R1 + R2

if we set E1 to 1V we'll have:

1V * 1E6R
E2 = ------------- = 0.999V
1E3R + 1E6R

which is an error of:

1.0V - 0.999V
--------------- * 100 = 0.1%
1.0V

of the reading, which isn't bad and, since the meter accuracy is +/-
0.5% should be fine.