# current measurement

Discussion in 'Electronic Design' started by Aman Singh Bindra, Sep 22, 2004.

1. ### Aman Singh BindraGuest

I have a circuit which has a 48V supply and a water conductivity cell
having a resistance. The current in the circuit is of the range 0-2
mA. I want to measure the current in this circuit accurately. The
easiest solution is to measure the current across a resistor by adding
a shunt in series with the 48V battery. But the resistor may change a
little bit with time and temperature. Can somebody recommend a
solution for measuring current at this high voltage of 48 V. I have
gone through some off the shelf IC's but they are not for this voltage
range. I need a reliable solution for the same. Can somebody help.

2. ### Mike MonettGuest

Normally, you would avoid the problem and place the sense circuit in the
negative lead of the cell. But if you want to measure water conductivity
accurately, you can't use DC.

The trace cations and anions in the solution will migrate to their
respective electrodes and form a Nernst diffusion layer. The readings
will be highly erratic and useless for anaysis. Use an instrument
designed for this, such as a Hanna

http://www.hannainst.com/index.cfm

The conductivity seems strange. At 48V and 2mA, it is about 42
microSiemens. The actual concentration of ions depends on the wetted area
of the elecrodes, which you didn't state. But this seems very high for
pure water applications, and very low for normal water processing.

What is your application? Are you doing electrolysis, and if so, what is
the electrode material?

Best Wishes,

Mike Monett

3. ### CFoley1064Guest

Subject: current measurement
The high voltage is not developed across the resistor if it's a shunt. Your
voltage will be proportional to the current across it (Ohm's Law, V = I * R).

The most precise way to measure DC current is with a wirewound shunt resistor.
If you want to minimize self-heating, you just make the resistor a large
wattage one. You're right in saying the resistance will change with
self-heating and changes in ambient temperature, but the change really won't be
that much (less than 1 mW power dissipation at 2mA), and it will be far less
than other methods. If you use an Ohmite 45F100 100 ohm, 5 Watt, 1% resistor
(Mouser 588-45F-100, \$1.14 ea. in single quantities), you're guaranteed a 20ppm
per degree C temperature coefficient. That's pretty good. You'll have 0 -
200mV developed across the resistor, ready to go.

If there's a potential for a short circuit, that could damage your resistor.
Use a fast blow 1/8A fuse in series with the load.

http://www.mouser.com/

Good luck
Chris

4. ### TweetldeeGuest

The problem is not the voltage across the resistor, but the common-mode
voltage seen by the opamps that handle the voltage drop across the shunt.
At 48 volts, that's way out of the range of available opamps without
resorting to more complex design.

A previous post had a more sensible approach.. put the sense resistor in
the low side of the loop. That has its own set of problems, but are more
easily solved than sensing in the high side.
--
Dave M
MasonDG44 at comcast dot net (Just subsitute the appropriate characters in

Never take a laxative and a sleeping pill at the same time!!

5. ### CFoley1064Guest

Subject: Re: current measurement
Hi, Dave. I thought that's what I meant (view in fixed font or M\$ Notepad):

.--------------.
| |
| .------o-----.
| | |
- | |
| '------o-----'
| |
| ,-.
| 3AG |||
| 1/8A|||
| '-'
| | ___
| o-|___|-- >
| | To
| .-. OpAmp
| 100R| |
| 1% | |
| '-'
| |
'--------------o
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

I guess I could have been a little more clear. Thanks.

Chris

6. ### Tony WilliamsGuest

Something like this maybe.

+-|>|-|>|-|>|------------+
| D1 D2 D3 R1,270 |
+48V-----+-----+---------------+--/\/\--+---->0-2mA
| | | 0.1% |
\_|_ \ | |
12V /_\ /R2,270,0.1% | |
| \ | \
| +------------+--------+ /1k
| | | | | \
| |s === | _ | |
| +--| | |/-|--+ |
| ptype||--/\/\--+--< | |
| +--| |\+|-----+
| |d | TL061,
| |
\ \
3k3,1W / /8k2
\ \
| |
+---------------+
| | |
| | +--------+-----> Vout, 1V/mA
\_|_ | | _ | |
5v6 /_\ | +--|-\| |
| | | >--+
| +------|+/| |
| | | |
| \ | \
| /R3,1k | /1k
| \0.1% | \
| | | |
0v------+-----+---------+--+------------

As discussed in a previous thread, the TL061/071 CMV
includes the +ve supply rail. So it can be used to
sense supply rail current and send a matching current
down to a resistor sitting on 0v. The voltage across
that resistor is buffered by a 1:1 single rail opamp.

7. ### Tom SeimGuest

You didn't say how accurately (makes a big difference). There are
op-amps that handle a high common mode voltage (TI/Burr-Brown). If
that isn't good enough you can have your circuitry common elevated to
35-40V, convert the signal with an ADC and couple the output with an
opto (serial ADCs are ideal for this).

Tom

8. ### JeffGuest

How about a switched capacitor OP amp?

9. ### TonyGuest

Is there a new switched capacitor opamp topology that extends input
common mode range?

Tony (remove the "_" to reply by email)