I was trying to measure the DC current of a circuit by inserting a DMM in series with the positive 3.3v supply terminal so that the circuit's positive supply was being fed from the DMM lead instead of the bench power supply.

I noticed the voltage was only 2.9 volts coming out of the meter while measuring the 220mA of current, so my circuit was not receiving the full 3.3 volts.

I duplicated the observation by trying another different type of DMM as well as replacing my circuit with just a load resistor to draw 220mA and there is always a loss of up to half a volt when I insert the DMM.

Is this normal and if so, is this current measurement even valid because my circuit is only getting 2.9 volts instead of 3.3 and its operation is impacted by supply voltage (RF transmitter).

How do I properly measure the current?

 

I can only speak to my own experience (electronics is a hobby). On the larger ranges, with DMMs that I've inspected internally, there is a wire shunt for the 10 amp or 20 amp range. With an arrangement like this one would expect there to be a voltage drop as the shunt is in series with the supply to the device under test.

__________________
stevez

 

Hi,


Various DMMs have different internal resistances for measuring current.
The resistance is put in series with the load so there is always some
voltage drop.
The way around this is to use your own 'current shunt', or buy one.
You can get away with a 0.1 ohm resistor sometimes, or a 0.01 ohm
resistor. You measure the voltage drop and calculate the current using
the formula:
I=V/R
where
V is the voltage drop
R is the resistance of the resistor or shunt
I is the measured current

They sell shunts on the web if you want to go that way. They are
calibrated.
You can also make your own shunt with a brass strip from a hobby shop.
You have to calibrate it yourself though.

 

My Fluke 87 would drop 0.396V when measuring 220mA.

 

To minimize the voltage drop when measuring current with a multimeter, use a higher range such as the 10A range, even for low current measurements. The internal meter shunts typically have a resistance which drops 200mV full scale for a meter with a typical minimum full scale voltage sensitivity of 200mV. Thus the 10A shunt typically has a resistance of 10mΩ (the meter measures 100mV across the shunt with 10A through it). This would reduce the 220mA voltage drop to 2.2mV, for example. Of course the current measurement resolution and accuracy is not as good as the lower ranges but it does reduce the voltage drop from the shunt.

Another way to compensate, while doing the current measurement, is to readjust the supply voltage to give the correct voltage at the circuit. Remember to adjust the voltage back to its normal value after removing the current meter.

__________________
Carl
Curmudgeon Elektroniker

 

I thought about increasing the supply voltage until I get 3.3v again restored at the circuit's + rail, but I wondered is the current reading going to tell me what current I'm getting "at 3.3 volts" in that case, since the main supply to the meter is of course higher...

 

Hi again,


Yes that works but you do risk overpowering the circuit a little.
This may or may not cause damage if the circuit draws a pulsed
current instead of continuous. It's not much, so it's only a little
risk, but a risk just the same.

You dont have any small value resistors laying around?

 

The only one I have is a 1 ohm that measures 1.6 ohms (assuming that measurement from the same Fluke meter is good).

For now I'll just take the measurements at face value from the meter as I've always done. I did a test by saying if each DMM causes a drop that I fear may interfere with my current measurement, let me see if I put 2 DMMs in series do I get an increased error.
I saw about a 2% drop in the current reading of the fluke when I added another cheap current meter in series (each meter on its own caused the 0.4v drop) so it looks like my best bet for now is just take the readings and not worry about how it works.

The meters wouldn't be very useful if they didn't give good readings at low scale so I have to hope I'm just thinking about it too much.

 

If you set the circuit voltage to 3.3V then the current will be an accurate value of the circuit current at 3.3V. The excess voltage and power is being dropped in the multimeter shunt resistor.

__________________
Carl
Curmudgeon Elektroniker

 

Most of that 0.6 ohms is probably in the meter leads.

Short the probe leads together and see what the resistance is. Then subtract that value from the measured resistor value to get the actual resistor value.

__________________
Carl
Curmudgeon Elektroniker

 

Hi again,

Oh i see, ok, well you know you can get resistors at quite a few
places that sell electronic parts, or online in several places.

Another idea is to use linear interpolation...
With this, if you measure 220ma with 2.9v then you will measure
3.3/2.9*220 or about 250ma if the voltage was really 3.3v as it
normally has. This is a good approximation if the circuit load
behaves as a linear load, and probably not too bad even if it doesnt,
unless of course it draws constant power in which case it will draw
less than 220ma. If you put two meters in series and the current
drops even lower than the 250ma estimate probably isnt too far off.

There's also a 'compensated shunt' which maintains the same voltage,
but you probably dont need that much accuracy anyway.

So to recap:
If you measure 220ma at 2.9v when it should be 3.3v then the corrected
current value is probably close to 250ma.