Specification of current reading on multimeter

I am in the UK. I have a £20 digital multimeter and would like some advice. The meter is a Maplin model PG10B (order code GW18U)

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(Q1) Under "DC CURRENT" my user guide says, "voltage drop: 200 mV". Is this the disturbance in a circuit when measuring DC current?

(Q2) If so, then does that 200 mV figure mean that the potential difference for the rest of the circuit will be reduced by 200 mV once I have interrupted the circuit and inserted my meter for DC Current measurement?

(Q3) Isn't that figure of 200 mV rather high? I mean, if I use the 2 mA or 20 mA scale in an electronics circuit powered by a 1.5 volt cell then a 200 mV drop in the circuit is very significant.


My real need is to measure current while charging a AAA NiMH cell.

When I set my multimeter to measure DC Current and insert it in the circuit, the charger's light fails to come on!

It's as if the prescence of the meter is disturbing the circuit too much.

The voltage of my uncharged cell might typically be 1.1 volts and the charger says it delivers up to about 120 mA.

(Q4) Is my multimeter simply too low-spec to measure current in this application?


Reply to
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** Yep.

** Yep.
** Nope.

** Could be.

** Could very well be.

** Yep.

Maybe try using a Hall effect sensor - no voltage drop involved at all there.

Or just be a tad smarter bear and measure the current at a point in the circuit that is NOT so damn sensitive to small voltage drops.

.... Phil

Reply to
Phil Allison

Yep, basically the thing is operating as a voltmeter with 200mV full scale, measuring the voltage dropped across an internal resistor in series with the meter terminals.

Only when the meter is reading full scale. If you use the meter on its

10A range to "measure" 10mA, the voltage drop will only be 200uV... However, that isn't exactly useful, from the measurement pov.

The lowest dc voltage range of the meter is 200mV - which is where the

200mV drop when measuring current comes from.

Not when on the 10A range.. :)

Yep. However, add a little opamp x50 amplifier, a suitable series resistor and use it as a voltmeter monitoring the opamp output and it will be fine.

-- Sue

Reply to

Yes, at full scale. (You didn't say what the range was.)

The current ranges are probably implemented by switching the meter to its

200mV range and shunting the meter with a resistor that will produce a 200mV drop for that current range.

So, for example, the 10A range uses a 20 milliohm shunt, while the 2mA range uses a 100 ohm shunt. The actual drop varies linearly with the current, of course.

How much this disturbs the circuit depends on the circuit's impedance. If the charging light comes on when you're not using the meter, but doesn't come on when you are, then, by definition, you're "disturbing" the circuit.

If you charge a lot of nicad and NiMH cells, get a MAHA (Powerex) MH-C9000 charger. Not cheap, but very convenient.

Reply to
William Sommerwerck

It is the volt drop across the meter at "full scale deflection".

If you measure 2mA on the 2mA range it will cause a 200mV drop. If you measure 2mA on the 20mA range it will cause 20mV drop.

Depends on the current you are measuring and the range you have selected - see above

200mV is a common figure for the DVM chips used in such meters. The chips measure voltage. A current is measured by passing it through a resistor (shunt) within the meter and measuring the volt drop across it
Reply to


Only at full scale reading. Adjust pro-rata.

Pretty average.


No, but you could measure the current further 'upstream' if possible to avoid the problem.


Reply to

This probably means that that the 200 mV corresponds to the drop across the meter required to give a full reading. If you do not need accurate current measurement, increasing the current range setting will reduce the voltage drop for the same current.

Find the impedance of the meter by dividing the 200 mV drop by the full scale current. Use that to cvalculate the drop for measured current. Subtract that off of the applied voltage. That is then what is left for the rest of the circuit.

Not necessarily. Higher resistance may be a way to increase sensitivity for a low cost,

Could be.

Because electronic meters can vary all over the place, there can be all kinds of peculiar nonlinear behavior that MAY kick in. That is not likely, but possible.


Reply to
Salmon Egg


As Bill mentioned, switching to a higher current range may lower the meter's resistance to the point where the charger's light comes on.

Even though the charger's light fails to come on, do you read any current at all on the meter? If not, you may have blown a fuse that protects the meter's current ranges. If the fuse is only in the current circuit, the meter may function normally for all other functions.


Reply to
Fred McKenzie

On Sat 03 Jan 14:16, Stuart wrote

Oh! Now that isn't what I was expecting.

I read what others have posted about the meter being based arounf a voltmeter which reads 200 mV at full scale.

I had inferred that the full scale deflection reading was dependent on the range chosen.

So, for example, the full scale reading on a 20 mA range would be

20 mA. (On the 2mA range it would be 2 mA.) And I then had the impression that when reading almost 20 mA on the 20 mA range, I would get the 200 mV drop mentioned in the specs.

Is that understanding wrong?


Reply to

No, but you need to re-read what the PP wrote.

When on the 20mA range, a current of 2mA would only cause a 20mV drop, as the meter would only be reading a tenth of full scale. As the current increases, on the same range, so will the volt drop - when 20mA is reached, the voltage drop will be 200mV.

When on the 2mA range, a current of 2mA will cause a full scale reading and a voltage drop of 200mV.

-- Sue

Reply to

No - it is correct.

The meter has a voltage drop of 200 mV at full scale - internally, it is a 200 mV full scale meter.

If you measure a current of 2 mA on the 2 mA scale, the meter will have a 200 mV voltage drop.

If you switch to the 20 mA scale, that 2 mA current will result in a

20 mV drop.

If you switch to the 200 mA scale, the 2 mA current will result in a 2 mV drop.

For any scale and current, the meter's voltage drop is: (current/full_scale) * 200 mV

Reply to
Peter Bennett

One way round this is to use a stable adjustable PSU and adjust for the required voltage *after* the current meter. If your circuit draws large pulses, pad the supply rails with a hefty electrolytic.

Reply to
ian field

Ask in store to be put on the mailing list and wait for a special offer better spec meter to be discounted.

Better yet look elsewhere for a better quality instrument and grin & bear the price, the last meter I bought from Maplin a couple of years ago worked ok for a while then some of the segments in the display failed, this turned out to be migrating flux on the LCD contact pads so I was able to fix it reasonably easily, but this seems to be a bit of a trend with stuff I've bought from Maplin.

Reply to
ian field

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