Multimeter's ammeter accuracy

[lynx]

Hi

in a simple circuit consisting of an exactly measured 100Ohm/5W resistor as a load
(double and triple checked with different multimeters) a stabilized power source and two
multimeters (one continuesly monitoring the voltage and the other measuring the current passing
through the resistor) at various voltages raging from 1.5VDC to 3.3VDC i've noticed that my calculations
for current had nothing to do with what the multimeter measures, there is a 18%difference.

then i got 5-6 different multimeters from different brands not expensive models but
all common ones, analog and digital and i've also noticed the same thing when measuring current... some
where more accurate and some other less accurate but none of them was less than 10%

seems the problem only exists when measuring current..while resistor and voltage measurment appears
to be very accurate i don't know if such thing is reasonable...

 

[Reloadron]

You are measuring the current placing the meter in series with and not across the load correct?

<EDIT> Like the attached image? </EDIT>

Ron

 

[JimB]

Also, you are leaving the ammeter in circuit while you measure the voltage aren't you?

If you take the ammeter in/out of circuit, that can change the current and you will get odd results.

JimB

 

[Ziddik]

i have also noticed those problems, and still don't know why is the circuit gets short circuited when measuring current

 

[lynx]

You are measuring the current placing the meter in series with and not across the load correct?

<EDIT> Like the attached image? </EDIT>

Ron

yes i do..

Also, you are leaving the ammeter in circuit while you measure the voltage aren't you?

If you take the ammeter in/out of circuit, that can change the current and you will get odd results.

JimB

yes...i continuesly monitor the voltage, also i think the accuracy gets
worse when using lower value resistors at the same voltage... (i tried that
too but i'm not sure yet if it is true..)

 

[audioguru]

A Current meter has a resistance. When it is in series with a load then its series resistance reduces the current.

 

[audioguru]

i have also noticed those problems, and still don't know why is the circuit gets short circuited when measuring current

A current meter is supposed to be connected in series with the load. It shorts the load (or power suopply) if it is parallel.

 

[lynx]

A Current meter has a resistance. When it is in series with a load then its series resistance reduces the current.

yes but in a modern digital multimeter it's internal resistance at which the current get measued it should be taken into account by it's micro controller when it calculates
the measurement result.. right?

also in analog multimeter i think it is already calculated when the instrument's
tablet gets printed by the factory...

 

[Nigel Goodwin]

yes but in a modern digital multimeter it's internal resistance at which the current get measued it should be taken into account by it's micro controller when it calculates
the measurement result.. right?

also in analog multimeter i think it is already calculated when the instrument's
tablet gets printed by the factory...

No to both, inserting the meter increases the circuit resistance, and thus reduces the current somewhat - the meter will read what the current IS, not what it might possibly have been if the meter wasn't there (which would be impossible).

To recheck what others have asked, and to which your answers haven't really been clear enough, you do have TWO meters connected at all times? - an ammeter in series with the load, and a voltage directly across the load (and only the load).

Incidently, this check is how you would accurately measure the resistance of the load.

 

[audioguru]

My fairly expensive Fluke digital multimeter calls the voltage loss caused by its current-sensing resistors its "burden voltage". It is 11mV/mA for low currents up to 40mA and is 30mV/A for high currents up to 10A. It can measure up to 20A for 30 seconds max.

EDIT: I forgot to say that the test leads and probes add additional resistance that reduces the current.

 

[Reloadron]

**** Ron raises his hand with another question****

When you measure the voltage exactly where are you measuring it? Upstream of the current meter at the actual source or downstream of the current meter as in right across the load? Try measuring the voltage both ways, across the source and across the load, let's see if there is a difference?

Ron

 

[lynx]

the meter will read what the current IS, not what it might possibly have been if the meter wasn't there (which would be impossible).

i've noticed one of my digital multimeters at the mA range uses 2 smd 1206 resistors in parallel which make aproximately an 8ohm load isn't it possible the MCU to somehow exlude
the value of the resistors from the calculation of the current??.

To recheck what others have asked, and to which your answers haven't really been clear enough, you do have TWO meters connected at all times? - an ammeter in series with the load, and a voltage directly across the load (and only the load).

the voltage meter initialy wasn't connected, but after few measurements
of the current i needed to know if the power supply is steady enough while
i'm writting down the current measurment.

**** Ron raises his hand with another question****

When you measure the voltage exactly where are you measuring it? Upstream of the current meter at the actual source or downstream of the current meter as in right across the load? Try measuring the voltage both ways, across the source and across the load, let's see if there is a difference?

Ron

i think i had the probes placed correctly.. at the power supply

Audioguru said something about probe's resistance...that's something i didn't hand in mind when measuring, but that's negligible since the probes add less than an ohm
to the circuit.. right?

 

[crutschow]

i have also noticed those problems, and still don't know why is the circuit gets short circuited when measuring current

To measure current the ammeter must be in series with the load. If you connect the ammeter in parallel with the load, the obviously you will short the load.

 

[crutschow]

All ammeters have some resistance with the resistance being higher when measuring smaller currents. This resistance will affect the load current, depending upon the relative value of the ammeter resistance to the load resistance.

You can place a voltmeter across the ammeter terminals to measure this voltage drop and compensate for that in your calculations.

 

[picbits]

As above - I have a meter with a 10A range down to a 40ma range. The 10A range puts very little resistance in series with the load while the 40ma range puts quite a lot of extra resistance in series with the load.

If you are measuring a 100R resistor with a 10 volt regulated supply, a 0.1R shunt ammeter will give you a much different reading than say a 10R shunt ammeter.

 

[Reloadron]

i think i had the probes placed correctly.. at the power supply

No what I asked you to try (think experiment here) is measure the voltage both at the source and at the load. Think about what AG mentioned about the burden resistance of the ammeter, also if you look back at the image I posted the voltage is measured downstream of the ammeter. There is actually a lesson in this and the numbers will work out correctly. Measure the voltage across the supply and then the load and note the difference.

Ron

 

[audioguru]

An amp-meter is a resistance in series with the load so it reduces the current because it also reduces the voltage at the load (due to Ohm's Law).

 

[KeepItSimpleStupid]

OK guys. You must measure the VOLTAGE across the device and the CURRENT THRU the device, that way the resistances of the ammeter, voltmeter and leads are compensated for, for low values of R. Measuring high resistances > 1 M requires different techniques such as a feedback ammeter.

Ever seen a 4-terminal resistor? They exist.

Setup

The ammeter goes in series with the power supply. The voltmeter goes across the resistor leads. If the ammeter has a resistance of 10 ohms, the current is the same through the resistor and the ammeter. If the voltmeter has a resistance of 10 Meg ohms then 1K || 10 Meg is negligible. || is read as "in parallel with".

There are two types of ammeters and they are SHUNT and FEEDBACK. Feedback ammeters have a nearly constant voltage drop below 1 mV or so is common.

Usually there is a max voltage burden listed for a shunt ammeter. Typical values are below 0.6 V because diodes are used for protection.

 

[lynx]

the problem got corrected by measuring the voltage across the resistor... exactly as
it mentioned above and not at the power supply, now the results from most multimeters
match the result from my calculations.

it was so simple...

thanks!

 

[KeepItSimpleStupid]

Now try it with a 10 Meg resistor. All bets are now off.