Multimeter Question

[raphaelriv]

I have 3 multimeters here and I was trying to read current from a power supply to a mother board, so i noticed all 3 were giving me different readings.

I take a closer look at them and their specs

**broken link removed**

http://www.tek.com/Measurement/cgi-...ts/catalog/tx3tx1/eng/index.html&FrameSet=mbd

http://www.keithley.com/data?asset=359

i know my craftsman 19dollar meter is not as reliable as the keithley 2000 or the Tektronix multimeter but what is the key spec to look at that is making the difference?

my scenario is

craftsman meter is reading: 1.4Amps
Tektronix meter is reading: 348mA
Keithley meter is reading: 550mA

I noticed that using the tektronix meter by playing with the range button that display it reads 1.4Amps the same way as the craftsman meter... but when i increase the range to 500 from 10 the current reading goes to 348mA this makes me believe that is an accuracy range and the 348mA is a better reading, this is what i was trying to look for in the specs sheets for all three meters but i get lost in the information could anyone help?

regards,

el rookie

 

[Optikon]

I have 3 multimeters here and I was trying to read current from a power supply to a mother board, so i noticed all 3 were giving me different readings.

I take a closer look at them and their specs

**broken link removed**

http://www.tek.com/Measurement/cgi-bin/framed.pl?Document=/Measurement/Products/catalog/tx3tx1/eng/index.html&FrameSet=mbd

http://www.keithley.com/data?asset=359

i know my craftsman 19dollar meter is not as reliable as the keithley 2000 or the Tektronix multimeter but what is the key spec to look at that is making the difference?

my scenario is

craftsman meter is reading: 1.4Amps
Tektronix meter is reading: 348mA
Keithley meter is reading: 550mA

I noticed that using the tektronix meter by playing with the range button that display it reads 1.4Amps the same way as the craftsman meter... but when i increase the range to 500 from 10 the current reading goes to 348mA this makes me believe that is an accuracy range and the 348mA is a better reading, this is what i was trying to look for in the specs sheets for all three meters but i get lost in the information could anyone help?

regards,

el rookie

Just look at the accuracy specs. In this case for a DC current measurement (DCI). Once you figure out what the accuracy of the measurement is, that will be your uncertainty in the real value.

 

[Optikon]

The tek spec says this under DC current:

±(0.2% + 2 ct.).

So I take that as +/- (0.2% of range + 2 counts)

You are on the 500mA range for the measurement of 348mA.

SO, there is 0.01 mA per count on the 500mA range (in 50,000 ct mode)
(500/50000)

Your accuracy for this measurment is +/-( (.002)*500 + 2*(0.01) ) in mA
That works out to +/- 1.02 mA

So the real answer could be anywhere between 346.98 to 349.02 mA

It is very important the unit still have a valid calibration! When you see different measurements for the same quantity, then one or more of the meters is out of calibration or operating outside of the specified accuracy. Double check the conditions that the accuracies are specified for and make sure you meet those. Also take into accound any special footnotes or the like.

BTW, this calculation is for DC current. If your current has an AC component to it, the calculation chages.
If you use the AC current range for the measurement, it likely is giving you the RMS value. And that may not match the DC current value measurement.

 

[Hero999]

Are you sure you've connected them in seies with the load and not made the typical nube error of connecting them in parallel?

 

[audioguru]

They use a resistor in series with the load to measure the voltage drop then calculate the current. The resistance of the meter's leads are also in series with the load. So the total load resistance is the load in series with the measuring resistance and in series with the leads resistance so of course the reading is too low.

If the power supply is 100V then a load resistance of 10 ohms will be 10A. Then the measuring resistance and the resistance of the leads is neglegible.

 

[Optikon]

They use a resistor in series with the load to measure the voltage drop then calculate the current. The resistance of the meter's leads are also in series with the load. So the total load resistance is the load in series with the measuring resistance and in series with the leads resistance so of course the reading is too low.

If the power supply is 100V then a load resistance of 10 ohms will be 10A. Then the measuring resistance and the resistance of the leads is neglegible.

Almost right. The resistance of the leads does not add to the resistance of the amps shunt inside the meter. The voltage is measured right across the shunt and so it does not include the voltage dropped by the leads.

The voltage dropped by the leads does however degrade a specification called the voltage burden. Ideally, an Ammeter would have zero volts dropped across it. In reality, it is the voltage drop in the leads and shunt that render a non-zero drop and thus a voltage burden.

 

[Hero999]

Just use a hall effect sensor and you don't get any voltage drop.

 

[Optikon]

Just use a hall effect sensor and you don't get any voltage drop.

And you dont get any good accuracy either. The OP seemed interested/lost in the accuracies of the measurement.

 

[audioguru]

He didn't say, but assume he is trying to measure the current through a 2.5 ohm resistor connected to a 5V supply. Of course the current should be 2.0A.

His meters have a 200mV range and a 2A range so their shunt is 0.1 ohms.
The meter's leads measure 0.3 ohms. So the 2.5 ohm load is in series with 0.40 ohms and the current measures 1.72A instead of 2.5A.

 

[Optikon]

He didn't say, but assume he is trying to measure the current through a 2.5 ohm resistor connected to a 5V supply. Of course the current should be 2.0A.

His meters have a 200mV range and a 2A range so their shunt is 0.1 ohms.
The meter's leads measure 0.3 ohms. So the 2.5 ohm load is in series with 0.40 ohms and the current measures 1.72A instead of 2.5A.

He didn't say but he gave no indication of using an external shunt to measure the current. One can only assume he is using the meters current function to measure the current since all the meters he listed have that feature (why do otherwise?)

That being the case, the meter lead resistance is not part of his accuracy problem. Since he did mention his confusion with the datasheets, it seemed he was trying to figure out accuracy of current measurement.

But it would be great if he would post back to clear up all the speculation.

 

[audioguru]

The resistance of the shunt in the meter is in series with the load, reducing the current.
The resistance of the meter's leads is also in series with the load, also reducing the current.

A current meter with a shunt and resistive leads is a poor way to measure current when the supply voltage is low.

 

[raphaelriv]

He didn't say but he gave no indication of using an external shunt to measure the current. One can only assume he is using the meters current function to measure the current since all the meters he listed have that feature (why do otherwise?)

That being the case, the meter lead resistance is not part of his accuracy problem. Since he did mention his confusion with the datasheets, it seemed he was trying to figure out accuracy of current measurement.

But it would be great if he would post back to clear up all the speculation.

this is correct thanks for the responses!, I used the meter's current function. I just cut the the + wire and hooked a meter in series.

 

[mcs51mc]

That's what calibration labs are for

Why not testing them with a known load instead of motherboard?
Or even better use a current source, like the Keithley 2400 for example?

 

[Rodney2]

To simply compare meters, hook them all in series and drive a current thru the string with an external power supply. If you have an accurate voltmeter, also put a precision resistor in series with the meter and measure the voltage drop across the series resistor so you know the value of the current being sent thru the loop.

 

[Leftyretro]

Compairing meter readings reminds me of the saying;

Man with one watch always knows the time,
Man with two watches never quite sure..........

Lefty

 

[mcs51mc]

That's why the the radio controlled watches where invented
One or more watches, always same time