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Autocorrecting multimeter?

Dear Newgroup,

I came across an interesting phenomenon:

In the lab we placed a "100M" Ohm resistor (whose resistance we couldn't measure directly with the multimeter) in series with a 10MOhm resistor (which we could measure). Building a basic voltage divider (3V battery) one should get 0.3V on the 10M resistor.

Now the internal resistance of both multimeters we have is 10M (Fluke

175, Amprobe 18-A), so we are placing two 10M in parallel. Doing the basics: the equivalent resistor combination is then 5M (sorry if that was insulting :) which would cause a 0.15V drop to be measured.

Instead, on the Amprobe 18-A I read the 0.3V, while on the Fluke I get

0.15V. So apparently the Amprobe somehow realizes it is changing the voltage and corrects for it.

Has anyone ever come across this before?

Any ideas on how the meter does this?

LabMonkey

Reply to
LabMonkey
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I have done such experiment.this depends on the precesion you want,and the multimeter's internal resistance.

Reply to
eehinjor

Once we know the resistance of the "100M" resistor we then place it in series with the multimeter and the battery. In essence we are replacing the known 10M resistor with the unknown internal resistance of the multimeter.

For both multimeters we are getting a votlage reading of 0.3V - which means 10M internal resistance.

As for the precision of the reading, both multimeters give us three digits after the decimal point. So it isn't a problem of rounding.

Any ideas of how to verify the internal resistance of the meters another way?

LabM> I doubt there is any autocorrecting going on - I suspect the input impedance

Reply to
LabMonkey 

--
Measure the current through them with a known voltage across them.
Reply to
John Fields 

--
    E1
    |
   [R1]
    |
    +---E2
    |
   [R2]
    |
   GND

            E1R2      3V * 10M
     E2 = ------- = ------------ = 0.2727... V
           R1+R2     100M + 10M
Reply to
John Fields

Thanks for all your help. I will try and find the parts to build the circuit and compare it with the capabilities of the meter.

Reply to
LabMonkey

I doubt there is any autocorrecting going on - I suspect the input impedance is just much higher than speced - may be spec is really "greater then 10M"

Dan

-- Dan Hollands

1120 S Creek Dr Webster NY 14580
Reply to
Dan Hollands

--- OK, but be aware that (in the meter) R1 and R2 will, in all likelihood, _not_ be 5 megohms each since the input of the ADC (ICL7135?) wants to see 200mV full scale no matter what range the meter's on.

-- John Fields Professional Circuit Designer

Reply to
John Fields

Sounds like the Amprobe actually has near-infinite resistance on its low range; that's not unheard of. It does *not* correct for source resistance!

Put the Fluke and the Amprobe in series, measure the battery, and tell us what they report.

John

Reply to
John Larkin

What figures do you get if connect both voltmeters in series across the 3V supply?

I get the feeling that atleast one of the meters isnt 10M resistance.

for measuring the "100M" resistor how about connecting it in series with the meter and a 1M resistor and measuring how much current it passes when a known voltage is applied?

Bye. Jasen

Reply to
Jasen Betts

You were right (although it is a little trickier :)

Placing a 3V battery in series with the Fluke and Amprobe meter I get

1.5V on both IF I have the Amprobe meter on a range other than mV. If instead I have the Amprobe on the mV range the votlage it displays is naturally "Ol" while on the Fluke I read 0.095V.

Doing the math I get 300M internal resistance for the Amprobe on the mV range and 10M on any other range.

Oh the joy of undocumented features.

Thanks for all the input from all of you!

Reply to
LabMonkey

Actually I spoke too soon.

To answer your first question: when having the Amprobe meter on the Volt range both meters read 1.5V which is what you get when the internal resistance are the same. If instead I place the Amprobe meter on the mV range it reads Ol and the other meter reads 0.095V instead of

1.5V. This indicates that on the mV range the Amprobe meter uses a different resistance than on the Volt range. From the initial battery voltage, the votlage drop on the Fluke meter (which was in series) and the internal resistance of the Fluke (10M) I got 0.3GOhms.

I will try out your suggestion once I find a 1MOhm resistor.

Now onto the latest problem. When I place the Amprobe meter in series with the 100M resistor and apply the 3V battery to both I get 0.3V drop displayed on the meter when it is in the Volt range. If I place it in the mV range I yet again read Ol. But having an electrometer I used that to measure the votlage drop on the 100M resistor, which is 0.2V. Solving again the simple votlage divider equation I now get out an internal resistance of 1.4GOhms for the Amprobe meter (instead of the

0.3G).

I will do it again tomorrow just to verify the result.

Reply to
LabMonkey 

--
Try this:


     3V E1
     |
   [100M]R1
     |
     +------+--E2
     |      |
    [Rx]   [Rm]
     |      |
     +------+
     |
    GND

With Rm (your meter) on the millivolt range, adjust Rx until you get
some arbitrary voltage reading (E2) on the meter. Record the value
of Rx and E2. 

Now solve for the parallel combination of Rx and Rm:


           E2 R1
     Rt = ------- 
           E1-E2


Now, knowing Rx, solve for Rm:

           Rt R1
     Rm = -------
           R1-Rt
Reply to
John Fields

The Amprobe may be "infinite" resistance with some protection diodes, so things may not be linear.

Connect a good 1 uF film capacitor across the Amprobe input, set to mV range, apply 100 mV or whatever, disconnect source, and watch it discharge. Then you can compute internal resistance.

John

Reply to
John Larkin 

--
Oops...


           Rx Rt
     Rm = -------
           Rx-Rt
Reply to
John Fields

Ran into some problems:

the meter has an auto off feature which I can't turn off (at least I couldn't find anything in the instruction manual). So after 10 minutes it turns itself off. So with the capacitor experiment this only rules out internal resistance with time constants less than 10 minutes. If the internal resistance were 300M and the capacitor was 0.9mF, then the time constant would be 4.5 minutes - which is not what I saw. After ten minutes it had decayed to 0.200mV instead of 0.230mV.

As for the other experiment, the electrometer I use to measure the voltage on the 100M resistor has some problems. I consistently got out that the experimental equivalent parallel resistance was greater than Rx, which naturally can't be.

I will try and find an electrometer which is more reliable, but as it is the weekend I will not be able to get my hands on anything until Monday.

Thanks for all of the help :)

LabMonkey

Reply to
LabMonkey

I found out how to trick the meter in staying on: quickly press the "hold" button twice and it resets the clock for the 10 minute inactivity.

Only problem is, when I did that I didn't get an internal resistance of

1.4G. The time constant would have been about 20 minutes and I had it running for 4.5 hours with it only decreasing to 160mV from 257mV.

So either the capacitors were doing something funny, or the internal resistance some how changes between even higher resistors when on the "mV" range.

I am still trying to get a hold of an electrometer

Reply to
LabMonkey

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