| Author |
Message |
LabMonkey
Guest
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 Posted:
Tue Nov 15, 2005 9:35 am Post subject:
Autocorrecting multimeter? |
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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
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eehinjor
Guest
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| Posted:
Tue Nov 15, 2005 9:35 am Post subject:
Re: Autocorrecting multimeter? |
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I have done such experiment.this depends on the precesion you want,and
the multimeter's internal resistance. |
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Dan Hollands
Guest
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| Posted:
Tue Nov 15, 2005 5:35 pm Post subject:
Re: Autocorrecting multimeter? |
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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
585-872-2606
QuickScore@USSailing.net
www.QuickScoreRace.com
"LabMonkey" <bbiritz@gmail.com> wrote in message
news:1132035270.371978.34860@g44g2000cwa.googlegroups.com...
| Quote: | 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
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LabMonkey
Guest
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| Posted:
Tue Nov 15, 2005 5:35 pm Post subject:
Re: Autocorrecting multimeter? |
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|
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?
LabMonkey
Dan Hollands wrote:
| Quote: | 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
585-872-2606
QuickScore@USSailing.net
www.QuickScoreRace.com
"LabMonkey" <bbiritz@gmail.com> wrote in message
news:1132035270.371978.34860@g44g2000cwa.googlegroups.com...
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
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John Fields
Guest
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| Posted:
Tue Nov 15, 2005 11:53 pm Post subject:
Re: Autocorrecting multimeter? |
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On 15 Nov 2005 07:27:52 -0800, "LabMonkey" <bbiritz@gmail.com>
wrote:
| Quote: | 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?
|
---
Measure the current through them with a known voltage across them.
--
John Fields
Professional Circuit Designer |
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John Fields
Guest
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| Posted:
Wed Nov 16, 2005 12:14 am Post subject:
Re: Autocorrecting multimeter? |
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On 14 Nov 2005 22:14:30 -0800, "LabMonkey" <bbiritz@gmail.com>
wrote:
| Quote: | 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.
|
---
E1
|
[R1]
|
+---E2
|
[R2]
|
GND
E1R2 3V * 10M
E2 = ------- = ------------ = 0.2727... V
R1+R2 100M + 10M
---
| Quote: | 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.
|
E1
|
[R1]
|
+---E2
|
[R2]
|
GND
E1R2 3V * 5M
E2 = ------- = ------------ ~ 0.1428V
R1+R2 100M + 5M
| Quote: | 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?
|
---
Dunno, but I haven't.
---
| Quote: | Any ideas on how the meter does this?
|
---
Nope, but if it was me I'd have it do something like this:
Vbat
|
Vin+<---+ |
| |
[R1] [VR1]
| |
E1--+--[R3]--+--E2
| |
[R2] [R4]
| |
Vin-<---+--------+
<--I2
Where R1 and R2 are 5 megohms each, giving me the 10 megohms input
resistance I need for the meter, but upset by R3 + R4 being in
parallel with R2. After connecting to Vin+ and Vin- and and
detecting current in R3 and R4 I'd turn on VR1 (a FET, probably)
and ramp it through its resistance range until the current in R3
went to zero (or until the currents in VR1 and R4 were equal). At
that point the bridge would be balanced, the input resistance would
be R1 + R2, and I'd measure the current in R4 (or VR1) and
calculate:
E2 = I2R4 = E1
--
John Fields
Professional Circuit Designer |
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LabMonkey
Guest
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| Posted:
Wed Nov 16, 2005 1:35 am Post subject:
Re: Autocorrecting multimeter? |
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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. |
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John Fields
Guest
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| Posted:
Wed Nov 16, 2005 1:35 am Post subject:
Re: Autocorrecting multimeter? |
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On 15 Nov 2005 13:36:54 -0800, "LabMonkey" <bbiritz@gmail.com>
wrote:
| Quote: | 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.
|
---
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 |
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John Larkin
Guest
|
| Posted:
Wed Nov 16, 2005 9:35 am Post subject:
Re: Autocorrecting multimeter? |
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On 14 Nov 2005 22:14:30 -0800, "LabMonkey" <bbiritz@gmail.com> wrote:
| Quote: | 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
|
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 |
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Jasen Betts
Guest
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| Posted:
Wed Nov 16, 2005 3:20 pm Post subject:
Re: Autocorrecting multimeter? |
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On 2005-11-15, LabMonkey <bbiritz@gmail.com> wrote:
| Quote: | 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.
|
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 |
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LabMonkey
Guest
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| Posted:
Thu Nov 17, 2005 9:07 am Post subject:
Re: Autocorrecting multimeter? |
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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! |
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LabMonkey
Guest
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| Posted:
Fri Nov 18, 2005 9:35 am Post subject:
Re: Autocorrecting multimeter? |
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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. |
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John Fields
Guest
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| Posted:
Fri Nov 18, 2005 5:35 pm Post subject:
Re: Autocorrecting multimeter? |
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On Fri, 18 Nov 2005 06:35:40 -0600, John Fields
<jfields@austininstruments.com> wrote:
| Quote: | Try this:
3V E1
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[100M]R1
|
+------+--E2
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[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
|
---
Oops...
Rx Rt
Rm = -------
Rx-Rt
--
John Fields
Professional Circuit Designer |
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John Larkin
Guest
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| Posted:
Fri Nov 18, 2005 5:35 pm Post subject:
Re: Autocorrecting multimeter? |
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On 17 Nov 2005 23:02:24 -0800, "LabMonkey" <bbiritz@gmail.com> wrote:
| Quote: | 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.
|
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 |
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John Fields
Guest
|
| Posted:
Fri Nov 18, 2005 5:35 pm Post subject:
Re: Autocorrecting multimeter? |
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|
On 17 Nov 2005 23:02:24 -0800, "LabMonkey" <bbiritz@gmail.com>
wrote:
| Quote: | 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.
|
---
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
--
John Fields
Professional Circuit Designer |
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