Re: Autocorrecting multimeter?
- From: John Fields <jfields@xxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 18 Nov 2005 06:35:40 -0600
On 17 Nov 2005 23:02:24 -0800, "LabMonkey" <bbiritz@xxxxxxxxx>
wrote:
>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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