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V DC
When the power supply resistance level is high, it is necessary to
adjust the reading according to the following formula:
Actual voltage (in V) =
display voltage (V) x (1 + power source resistance (ohms)) / 10
[\\quote]
There must be a mistake: its input impedance is 10M, so the \"power
supply resistance\" would have to be of that order to affect the reading,
yes? They must mean a source resistance in M. And then greater than 10M.
Or am I just confused?
BTW - the meter is very nicely made. It\'s in their \"Bouncer\" line & the
case is a thick rubbery plastic. Made to bounce? Age unknown, but
PCB\'s are hand laid out & it uses a Maxim ICL7136 LCD driver that dates
from 1996.
I got an AEMC model 9100 at a swap meet. Google found nothing on it,
but I called the AEMC number and they dug into the paper files and
scanned a User Manual for me.
The instructions for DC voltage had the following:
V DC
When the power supply resistance level is high, it is necessary to
adjust the reading according to the following formula:
Actual voltage (in V) =
display voltage (V) x (1 + power source resistance (ohms)) / 10
[\\quote]
There must be a mistake: its input impedance is 10M, so the \"power
supply resistance\" would have to be of that order to affect the reading,
yes? They must mean a source resistance in M. And then greater than 10M.
Or am I just confused?
Many of the tube radios have resistors of 1 to 2 megohms in the circuit.
That could throw off the measurements by a bit. It all depends on how
accurate you need to be. Some of the older schematics had several
voltages shown on them. It depends on if a 1000 ohn/volt or 20,000
ohm/volt or a VTVM which usually was 10 meg input impedance.
In article <rn9msu0dr3@news3.newsguy.com>, BobEngelhardt@comcast.net
says...
I got an AEMC model 9100 at a swap meet. Google found nothing on it,
but I called the AEMC number and they dug into the paper files and
scanned a User Manual for me.
The instructions for DC voltage had the following:
 First, a power supply with 10M source resistance measured with a 10MV DC
When the power supply resistance level is high, it is necessary to
adjust the reading according to the following formula:
Actual voltage (in V) =
display voltage (V) x (1 + power source resistance (ohms)) / 10
[\\quote]
There must be a mistake: its input impedance is 10M, so the \"power
supply resistance\" would have to be of that order to affect the reading,
yes? They must mean a source resistance in M. And then greater than 10M.
Or am I just confused?
Many of the tube radios have resistors of 1 to 2 megohms in the circuit.
That could throw off the measurements by a bit. It all depends on how
accurate you need to be. Some of the older schematics had several
voltages shown on them. It depends on if a 1000 ohn/volt or 20,000
ohm/volt or a VTVM which usually was 10 meg input impedance.
meter would read 5 V.
The formula as written does not work with that.
What is the correct formula?
                                         Mikek
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On 10/27/2020 1:40 PM, Ralph Mowery wrote:
....
When the power supply resistance level is high, it is necessary to
adjust the reading according to the following formula:
Actual voltage (in V) =
   display voltage (V) x (1 + power source resistance (ohms)) / 10
[\\quote]
 First, a power supply with 10M source resistance measured with a 10M
meter would read 5 V.
The formula as written does not work with that.
What is the correct formula?
                                         Mikek
Obviously
Obviously
Actual voltage (in V) =
display voltage (V)x(1 + power source resistance (MEGohms) / 10 MEGOhm)
peter