Think "Ohm's Law." The meter movement responds to a current. To read
voltage, you put a resistor in series with the meter movement so that
V(full scale) = R(total)*I(meter, full scale). R(total) is the sum of
the meter's internal resistance and the external series resistor. So
a 1mA meter movement always gives 1kohms/volt, and a 20uA meter
movement gives 50kohms/volt.

Cheers,
Tom

Uwe Langmesser wrote in message ...
Is there a simple relationship between a meters internal resistance and its
sensitivity (ohms per volt).

Maybe this is trivial but I don't see it.

Uwe

You are right Tom.

But my question was that when the sensitivity, for example 1 kohms/volt, is
known, can one then also know, without further measurement, the internal
resistance.

And from what I gather the answer to that question is NO.

thank you all

Uwe



in article , Tom Bruhns at
wrote on 2/13/04 1:26 PM:

Think "Ohm's Law." The meter movement responds to a current. To read
voltage, you put a resistor in series with the meter movement so that
V(full scale) = R(total)*I(meter, full scale). R(total) is the sum of
the meter's internal resistance and the external series resistor. So
a 1mA meter movement always gives 1kohms/volt, and a 20uA meter
movement gives 50kohms/volt.

Cheers,
Tom

Uwe Langmesser wrote in message
...
Is there a simple relationship between a meters internal resistance and its
sensitivity (ohms per volt).

Maybe this is trivial but I don't see it.

Uwe

You are right Tom.

But my question was that when the sensitivity, for example 1 kohms/volt, is
known, can one then also know, without further measurement, the internal
resistance.

And from what I gather the answer to that question is NO.

thank you all

Uwe



in article , Tom Bruhns at
wrote on 2/13/04 1:26 PM:

Think "Ohm's Law." The meter movement responds to a current. To read
voltage, you put a resistor in series with the meter movement so that
V(full scale) = R(total)*I(meter, full scale). R(total) is the sum of
the meter's internal resistance and the external series resistor. So
a 1mA meter movement always gives 1kohms/volt, and a 20uA meter
movement gives 50kohms/volt.

Cheers,
Tom

Uwe Langmesser wrote in message
...
Is there a simple relationship between a meters internal resistance and its
sensitivity (ohms per volt).

Maybe this is trivial but I don't see it.

Uwe

In article , Uwe Langmesser
writes:

But my question was that when the sensitivity, for example 1 kohms/volt, is
known, can one then also know, without further measurement, the internal
resistance.

And from what I gather the answer to that question is NO.

thank you all

In actual practice, d'Arsonval meters vary in internal resistance.
The "Ohms/Volt" rating is only approximate.

D'Arsonval meters consume DC power dependent on the meter
motor (the coil winding driving the needle) position. Miniscule power
to be sure but finite power demand. If the power of movement and
display is the criterion for "sensitivity," then the meter resistance
must be measured or known. That is imperative if the meter is to
be shunted for a higher actual current indication.

A millivoltmeter can measure the meter motor voltage drop at a
given current through it. Law of Resistance will then apply to find
meter motor resistance from current and voltage. To shunt a
meter for higher current the external shunt resistor must have a
value equal to the meter motor voltage drop divided by the total of
desired current minus the meter current. At high current full scales
the shunt resistance becomes quite low and the meter motor
resistance is negligible for accuracy of high current. At low current
full scales approaching that of a meter alone, the meter motor
resistance is an appreciable part of the shunt resistance.

In voltmeter applications, a series resistor is approximated by the
"Ohms per Volt" value on relatively high full-scale ranges. That series
resistor should be (for precision) the quantity total voltage drop minus
the meter motor voltage drop, all divided by the meter current.

Fortunately, the meter motor resistance is small and the voltage drop
is quite small compared to the total needed voltage drop. The difference
in voltmeter resistance values between approximate and precision
becomes less as the meter current becomes less for a given full scale.
Note: For full scale voltmeter readings of 100 V and up, the approximation
of "Ohms/Volt" is quite good...one needs the meter motor resistance for
an accurate voltmeter indication of 5 V and below to fit modern digital
logic rail voltages.

One can expect most d'Arsonval meters to exhibit about 50 mV motor
drop at full scale...but could be almost any value from 20 to 100 mV.

Len Anderson
retired (from regular hours) electronic engineer person

In article , Uwe Langmesser
writes:

But my question was that when the sensitivity, for example 1 kohms/volt, is
known, can one then also know, without further measurement, the internal
resistance.

And from what I gather the answer to that question is NO.

thank you all

In actual practice, d'Arsonval meters vary in internal resistance.
The "Ohms/Volt" rating is only approximate.

D'Arsonval meters consume DC power dependent on the meter
motor (the coil winding driving the needle) position. Miniscule power
to be sure but finite power demand. If the power of movement and
display is the criterion for "sensitivity," then the meter resistance
must be measured or known. That is imperative if the meter is to
be shunted for a higher actual current indication.

A millivoltmeter can measure the meter motor voltage drop at a
given current through it. Law of Resistance will then apply to find
meter motor resistance from current and voltage. To shunt a
meter for higher current the external shunt resistor must have a
value equal to the meter motor voltage drop divided by the total of
desired current minus the meter current. At high current full scales
the shunt resistance becomes quite low and the meter motor
resistance is negligible for accuracy of high current. At low current
full scales approaching that of a meter alone, the meter motor
resistance is an appreciable part of the shunt resistance.

In voltmeter applications, a series resistor is approximated by the
"Ohms per Volt" value on relatively high full-scale ranges. That series
resistor should be (for precision) the quantity total voltage drop minus
the meter motor voltage drop, all divided by the meter current.

Fortunately, the meter motor resistance is small and the voltage drop
is quite small compared to the total needed voltage drop. The difference
in voltmeter resistance values between approximate and precision
becomes less as the meter current becomes less for a given full scale.
Note: For full scale voltmeter readings of 100 V and up, the approximation
of "Ohms/Volt" is quite good...one needs the meter motor resistance for
an accurate voltmeter indication of 5 V and below to fit modern digital
logic rail voltages.

One can expect most d'Arsonval meters to exhibit about 50 mV motor
drop at full scale...but could be almost any value from 20 to 100 mV.

Len Anderson
retired (from regular hours) electronic engineer person