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

Uwe Langmesser wrote:

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

The higher the resistance, the lower the operating current, but there
is no simple formula.

You need a current limiting resistor in series with the meter, and a
known supply voltage. Start at the maximum resistance, and slowly reduce
it till you have a full scale deflection. Disconnect the voltage, then
measure the total resistance of the meter and adjustable resistor. Then
use Ohm's law to determine the meter's sensitivity. Its simple and only
takes a couple minutes to test. Use a low voltage, and it is a good idea
to have different variable resistors to put in series. Start with a
higher value resistor than you think you need, so you don't damage the
meter, and work your way down. I use an expensive, six decade resistor
with an adjustable regulated power supply. I can adjust the resistance
in steps, from zero ohms to one megohm in one ohm steps.

--
We now return you to our normally scheduled programming.

Michael A. Terrell
Central Florida

Uwe Langmesser wrote:

Is there a simple relationship between a meters internal resistance and
its
sensitivity (ohms per volt).

============================

Yes. Couldn't be more simple.

It is the internal resistance of the meter divided by the voltage range the
meter is set to.

When the meter is at full-scale deflection the current flowing through the
meter is always (Voltage Range) / (Ohms per Volt).

The internal resistance of the meter, on any voltage range, is the
resistance of the moving coil plus the series meter-multiplying resistance.
---
Reg, G4FGQ

Uwe Langmesser wrote:

Is there a simple relationship between a meters internal resistance and
its
sensitivity (ohms per volt).

============================

Yes. Couldn't be more simple.

It is the internal resistance of the meter divided by the voltage range the
meter is set to.

When the meter is at full-scale deflection the current flowing through the
meter is always (Voltage Range) / (Ohms per Volt).

The internal resistance of the meter, on any voltage range, is the
resistance of the moving coil plus the series meter-multiplying resistance.
---
Reg, G4FGQ

See URL:
http://www.tpub.com/neets/book16/68e.htm

Too lengthy to repeat here -- but gives the full explanation

--
Incognito By Necessity (:-(

If you can't convince them, confuse them.
- - -Harry S Truman




"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

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

No.

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

Not trivial at all. The internal resistance of a meter depends on the
type of meter (D'Arsonval, moving iron, electrodynamometer, etc.) and
its design.

As a simple example, imagine two identical 0-1 mA D'Arsonval meters.
Now replace the springs in one of the meters with new ones that
require more force. The meter with the "stronger" springs will need
more current for fullscale deflection even though it has the same
internal resistance.

---

There *is* a relationship between a milliammeter's fullscale reading
and the ohms-per-volt when it is used with a series resistor to make a
voltmeter.

Ohms-per-volt = 1/current for full scale deflection (in amps)

So, when used with the appropriate series resistor(s):

a 0-1 mA meter will give 1000 ohms-per-volt
a 0-100 uA meter will give 10,000 ohms-per-volt
a 0-50 uA meter will give 20,000 ohms-per-volt

etc.
73 de Jim, N2EY

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