Posted by Nikki on February 18, 2005, 12:24 pm
Thanks guys
If I have a circuit lets say 1.5 volt bat and a 1.5 volt lamp and I want to
measure the current I would put the meter parallel to a shunt resistor in
the circuit right. But does the shunt resistor not cut down on the voltage
to the lamp
>
>
>
> > Guys thanks for all your replies
> > I took the suggestion of Rich and found that with a 1.5volt battery and
a
> > resistance of 10400 ohms I can get the meter to read at its last
> increment.
> > The meter is marked in increments of 10 up to 60 how would I get a 2 amp
> > draw to read 20 where a 6 amp draw would be the last increment on the
> meter
> > Thanks
> > Nikki
> ---------------
> You will have to do a bit better. If you have a voltmeter, measure the
> voltage at the battery- it is unlikely to be 1.5V. You could also measure
> the voltage across the potentiometer and determine the actual current. It
> appears to be a 50mv meter which seems to be reading full scale at
> 120ma.giving a resistance of 0.417 ohms.
> If this is so -then for 6A full scale, the meter will need an external
shunt
> of 0.0085 ohms. You will need a wire large enough to carry 6A and long
> enough to get 0.0085 ohms (4 inches of #24 wire may do).
> However- check the actual current and voltage as above as the FS reading
may
> not be 120ma.
> I assume you are not looking for great accuracy.
> See calculations below
> V=meter FS voltage. Im=meter FS current. I=desired full scale current
> Rshunt =V/(I-Im)
> In your case V=50mv, I=6A and, if the assumption above is correct, Im
> =0.120A
> then Rshunt =0.05/(6-0.12) =0.0085 ohms
> #24 wire should carry 6A and has a resistance at 20degrees C of 25.7
> ohms/1000ft so 4 inches should do.
>
> Someone please check the above.
> --
> Don Kelly
> dhky@peeshaw.ca
> remove the urine to answer
>
>
> > > Nikki wrote:
> > >
> > > > On closer inspection of the meter I have taken the cover off and the
> > needle
> > > > will not move so I think the meter is no good. I have a second I
> would
> > like
> > > > to use this one goes from 0 to 60 On the back it has a + on one of
the
> > posts
> > > > and printed on the back is 120 on the front of the meter is says
(its
> > very
> > > > small print) FS-50mvDC and it also says D C Amperes. Now I know
this
> > came
> > > > from the Phone company and it was on a 48 volt circuit.
> > > > Nikki
> > > >
> > > you will need to determine the coil R (resistance), 120 maybe it.
> > > lets assume for the moment that it is.
> > > your scale is 0..60, we will use that for amp scale.
> > >
> > > Imeter = 0.050/120 = 416 uA ( does not seem likely etc).
> > >
> > > Rshunt = 0.050/(60.0 - 0.000416) = 0.00083 ohms
> > >
> > > so this would mean you need a shunt resistor on the
> > > meter terminals of 0.00083 which is more like a bar
> > > of copper across the terminals of the meter.
> > > if the 120 number you are reading is actually indicating
> > > I of the meter then you use 120 uA or 120 mA instead of the
> > > 416 uA in the first calculation.
> > >
> > > in either case, you will need to refer to a wire&meteral
> > > chart to get teh ohms per 1K' (per 1k feet) to determine what
> > > you need to create a shunt .
> > >
> > > these are only my thoughts off the top of my head going by the
> > > info your passing.
> > >
> > >
> > >
> > >
> >
> >
>
>
Posted by Jamie on February 18, 2005, 5:16 pm
Nikki wrote:
> Guys thanks for all your replies
> I took the suggestion of Rich and found that with a 1.5volt battery and a
> resistance of 10400 ohms I can get the meter to read at its last increment.
> The meter is marked in increments of 10 up to 60 how would I get a 2 amp
> draw to read 20 where a 6 amp draw would be the last increment on the meter
> Thanks
> Nikki
that looks to be a 150 uA meter at full scale.
if memory serves,.i think you said the meter was a 0.050 mv?
to get a 6 amp scale.
6.00 - .000150 = 5.999850
Shunt = 0.050 / 5.999850 = 0.0083 ohms
i think that is correct,i normally measure the
coil R and use the I (current) to do all of this
how ever, this should still work.
this is just off the top of my head, i am sure some
one will correct me.
Posted by Roger Johansson on February 18, 2005, 11:38 pm
> that looks to be a 150 uA meter at full scale.
> if memory serves,.i think you said the meter was a 0.050 mv?
>
> to get a 6 amp scale.
> 6.00 - .000150 = 5.999850
>
> Shunt = 0.050 / 5.999850 = 0.0083 ohms
>
> i think that is correct,i normally measure the
> coil R and use the I (current) to do all of this
> how ever, this should still work.
> this is just off the top of my head, i am sure some
> one will correct me.
>
I think you are right.
In practical terms: Take a piece of copper wire, diameter 1-2mm.
Send 1 Amp through it. Put the wires from the meter together on that
wire. Pull them slowly apart until the meter reads 1 Amp. Mark these
measuring points and solder the instrument wires to them. Cut off the
thick wire outside the measuring marks and attach the current connections
to the ends of the wire.
--
Roger J.
Posted by Roger Johansson on February 16, 2005, 5:20 pm
> On closer inspection of the meter I have taken the cover off and the
> needle will not move so I think the meter is no good. I have a second
> I would like to use this one goes from 0 to 60 On the back it has a +
> on one of the posts and printed on the back is 120 on the front of the
> meter is says (its very small print) FS-50mvDC and it also says D C
> Amperes. Now I know this came from the Phone company and it was on a 48
> volt circuit.
It sounds like the meter mechanism inside is a 50mV full scale
instrument, but it is very likely that it has a series or parallell
resistor inside the cover to make it suitable for 48 Volt or a few
amperes.
Open the cover and try to find a series or parallell resistor. Or simply
test the meter very carefully, using a battery and a high value resistor
in series, to limit the current to less than 10mA to begin with. If there
is no reaction change to a ten times lower resistor until it works.
--
Roger J.
Re: Amp Meter
Re: Amp Meter
>
>
> > Guys thanks for all your replies
> > I took the suggestion of Rich and found that with a 1.5volt battery and