ammeter help

IMHO you'd be better off shunting than attempting to mod.

In message , Roland Craven writes

You could use an external shunt. If you accurately measure the resistance of the ammeter and find it to be, say, 0.3 ohms, then Ohm's Law (V = I x R) shows that you would get 3 amps to flow for a potential difference of 0.9 Volts. If you then connect an external resistance of suitable power dissipation across the meter contacts (thus in parallel with the coil) and arrange its resistance to be one-ninth of the coil resistance (0.027 ohms in the example above) then one tenth of the current flowing will pass through the meter and nine-tenths will pass through the shunt. Your meter will effectively have an FSD of 30 amperes. For resistances in parallel, the total resistance can be calculated with the formula below (best in fixed-pitch font):

- = - + - + ... + - R(Total) R(1) R(2) R(n)

The power rating for the shunt can be calculated as I x I x R, so for the example above, 27 x 27 x 0.027 = 19.83 Watts

Regards

Pete

A shunt it is and whilst I remember the formula I'm indebted for the clear simple explanation. regards Roland

That was clear and simple! My head hurts :-))

Mart> A shunt it is and whilst I remember the formula I'm indebted for the

The easiest way to make such a low value shunt is to connect a length of thick wire across the ammeter and progressively move the connection points further apart while checking the reading. Another ammeter in series gives the actual current. If the length of the shunt wire gets unreasonably long, stop and do it again with a thinner wire. Once the length is known, replace the trial wire with a new lacquered or insulated piece to prevent corrosion.

John

Good plan, sounds like this also gives a four terminal shunt to avoid contact resitance problems. AVO used copper strips trimmed by filing a notch.