|
Shunts can be used to measure high DC currents. They are basically very low
value resistors (ie. 0.0005 OHM) that can take a high current because of their
low resistance and large size. The formula V = IR is used to design the shunt.
The shunt should have a 50mV or 100mV drop at its maximum rated current. So for
designing a shunt capable of measuring 100Amps:
50mV = 100A*resistance
Solving for resistance we get 0.0005 Ohms. So the shunt has 50mV across it at 100A current.
Stainless steel is a good shunt material because it has a moderate resistance,
so we only need a short length of it to get our required 0.0005 Ohms.
If we were using copper the shunt would be much larger.
Using matweb (www.matweb.com) I got a value of 0.74 ohm*mm^2/meter for stainless steel.
This means that a 1mm by 1mm rod of stainless steel 1 meter long would have a resistance of
0.74 ohms. We had some 3/8" stainless rod laying around so we used this.
3/8" stainless rod (9.525mm) has a cross-section area of 71.14mm^2 as:
A = pie(r^2) so A = 3.14*(4.76*4.76) = 71.14mm^2 cross section area of 3/8" rod
For the desired shunt resistance of 0.0005 ohms we needed a length of rod approx 48 mm long:
cross section resistance ratio = 0.74/0.0005 = 1480X
length ratio: 1480/71.14 = 20.8X
1meter = 1000mm/20.8 = 48.08 mm shunt length
We cut a piece of stainless rod and two pieces of aluminum end-pieces and hammered the rod into the
tapped aluminum end-pieces until there was an approx 48mm length of stainless rod between the two
aluminum pieces. Four holes were tapped into the aluminum ends to allow for the in circuit connection
as well as the voltage measurement connection.
|
