for around 80-100A
hunt, anybody made one of these ?
Stable R metals like constantan wire have too much R for this. Low resistan= ce metals like copper change R as they heat up. A simple diy solution when = making a shunt from something like copper is to dip it in water, then temp = changes are far smaller. In principle you could probably also use a cup-lik= e shape.
NT
Or even micro ohms
around 80-100A
shunt, anybody made one of these ?
page:
Not if they are solid bars, which many ammeter shunts are.
Low resistance metals like copper change R as they heat up.
A simple diy solution when making a shunt from something like copper is to dip it in water, then temp changes are far smaller.
In principle you could probably also use a cup-like shape.
Hadn't done the maths yet ... I used 1 ohm as that was the example in the Doc I posted
Agree ...That was the problem I saw with a really low value shunt ... leads are going to have appreciable resistance
I don't have such a clamp meter, and impractical to buy one for just one set of measurement.
around 80-100A
shunt, anybody made one of these ?
page:
Never used these ... are they passive and direct interfacing to a DVM, or do you need driver circuitry ?
page:
Now one of those would be good ... but not at that price :-)
obviously, I was addressing the idea of using wire as a homemade shunt
As long as the connecting leads can handle the current and are "low" resistance, the resistance of the leads makes no difference to the measurement.
The voltage drop is only measured across a fixed known resistor in series with the connecting leads. The voltage drop in the connecting leads carrying, say, 100Amps is not being measured.
Cheap and commercially available low value shunt resistors (milli or micro Ohms) come with 4 terminals. Two terminals will handle the 100Amps and the other two connect to a high impedance volt meter. The current in the connecting leads to the voltmeter is vanishly insignificant.
anybody made one of these ?
You don't "make" a shunt, you find or acquire one, then you calibrate it, according to whatever resistance it turned out to have. Otherwise it's a right sod of a job to adjust one.
If you want one calibrated to give easy round numbers as an output, just use a potentiometer across its output. If your display meter has a modern high impedance, this is easy.
Swapping leads on my various DVMs - using what appears to be decent quality leads - will give different resistance readings for very low values. Try it.
The connecting leads referred to are clearly those to the shunt, hence the reference to the current rating. They make no difference to the reading across the shunt for a given set of test leads and DVM.
MBQ
Well, yes. But my point is in practice they will make a difference.
It doesn't matter if the measuring leads are 0.01ohms or 1 ohm when measuring the _Voltage_ across a shunt resistor with a DVM (or voltmeter) with a high impedance input. There may be 100s of Amps flowing through the shunt resistor but microAmps in the measuring leads.
But those connecting leads I was referring to will make a difference to the results. In the same way as test leads do when measuring very low resistance. As when doing that you're actually measuring voltage drop across a resistance.
If you're measuring the voltage drop due to 100 amps across a few milliohms as in the case proposed, then the error due to probe resistance should be negligible, and well within the tolerances of anything outside a standards lab. Assuming 2kpv for the meter, (500 microamp/ volt FSD), the meter impedance will be on the order of dozens of ohms, with a current of less than 100 microamps. If the meter is the now standard? (Electronic multimeter) 10 Megohm input impedance, then the current in the leads will be hundredths of a microamp. The resistance of the leads and their connection to the shunt should be hundredths of an ohm at most, given clean contacts and a steady hand. The major errors are going to be bad contact between the probes and the shunt, thermal noise in the shunt, thermal noise in the leads ands meter, and induced currents from nearby AC circuitry, especially stuff like switch mode supplies, which can generate some very strange induced waveforms.
The biggest error in this measurement will probably come from the change in resistance of the circuit being measured by adding and removing the shunt, unless the shunt is permanently in circuit, with a proper meter connection for it. In this case only, the effect of the meter being connected or not on the circuit being measured will be minimal. Now, how accurate is the meter, and will these effects be swamped by meter error?
One way round this problem is to use the supply lead as the shunt resistor, and calibrate the voltmeter accordingly.
around 80-100A
At the price these are, with shunt, I don't see the point of pissing about, although making your own is good for a laugh just out of principle.
Irrelevant. The meter is measuring the *VOLTAGE* dropped across the reference resistor. No current at all flows in the meter leads.
OK a few fA or less must flow but with modern FET front ends even that is probably an exaggeration. The voltage dropped in the meter leads would not be measurable even with state of the art kit.
But there the meter leads are both supplying the current and trying to measure the voltage drop in the load. Four terminal measurement does not have that problem - that is exactly why it is done that way!
No it shouldn't if the DVM is measuring voltage. Modern meters have such a high input impedance that they do not affect the measurement.
Exactly where you measure the voltage between could do, but provided that the system is a four terminal shunt then the measured voltage should be stable and reproducible no matter what test leads are used on the meter.
All measuring instruments change whatever they are trying to measure. Can't be helped. In the case of voltmeters, the higher ohms per volt the better as it draws less current.
Have something to add? Share your thoughts — no account required.
Ask the community — no account required