Hello all,

I found this article and i was wondering if someone could explain a little bit better.

http://lygte-info.dk/info/How do I test a charger UK.html

Are the resistors effectively in parallel with the shunt resistor inside the dmm, is that how this works? If i am thinking correctly when you measure the current you could draw a current source in parallel with the network of .1 ohm resistors, but how does that "cancel" the lead resistance? Does it have to do with the voltage drop caused by the network of .1 ohm resistors somehow making up for the voltage drop in the leads?

 

Are the resistors effectively in parallel with the shunt resistor inside the dmm,

That is the root of your misconception. The pile of shunt resistors in parallel with each other develop a voltage when the current passes through. That voltage is measured on the voltage scale of the meter. A Fluke meter with a 10 megohm input compared to a 0.01 ohm shunt creates an error at about 10^-7 as a percentage of the reading, or, 1 part per billion of error.

The method of eliminating the effect of voltage drop in the leads is called a Kelvin connection. This is what the dual sided circuit board is doing. It is providing a place to measure the voltage across the shunt resistors which is physically outside of the current path.

 

That is the root of your misconception. The pile of shunt resistors in parallel with each other develop a voltage when the current passes through. That voltage is measured on the voltage scale of the meter. A Fluke meter with a 10 megohm input compared to a 0.01 ohm shunt creates an error at about 10^-7 as a percentage of the reading, or, 1 part per billion of error.

The method of eliminating the effect of voltage drop in the leads is called a Kelvin connection. This is what the dual sided circuit board is doing. It is providing a place to measure the voltage across the shunt resistors which is physically outside of the current path.

Ok so he's just measuring voltage drop across the .01 ohm resistor and doing the math I = V/R I knew I was missing something.

Thanks

 

Are the resistors effectively in parallel with the shunt resistor inside the dmm, is that how this works?

Yes and no.

He's using the multimeter in its voltage range to test the voltage drops over the resistors. However, the meter does have limited input resistance so there is an effect of "shunt" but the typical input resitance in the voltage range is in excess of 10Mohm, vs. 10mohm so unless you are into extreme measurement errors, you can ignore such effect safely.