[SOLVED] Shunt Resistor Through PCB trace

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poojangarg

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Hi,

I need to make a shunt resistor of 300uohms on my board but we cannot use actual shunt resistir due to cost issue so we have decided to make it on PCB by copper trace.

Could you please guide me to make it on PCB.

Thanks & Regards,

Poojan
 

Hi,

Where do you see the problem?
Calculation of trace resistance? length, width, ...
Calculation of resistance tolerance, temperature drift?
Connection of sense wires?
Designing the amplifier circuit?

Generally:
Calculate width and length.
Place the trace on the PCB, use two vias to connect the sense wires on another layer.

Klaus
 

it a very high current tarce of around 80 A if route in multiple layers than reistnace incearses and if I rout in single layer than trace will burn out.

also with multilayer routing in think via impedance will also be added.

Poojan
 

You need to use many traces in parallel; 80A is a lot of current but if your target resistance is 300uohms, it may be possible. The key problem will be how to control the width and thickness of individual traces and get identical resistances in many traces.
 

Hi,

It will be hard to help you, as long as you don´t give the informations.

***
The only informatoin you gave is "80A".
But we don´t know if it is pulsed peak or RMS.
We don´t know your layer setup.
We don´t know the Cu thickness.

***
BTW: I never mentioned to route the power path to multiple layers. I recommended to route the SENSE wires on a different layer. Then the via resistance plays no role, that´s the benefit.

Klaus
 

it is pulsed peak of 80A , a 6 layer stackup and oueter layer copper thickness is 1 oz and inner layer is 2 oz.

I understood sense wire concept but my concern is how will maintain 80A current if i route in single layer.\

thanks
 

 
For 80A measurement a current transformer is the usual method...
2oz copper is to thin for 80A PCB traces you need to use 6oz at least and even then you need a 30mm trace for 80A constant. For pulsed current you have to do some more leg work...
As said for such currents a CT is normally used...
 
as mentioned earlier we want to save cost on this so we cant use CT.
is there any link or doument pertaining to uage of PCB trace as resistor.
 

marce said:
2oz copper is to thin for 80A PCB traces you need to use 6oz at least and even then you need a 30mm trace for 80A constant.
Click to expand...

A 800W PC SMPS usually supplies more than (or close to) 100A on the 5V rails and the PCB is single sided and I do not see any trace that is 30mm wide.

The trick is to keep a part of the trace (central line) exposed and that picks up molten solder and that carries most of the current (hopefully). It is equally good to solder a copper wire on the trace itself.

Anyway, shunts are common for DC currents only.
 
Hi,

Electrical resistivity and conductivity: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity
There you find all you need to know.

The active length of the trace is the distance between the vias. Between the vias I recommend to use a uniform trace.
At both ends consdider to extend the trace (length at least 1 - 2 x trace width) to get uniform current around the vias.

Copper has large thermal drift in resistance. The current measurement is not very precise.

If you need further assistence, then please say what EXACTELY you need to know.

Klaus
 
The trick is to keep a part of the trace (central line) exposed and that picks up molten solder and that carries most of the current (hopefully). It is equally good to solder a copper wire on the trace itself.
Click to expand...

No, solder has about tenfold resistivity of copper and doesn't help much to reduce the trace resistance.

PCBs designed for high current will probably use copper thickness of 70 µm (2 oz) or more. 80 A is no problem then.

Regarding trace copper as shunt, if we presume that the OP knows about about basic high current PCB design, there are only two problems:

- initial tolerance due to copper thickness variations, I would expect a few 10 percent. Can be adjusted.

- temperature coefficient. As most pure metals, copper has a resistivity proportional to absolute temperature. So you should expect several 10 percent resistance change, depending on environmental temperature range and self heating. Might be compensated with a temperature sensor.
 

Nope wrong, the extra solder adds not a lot of current capacity, maybe this is why I do high power supply layout for critical projects because I know what I am doing and I am au fait with all the regulations.... and my designs work....
Use some on line calculators....
Its all down to thermal management...

The last SMPS I did that would do 60A was done using 4 and 6oz copper...... Go figure why....
Try Saturn PCB toolkit for your calculations.
 
poojangarg:

Rather that do it on board, the cheapest would be to use an external solid bare-copper wire.

Google AWG wire tables. They will list resistivity in ohms per unit length for different wire gauges.

The beauty of this is that not only is very inexpensive, but you can make a 4-wire resistor easily.

Let's assume (as an example) that your calculations yield a wire length of 3.5 cm. Make it slightly longer, let's say 4 cm.
Then, at exactly the 3.5 cm points you solder a pair of thin wires...those will be your sense terminals, whereas the main (thick) length will be your power terminals.

The only problem is (as FvM explained) copper has a lot of temperature coefficient, in the order of 4% per 10 degrees C. You will have to compensate for this effect, since the wire will heat perhaps 40 or more degrees.

I've done this trick and works reasonably well considering its low cost.
 
The advantage of schmitt's advice is that, not only the temperature effect is less present, but also trimming resistance is easily done just by cutting the bar/wire a little, or fastening terminals closest to the middle of the piece.
 
Dear All,

Thank for all your help.

What I have done is I made a trace resistance of 300uohms on external layer and routed the sense trace in taht layer only

For thermal, I have routed traces in inner layer(2oz)

Poojan
 

Hi,

I'm confused. A picture / screen shot will be more descriptive.

Klaus
 

Depending on the thickness, the copper on your PCB has a specific impedance PER SQUARE, from which you can calculate the number of "squares" you need to make a 300 uOhm trace segment. The thickness of the squares can be found in trace width in function of current tables.
 
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