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So I have this shunt in my power system connected just a few inches from the battery bank's final outgoing (to system) negative terminal. But I can't seem to find any data on whether or not these devices actually lower the final output voltage to the system by the stated 50mV drop. I did find some articles that seem to suggest higher current shunts have the ammeter wired in parallel to the shunt resistor such that only a very small percentage of current flows through the ammeter. That would seem to suggest that the voltage drop may only be experienced by this small parallel circuit, and the main system voltage would be unaffected.

This type of shunt is widely used, so I can contact the manufacturer, but I suspect the answer applies to nearly all of the major battery shunts commonly available for purchase at RV and Marine stores, inverter/solar shops, Amazon, etc. Does anyone know for sure how this works?

Thanks!

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    \$\begingroup\$ The meter that's connected to the shunt reads the voltage across it, this voltage is then scaled to read the current flowing through the shunt. At 500 A, the shunt will drop 50 mV, at 10 A it will only drop 1 mV, the shunt has a resistance of 100 uohm. The voltmeter will have a much higher resistance than the shunt so will take negligible current, which means it can use a long cable with connectors at the end, and still have little reading error. \$\endgroup\$
    – Neil_UK
    Oct 18 '21 at 13:19
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    \$\begingroup\$ The shunt is a resistor. It obeys Ohm's law. The meter is a resistor of much higher resistance. It also obeys Ohm's law. The overall effect is determined by the parallel resistance of the pair. \$\endgroup\$
    – Transistor
    Oct 18 '21 at 13:19
  • \$\begingroup\$ Excellent! That explains how the meter can have its display on a 25ft cord. \$\endgroup\$ Oct 18 '21 at 13:22
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Yes, with the caveat that for a 500A/50mV shunt, the output will drop 50mV only with 500A flowing.

For 250A, the drop will be less, i.e. 25mV.

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  • \$\begingroup\$ Okay, and you're sure of this? Again, I just read some article saying only a very small amount of current flows through the ammeter, but is that just for its measurement circuitry, and the actual shunt resistor is still dropping by 50mV at 500A? Very helpful to know it's current related. I think I had even read that at one point and forgot. Makes perfect sense. My current will often be 10A discharging, and 100-150 charging, so I think I'll only see 2mV discharging, and 20'ish mV charging. That's very good to know. Now I can translate what I observe at the pack level into cell voltages. Thanks! \$\endgroup\$ Oct 18 '21 at 13:16
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    \$\begingroup\$ Yes, absolutely sure. The shunt resistor, to dissipate 50mV at 500A, is implying that it has a resistance of R=V/I = 0.05/500 = 100µΩ which is a very low resistance. The buss bars / tracks / cabling are likely to have more than this. A very robust mechanical and electrical connection will be needed. \$\endgroup\$
    – rdtsc
    Oct 18 '21 at 16:50
  • \$\begingroup\$ Thank you! Will just a few feet of quality 2/0 copper wire and lugs really drop it by more than 50mV? I guess I need to go pull some wire voltage drop charts and run the math. I figured it would be negligible at 100A and such short runs. \$\endgroup\$ Oct 19 '21 at 0:37
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Resistance is simply a measure of how much voltage is developed across a resistor when 1A of current flows through it.

Current measurements that use shunt resistances measure voltage directly, not current. They use the shunt resistance as a transducer to convert current into that voltage. The voltage appears directly across the shunt resistance, and thus constitutes a voltage drop. A current-measuring shunt is just a fancy name for a resistor.

A 500A/50mV shunt is really a 5E-2 V/5E2 A = 1E-4 Ohm resistor, or 0.1mOhm resistor. Any resistor with such resistance will drop 50mV if you pass 500A through it. Not every resistor will survive very long with 500A flowing through it. Such "resistors" look more like chunks of metal than resistors :)

You would be surprised how much voltage all the cabling and connectors between the battery and the load(s) will drop: I bet it'll be much more than 50mV at 500A.

So if you worry that this voltage drop is significant, the shunt is probably the last thing on your list of things to worry about. Most of the voltage drop will be elsewhere.

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  • \$\begingroup\$ With only a few feet of quality 2/0 copper wire between the battery and the various devices, will that cabling and 2/0 lugs really drop the voltage by 50mV or more, even at say 100A max? \$\endgroup\$ Oct 19 '21 at 0:34
  • \$\begingroup\$ @BeachInMexico Measure it. The quality of connectors and cleanliness of everything will matter a lot. \$\endgroup\$ Oct 19 '21 at 20:06

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