Aggregating Current Measurements from Two Different Shunts

[goboatingnow]

Originally Posted by botanybay

I have considered the following:



I have a leftover double shunt where it is two shunts with a common bar for the DC buss and two connections, one for each battery. It came off of a Link 2000R which monitored and adjusted the alternator based upon state of charge, but that was a long time ago on a different boat.



So two 50mv shunts, one for each battery with the side opposite the battery being a common copper block. I have two Victron BMV 701s (Amp Hour Monitors from Victron), one on each shunt which works nicely but I have to multiply all of my currents and amp hours by a factor of two assuming both sides are active.



Note that on the positive side I can isolate either of the two batteries and the system works at half capability.



So, to make the system work better how about a third BMV 701 which is wired with two 100k Ohm resistors in series between the battery side of the two shunts. These are both 50mv shunts @ 500A have a resistance of E=I/R --> R=E/I = 50e-6V/500A = 0.0125 Ohms



I have attached a GIF of the circuit.



Essentially if both shunts are reading the same value then there is no current through the 100k ohm resistors and the voltage is the same as either shunt:

10mv on shunt 1 and 10mv on shunt 2, no current through the 100k ohm resistors so midpoint is at 10mv.



With only shunt 1 active the midpoint of the two series 100k ohm resistors is half the voltage on shunt 1 as shunt 2 has no current and so the battery side is approximately 0 (yeah, a small current flows I = E/R = 10mv / 200k ohm = 0.00000005 amps :-) )



So the midpoint behaves as if the battery sides of the shunt has been shorted together and acts like a 50mv @ 1000A shunt.



So, if I program the midpoint BMV 701 to have 50mv = 1000A it should properly measure the overall consumption on the two banks without impacting the other two shunts.



Why pick 100k Ohms for the resistors? I am guessing that the input impedance of the BMV is on the order of 1Meg to 10Meg Ohm, by making the series resistors an order of magnitude less the mid point voltage is not impacted significantly by adding the BMV between the midpoint of the series resistors and the common side DC negative bus.



Just remembering back to basic circuit theory and I don't think I have messed anything up. Since the shunts are a single combined block there is no risk of lifting one side of the Shunt to DC common but even if I did the resistance is such that the current flow would be tiny.



Any reason this simple solution is a bad idea.



(I do like the other devices which can perform the summation correctly but I also like that my Color Control GX is uploading this data to the cloud all of the time and I can check on the boat from anywhere on the planet :-) )

Your circuit makes no sense the centre bmv is expecting to see shunt currents but the 100k will throw everything off

[goboatingnow]

Originally Posted by hzcruiser

From what I gathered in your diagram, both shunts are in series, yes, but the whole current from both panels is going through both shunts?

Still doesn't add up in my head, sorry.

Two shunts in series is meaningless , it’s just one shunt with twice the sensitivity, shunts are just resistors so it’s just two resistors in series resulting in a combined resistance of 2x , the result turns a 50mV shunt into a 100mV shunt that’s all.

[hzcruiser]

Originally Posted by goboatingnow

Two shunts in series is meaningless , it’s just [...].

That's what I meant by "it doesn't add up".

[s/v Jedi]

It’s

So who is going to post how to sum values from two BMV’s on the Cerbo display?

[botanybay]

Originally Posted by goboatingnow

Your circuit makes no sense the centre bmv is expecting to see shunt currents but the 100k will throw everything off

My drawing did not match my text, I forgot to show the battery connections:

So, to the left is the connection to the negative buss bar which feeds everything on the boat, on the right are the connections to the two independent battery banks to be monitored.

The top BMV is across the normal shunt for Bank 1
The bottom BMV is across the normal shunt for Bank 2

The Middle BMV will measure a voltage which is half way between the voltage across shunt 1 and the voltage across shunt 2.

No significant current flows due to a maximum difference of 50mv if Bank 1 was drawing 500A and Bank 2 was not providing any current (it was switched off on the positive side for example)

E=IR --> I = E/R = 50mv / 200K ohm = 0.00000025 Amps

That is well below the measurement uncertainty of the system.

The solution only works because the negative buss side of the shunts is exactly the same for both shunts (not two independent shunts with two different lengths of wire, two different contact resistance values for the lugs).

So in this case, finding a voltage which is half way between the battery sides of the two shunts results in a value which is linear with the total current flowing through both shunts and is half the mv/amp of the independent shunts (25mv/500A or 50mv/1000A).

The resistance of the two resistors is large enough that the current is very low, thus the power (heat) is essentially zero.

Does this help explain what I am considering?

[hzcruiser]

Originally Posted by botanybay

So, to the left is the connection to the negative buss bar which feeds everything on the boat, on the right are the connections to the two independent battery banks to be monitored.

The top BMV is across the normal shunt for Bank 1
The bottom BMV is across the normal shunt for Bank 2

The Middle BMV will measure a voltage which is half way between the voltage across shunt 1 and the voltage across shunt 2.

No significant current flows due to a maximum difference of 50mv if Bank 1 was drawing 500A and Bank 2 was not providing any current (it was switched off on the positive side for example)
[...]

The solution only works because the negative buss side of the shunts is exactly the same for both shunts (not two independent shunts with two different lengths of wire, two different contact resistance values for the lugs).

The pic is very small, even after clicking on it, but here we go:
Apart from the mistake that the shunt cannot be 12.5 mOhm but is 0.1 milliOhm based on your numbers: R = 50 mV / 500 A, your solution
won't work at all, I'm afraid.

Consider this: Whenever the current to Bank 1 is similar to the current to Bank 2, the voltages on the two shunts are identical and hence the difference (which is what you would measure in the middle) is zero V.
What you have designed is somewhat like a Wheatstone bridge.

Adding voltages in the analog world can be done with OpAmps, but in many cases the simplest approach would be to use two current monitors (like the INA219 I'm using quite often), one across each shunt, and then add those two values in a microcontroller.

[hzcruiser]

Originally Posted by botanybay

[...]

E=IR --> I = E/R = 50mv / 200K ohm = 0.00000025 Amps

[...]

Oh, and, no offence, but I don't know any EE (electrical engineer) who counts zeros past the decimal point. It's very likely to make a mistake there and just tedious.

Try: 50 mV / 200 KOhm = 50/200 uAmps (microAmps) = 5/20 uA = 0.25 uA or 250 nA.

[Dockhead]

The EE discussions are fascinating; carry on guys. We sure have a lot of expertise on this forum.


FWIW, I got a reply from Balmar that their battery monitors will not aggregate current from two shunts.


So right now it looks like there are two commercial solutions -- (1) Pico; and (2) Victron Venus. And a possible third -- Mastervolt.

[goboatingnow]

Originally Posted by botanybay

My drawing did not match my text, I forgot to show the battery connections:



So, to the left is the connection to the negative buss bar which feeds everything on the boat, on the right are the connections to the two independent battery banks to be monitored.



The top BMV is across the normal shunt for Bank 1

The bottom BMV is across the normal shunt for Bank 2



The Middle BMV will measure a voltage which is half way between the voltage across shunt 1 and the voltage across shunt 2.



No significant current flows due to a maximum difference of 50mv if Bank 1 was drawing 500A and Bank 2 was not providing any current (it was switched off on the positive side for example)



E=IR --> I = E/R = 50mv / 200K ohm = 0.00000025 Amps



That is well below the measurement uncertainty of the system.



The solution only works because the negative buss side of the shunts is exactly the same for both shunts (not two independent shunts with two different lengths of wire, two different contact resistance values for the lugs).



So in this case, finding a voltage which is half way between the battery sides of the two shunts results in a value which is linear with the total current flowing through both shunts and is half the mv/amp of the independent shunts (25mv/500A or 50mv/1000A).



The resistance of the two resistors is large enough that the current is very low, thus the power (heat) is essentially zero.



Does this help explain what I am considering?

Still makes no sense

The bmv is designed to measure a small shunt voltage and convert that to amps

Since we can assume no current flows into the middle BMV , therefore the 100 K resistors serve no function as far the the middle BMV is concerned, they merely serve to create a potential divider between the two battery negatives

Hence the bmv will indicate positive or negative currents flowing , ie -10 A would mean one battery was seeing bigger return currents then the other.

But this information is already effectively being supplied by the two BMVs anyway.

The middle bmv does not show cumulative current which is what the OP wanted.

[botanybay]

Originally Posted by hzcruiser

Oh, and, no offence, but I don't know any EE (electrical engineer) who counts zeros past the decimal point. It's very likely to make a mistake there and just tedious.

Try: 50 mV / 200 KOhm = 50/200 uAmps (microAmps) = 5/20 uA = 0.25 uA or 250 nA.

No offense taken, not being an electrical engineer and being in a hurry (sloppy) I normally work purely in scientific notation rather than uA, mA, KA...

I was just cutting and pasting from the calculator on the computer and not double checking :-)

The only point was that the current through the 100k ohm resistors was so small that there was no need for high wattage devices.

[botanybay]

Originally Posted by goboatingnow

Still makes no sense

The bmv is designed to measure a small shunt voltage and convert that to amps

Since we can assume no current flows into the middle BMV , therefore the 100 K resistors serve no function as far the the middle BMV is concerned, they merely serve to create a potential divider between the two battery negatives

Hence the bmv will indicate positive or negative currents flowing , ie -10 A would mean one battery was seeing bigger return currents then the other.

But this information is already effectively being supplied by the two BMVs anyway.

The middle bmv does not show cumulative current which is what the OP wanted.

The middle will be half way between the voltage difference between the two battery negative terminals relative to the DC negative buss.

Consider the DC negative buss the "common reference point" rather than the negative terminals of the battery (which are offset from the DC negative buss by the voltage across the shunt.

In the extreme case of using 0 ohm resistors (which makes no sense for our purpose but is the illogical extreme) the pair of 50mv @500 amp shunts becomes a single 50mv @ 1000 amp shunt.

If the voltage at the two battery terminals are both 10mv relative to the DC buss then the mid point between them will be 10mv relative to the DC buss, independent of the resistance of the two resistors making a voltage divider between the two negative terminals.

Just thinking through the problem.

[hzcruiser]

Try re-reading post #52...

[RaymondR]

Originally Posted by hzcruiser

Oh, and, no offence, but I don't know any EE (electrical engineer) who counts zeros past the decimal point. It's very likely to make a mistake there and just tedious.

Try: 50 mV / 200 KOhm = 50/200 uAmps (microAmps) = 5/20 uA = 0.25 uA or 250 nA.

Whilst this is factual, nothing visually represents the magnitude of a quantity quiet as quickly and effectively as a long line of zeros whether left or right of the decimal point.

[goboatingnow]

Originally Posted by botanybay

The middle will be half way between the voltage difference between the two battery negative terminals relative to the DC negative buss.

Consider the DC negative buss the "common reference point" rather than the negative terminals of the battery (which are offset from the DC negative buss by the voltage across the shunt.

In the extreme case of using 0 ohm resistors (which makes no sense for our purpose but is the illogical extreme) the pair of 50mv @500 amp shunts becomes a single 50mv @ 1000 amp shunt.

If the voltage at the two battery terminals are both 10mv relative to the DC buss then the mid point between them will be 10mv relative to the DC buss, independent of the resistance of the two resistors making a voltage divider between the two negative terminals.

Just thinking through the problem.

The “ sense “ voltage at the junction of the 100k resistors , will sit halfway between difference between the two sense voltages from the battery shunts , I’m not arguing that

But let’s say there’s 20 mV on one shunt and 30mV on another hence the mid point will be 25mV , for say 50 mV/500A shunts this means you read 200A from one , 300A from another and the mid one reads 250A . It’s a meaningless figure

Great stuff but the OP wanted a cumulative reading ,