2
\$\begingroup\$

I really need your help in order to figure out a solution to detect a blow fuse in a 16S parallel connection.

The issue is: Despite a fuse has been blown, the voltage at the other side of the fuse is going to be the same just because it comes from the rest of the parallel batteries.

I would really appreciate your help to figure out a solution to send a signal when a fuse has been blow.

This is my battery project and the PCB:

The circuit diagram: Parallel Diagram

\$\endgroup\$
11
  • \$\begingroup\$ Your circuit seems to be fundamentally flawed. THe diagram shows each cell at a given level connecting via a fuse to a common point for that level. However, the load is taken only at the top and bottom of a 16s stack. As shown a blown fuse would prevent balancing that cell BUT the 16S stack would be connected without fuses and still provide power. Unless power is drawn via a fuse per cell it does not protect that cell. Perhaps draw an eg 4S2P stack and show what happens when you blow one fuse and then draw power. \$\endgroup\$ – Russell McMahon Mar 29 '20 at 10:56
  • \$\begingroup\$ If you blow a fuse where current WOULD have flowed then I'd expect a series LED + resistor acxross the fuse would be enough to show where the blown fuse was. \$\endgroup\$ – Russell McMahon Mar 29 '20 at 10:58
  • \$\begingroup\$ Thanks for your comments, yes thats is the reason I trying to get a signal in each module to be able to trigger a contactor in order to disconect the whole module. So far I am able to show a led is some fuse have been blown but I need to disconected the parallel with the previous and the next. As you have said there is no security cut when that happens. Please could you help me out to figure out a solution. \$\endgroup\$ – BatteryDIY Mar 29 '20 at 11:13
  • \$\begingroup\$ I suggest that you draw a small subset diagram with all components shown. eg a say a 3S2P arrangement with middle fuse on one series string blown. || WHY should the fuse blow as shown? The fuse ONLY gets involved in balancing - not in power transfer. || Proper fusing connects the say V+ from each cell to the bus common V- for the next layer of cells above. || Q1: When a fuse blows does the voltage of the cell with the blown fuse rise above that of the other cells in the group? - or is it lower. I'd expect it probably rises. If so a 3 input comparator will allow "signalling". \$\endgroup\$ – Russell McMahon Mar 30 '20 at 0:44
  • \$\begingroup\$ Wires used are really thin, at the begining (when battery is new) almost no current will go through those wires, when batteries get older their capacity change and so The rest of the battery are going to transfer current to the weak one. At some point that transfer will be more than 3a or 5a. Or even worse, some battery can get inverted polarity or simply die therefore we need to cut the whole module down. I hope it make sense my explanation to you all. \$\endgroup\$ – BatteryDIY Mar 30 '20 at 23:02
0
\$\begingroup\$

Something like this should work.
Comparators are usually slightly biased (biasing not shown - could perhaps just be drop across wiring with proper design) such that when fuses are not blown comparator outputs are high and LED not lit.

If say battery B12 blows fuse 4 then B12 voltage rises under sudden no load.
Comparator 2 in- rises above its in+ and comparator output goes LOW and lights LED D2.

schematic

simulate this circuit – Schematic created using CircuitLab

___________________________________________

WINDOW COMPARATOR MONITORING OF BLOWN BALANCING FUSE.

Here is a conceptual window comparator system that lights an LED if a balance fuse is high or low.
I'm not convinced that this is useful - but, it may be.
I expect that being able to monitor whole string current contribution would tell you if a string was bad - but with a lot less work.
However:

CP is the common point for all 3 cells B4 B5 B6.
Say F3 blows.
CP is now the common +ve voltage of BAT4 and BAT5 but NOT BAT6.
CMP1_+ is slightly above CP due to R3/R1. If BAT6 +ve is still AT CP or below it then CMP1 output will be high.
If BAT6 output rises slightly above CP then CMP1 output will go low signalling a fault.
Similarly
CMP2_- (opposite of for CMP1) is slightly above CP due to R3/R1. If BAT6 +ve is still AT CP or ABOVE it then CMP1 output will be high.
If BAT6 output falls slightly below CP then CMP2 output will go low signalling a fault.

So, CMP1 + CMP2 form a window comparator so that if BAT6 with blown F3 remains very close to CP then no alarm is given. BUT if BAT6 Vout deviates more than very slightly from Cp an alarm will be given.

You need two comparators and 4 resistors per battery.
Comparators can be very low cost. I've assumed an open collector output (eg ye olde LM339, LM393 etc).

BUT - is this really meeting your need? It MAY be.

If

schematic

simulate this circuit

\$\endgroup\$
1
  • \$\begingroup\$ Many thanks for your proposal, I need to analize it. I will reply to you if I have some doubts. Again, thanks. \$\endgroup\$ – BatteryDIY Mar 30 '20 at 22:58
0
\$\begingroup\$

All the answers assumed an external fuse with easy access to both sides. However, the lithium battery pack you are using is probably made of three or four lithium batteries in parallel. The final output voltage is created by repeating so many of these groups of 3 or 4 parallel batteries, each connected in series. The fuse is inside each lithium batteries, at the very tip of the positive side. That fuse will blow when current is too high or voltage drop too low. The battery then become permanently disconnected and we will call it _"dead".

How to detect one dead batteries?

Method #1: charging and discharging time is proportional to the number of batteries in parallel. When you connect the charger to the pack, the voltage will rise faster on the group that has one or more dead batteries. Starting from fully charged state, the voltage across the group with dead batteries will fall faster. Method #2: Insert a small resistance in series with every single lithium batteries on the negative side. The smaller the resistance, the lower the loss but the more difficult it is to distinguish from noise. Method #3: The best but most expensive method is too use connectors such that you can disconnect any single batteries to test voltage and load current as needed.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.