0
\$\begingroup\$

I have a circuit where a charger loads a 24 V battery. When the charger is not plugged, there is a risk that of short circuit and the battery will then explode. To avoid this I thought about using transistor.

The MOSFET was my primarily choice but the problem with the MOSFET is that the gate source voltage will be too high (>20 V, which is the upper limit).

A bipolar transistor is another option, but I am also concerned about the voltage dropout. If the dropout is too high, then the battery will never get fully charged when the charger is put on.

Is there another simpler option that I haven't thought of?

\$\endgroup\$
7
  • \$\begingroup\$ "When the charger is not plugged, there is a risk that of short circuit and the battery will then explode. Describe this more thoroughly. What exactly is causing the risk of short-circuit? Because transistors fail short so depending on the details they may not be any good. Also MOSFETs only block current in ONE direction and let it freely flow in the opposite direction and since a charger is involved it may not do what you want at all. Fuses are often used for shorts, but may not be suitable depending on the details of the scenario you are trying to protect against. \$\endgroup\$
    – DKNguyen
    Commented Feb 19, 2023 at 18:01
  • \$\begingroup\$ There are two wires connecting the battery to the charger. When fully charged, the charger would be taken off, leaving the wires 'open'. My project is to be use outside, and I fear that water for example, or something else, could close the connection and make a closed loop with the battery without any other component. In this scenario, I presume the current will be too high, and the battery will explode. \$\endgroup\$
    – cobdmg
    Commented Feb 19, 2023 at 18:34
  • \$\begingroup\$ I would check to see if it's okay to leave the battery connected to the charger when the charger is unpowered. Often you can't do that with components and circuits since the construction requires input power to prevent a power source like a battery on the output from pushing current backwards. With regards to your concern, I would just use a fuse to protect the battery from overcurrent if something shorts it. \$\endgroup\$
    – DKNguyen
    Commented Feb 19, 2023 at 18:36
  • \$\begingroup\$ Sorry, I might have not been clear. The charger will not be physically present anymore once it's charged. So I just get two wire 'in the air', not joined. Do you mean something like putting a diode ? \$\endgroup\$
    – cobdmg
    Commented Feb 19, 2023 at 18:39
  • \$\begingroup\$ Diodes don't stop short-circuits. In your case a fuse would technically work but still be sloppy. The correct approach is to use proper connectors that simply combine house both wires together so they aren't loose. Preferably a housing that can also be capped. Something like AndersonPowrepoles or any other number of similar connectors. A distant second is having two loose wires that can be capped. \$\endgroup\$
    – DKNguyen
    Commented Feb 19, 2023 at 18:54

1 Answer 1

Reset to default
1
\$\begingroup\$

In your case a fuse would technically work but still be sloppy. The correct approach is to use proper connectors that simply combine house both wires together so they aren't loose. Preferably a housing that can also be capped. Something like AndersonPowrepoles or any other number of similar connectors. A distant second is having two loose wires that can be capped.

\$\endgroup\$
2
  • \$\begingroup\$ Correct me if I'm wrong, but fuse interrupt very high current. If I know that my pcb would never have current higher than 5A, then I would like the fuse to explode at 10A. But when looking for fuse, their interrupting ratings exceeds the 100A, making me wonder whether they suit my application. \$\endgroup\$
    – cobdmg
    Commented Feb 20, 2023 at 8:33
  • \$\begingroup\$ @cobdmg A fuse interrupt ratings is not the current that the fuse blows at. It is the current that the fuse is able to stop when it does blow. \$\endgroup\$
    – DKNguyen
    Commented Feb 20, 2023 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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