Hello,

I have a battery-powered PCB that stays on even after the battery is removed. This is because a super cap (250 F) is placed in parallel to the battery. I am trying to come up with some rough ideas on a circuit that could discharge the super cap once the battery is removed. The battery is 2-lead. I am thinking a transistor that is placed in parallel to the battery/capacitor with a series, current-limiting resistor.

The problem is I'm not sure how I would "turn on" the transistor (make it conduct) only when the battery is removed. I'm not sure what I would connect the gate/base to. I can not connect it to the battery rail, as when the battery is unplugged, the voltage remains due to the super cap.

This issue could be solved with a mechanical solution. It would be something to do with the connector, maybe - physically removing the battery would short the super capacitor's terminals (with a series resistor between, of course). Or something like this.

However, I am curious if there could be a solution to this in the form of a circuit. Has anyone come across a similar problem before?

Does anyone have any ideas?

 

Does anyone have any ideas?

Nothing very sophisticated needed. https://en.wikipedia.org/wiki/Bleeder_resistor

 

Hello,

I have a battery-powered PCB that stays on even after the battery is removed. This is because a super cap (250 F) is placed in parallel to the battery.
Does anyone have any ideas?

250 Farad, seriously?
Do you need the super cap?
If not just remove the cap

 

Why do You want to discharge the Super-Cap ?
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How far do you want the capacitor discharged?

 

A super-cap is typically used to maintain a voltage if the normal power supply is removed. As per post #4, why would you want to deliberately discharge it?

 

Or another question, why do you have a super cap in parallel with the battery in the first place?

 

Hey folks,
The capacitor needs to be in parallel with the battery because it aids the battery's lack of electric current supply capability. The PCB has a transceiver that demands approximately 2A when it is transmitting (transmission time is very brief, but it is enough to reset the board). The battery is not suit for delivering that much current. I am not able to change the battery, unfortunately.

I do not need to drain the super cap, I was just thinking that could be a solution. Here is a transistor network I have been thinking about. It essentially aims to connect/disconnect the super cap from the rest of the circuit. The circuit does have problems, though. If Q1 is N-type, VGS=0 which is not above a typical transistor's threshold voltage. If Q1 is P-type, the body diode will always conduct, leaving the capacitor connected to the load.

 

I can't think of any reason for removing the Super-Cap from the Circuit, ever.
I should be permanently soldered-in.
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I can't think of any reason for removing the Super-Cap from the Circuit, ever.
I should be permanently soldered-in.
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This is a device that needs to be placed in an "alarm" state. For safety reasons, the device can only be taken out of this state by rebooting the device (disconnecting power). This is done by removing the battery, then re-inserting it.

Unfortunately, the super capacitor keeps the circuit on even after the battery is removed. For this reason, I am trying to develop a circuit that disconnects the super cap from the entire circuit when batteries are removed.

 

2A at 3.6V. That is 7.2W!!!
What are you powering that takes 7.2W?

 

2A at 3.6V. That is 7.2W!!!
What are you powering that takes 7.2W?

During a transmission, correct. 7.2 W for approximately 100 ms. Actually it is 6.6W because the transceiver is powered by 3.3V (the battery is bucked down slightly).

 

What about a 2 step reboot?

Remove battery
Press button

 

What about a 2 step reboot?

Remove battery
Press button

This is my backup plan but I am trying to see if a transistor topology can achieve what I desire. But yes, this is definitely a valid solution that I am keeping in my arsenal should I not achieve a circuit that disconnects the super cap.

 

A 2A dummy load connected in place of the removed battery would drop the cap volts from 3.3 to less than 1 in 5 minutes.

 

How about using a Normally-Closed Push-Button, right next to the Re-Boot-Button ?
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If anyone is curious, the solution that is being implemented is a small signal latching relay. The device is battery powered, so a normal relay that requires a constant current flowing through a coil is a no-go. Telecom latching relays are pretty neat: just hit the coil with a pulse and the relay will latch to that position. Hit the coil with the opposite polarity to change the switch position. Here is a diagram: