Charge 18650 with BCAP0050 2.7V 50F is possible?

[fer344]

Can I put a resistor and control discharge of BCAP0050 2.7V 50F in order to charge 18650 battery?

 

[berkeman]

Can I put a resistor and control discharge of BCAP0050 2.7V 50F in order to charge 18650 battery?

Welcome to the PF.

It helps others a lot if you define your terms better so we don't have to go searching for them to decode your post. It looks like you want to use a 2.7V capacitor to charge a single Lithium Ion battery cell. Is that right?

First, you did not say what voltage your capacitor will be charged to. If it is rated at 2.7V, you need to stay below that voltage by some margin, or you can damage the capacitor.

Second, you would not use a voltage dropping resistor in a battery charging circuit. That is just too wasteful of the energy. You would use a DC-DC switching voltage regulator to transform the source voltage to the battery charging voltage.

Finally, as we know from many recent battery fires, charging a Lithium Ion battery is not something to be done by amateurs.

What is the application?

 

[Rive]

The charged capacitor will be full at 2.7V, while the empty battery will be at 3.3V (in case it is a standard Li-Ion cell). So whatever you do, the current will flow backward, from the battery to the capacitor if connected with a resistor.It'll overload the cap in no time.

The energy stored in the capacitor is around 180Joule, the energy stored in one 18650 cell is ~ 30kJoule (I hope I did the math right). Less than 1%. Even if you can push all the juice in the capacitor into the battery (you can't, and even partial discharge will require complicated electronics due the voltage difference) it won't do much.

 

[fer344]

The charged capacitor will be full at 2.7V, while the empty battery will be at 3.3V (in case it is a standard Li-Ion cell). So whatever you do, the current will flow backward, from the battery to the capacitor if connected with a resistor.It'll overload the cap in no time.

The energy stored in the capacitor is around 180Joule, the energy stored in one 18650 cell is ~ 30kJoule (I hope I did the math right). Less than 1%. Even if you can push all the juice in the capacitor into the battery (you can't, and even partial discharge will require complicated electronics due the voltage difference) it won't do much.

Welcome to the PF.

It helps others a lot if you define your terms better so we don't have to go searching for them to decode your post. It looks like you want to use a 2.7V capacitor to charge a single Lithium Ion battery cell. Is that right?

First, you did not say what voltage your capacitor will be charged to. If it is rated at 2.7V, you need to stay below that voltage by some margin, or you can damage the capacitor.

Second, you would not use a voltage dropping resistor in a battery charging circuit. That is just too wasteful of the energy. You would use a DC-DC switching voltage regulator to transform the source voltage to the battery charging voltage.

Finally, as we know from many recent battery fires, charging a Lithium Ion battery is not something to be done by amateurs.

What is the application?

Thanks for response! :)

My interest is on testing batteries, want to control charge-discharge but have not enough knowledge.

 

[berkeman]

My interest is on testing batteries, want to control charge-discharge but have not enough knowledge.

Well, especially with Lithium Ion batteries, it would be best if you could find a local person to Mentor you in your learning about battery technology. It can be a bit dangerous to try to figure this out on your own, especially if you don't yet have an EE degree.

 

[fer344]

My doubt born here: https://www.unitjuggler.com/convert-electriccharge-from-Fd-to-Ah.html

I see that I could transform between Farad and Ah, but not really... instead Farad was Faraday...