Need help on charging supercapacitor

[tattee]

Hello! I've been doing experiments with supercapacitor. I'm trying to integrate it to help prolong battery runtime.

I tried some scenarios like connecting the Supercap directly in parallel with the battery with no added components/circuit. The thing is it didn't help the battery at all rather it degrades it even more. After reading some articles i've found out that the in-rush current may have been the cause of my problem.

Now, im planning to add some components that would somehow limit the current draw from the battery using only simple components like resistor/transistor/mosfet/etc.. and at the same time charge the supercap fast. Ive seen some chips for supercap chargers but my budget is limited so I opt to use simple components if its possible.

From my experiments, (correct me if im wrong) i have noticed that the higher the current supplied to the supercap the faster it charges. So im planning to use darlington pair so that i can input small amount of current(therefore preserving the battery) and provide an ampified current at the output(thus, charge the Supercap fast).

Is my idea somehow effective or im not making any sense at all? I do appreciate any suggestions you want to share. This is my first time handling supercap and i dont have that much experience with electronics either.

Im using two AA rechargeable batteries(1.2V, 2100mAH) and 1 SuperCap (2.5V, 33F). My Load will take about 200mA-500mA. Im also planning to replace the SUperCAp with smaller value (0.1F, 2.5V) because it's taking too much time to charge this 33F supercap.

Thank you in advance!

 

[Torben]

From my experiments, (correct me if im wrong) i have noticed that the higher the current supplied to the supercap the faster it charges. So im planning to use darlington pair so that i can input small amount of current(therefore preserving the battery) and provide an ampified current at the output(thus, charge the Supercap fast).

Is my idea somehow effective or im not making any sense at all? I do appreciate any suggestions you want to share. This is my first time handling supercap and i dont have that much experience with electronics either.

Unless you have another power supply which you will use the Darlington to control (or I have totally misread your post), that won't work. A transistor cannot make current; it can only use a small current to control a larger one.

Here's some more reading on the topic, though:

**broken link removed**

As I understand it, supercaps are not generally used to prolong battery life, but rather are used to provide enough power to get a device through a short power outage.

Perhaps if you explained what your actual project is we could provide more help. Many times people come to the site looking for help with a part of a project, when the best answer would be to solve the problem using another method entirely.


Torben

 

[tattee]

I got your point on the darlington issue. ive been experimenting on it for days now and ive observed the same thing you just said. I thought darlington was use to amplify weak currents(my mistake).

Anyway, waht im trying to do is to compare battery runtime between two circuits. one with battery only and the other a battery parallel with supercap. both have the same LOAD. I was expecting the circuit with supercap will last longer but the opposite happpens.

I am concerned that my supercap (33F) is too big to charge for an AA 1.5V batteries(2pcs). In-rush current may be the issue.

 

[tattee]

here is what im going to do next:
1. replace supercap with 3F
2. use load which draws about 600mA

Im hoping the 3F supercap wont take too much time to charge if i place current limiting resistor to limit in-rush current. Also, i will try not to continously attach the load to the supercap to allow it to be charged by the battery. i will try to time it and pulse the circuit. I guess this time the 600mA pulsed circuit will hurt the battery only circuit and will benefit the battery w/ supercap circuit. waht do u think?

 

[ericgibbs]

here is what im going to do next:
1. replace supercap with 3F
2. use load which draws about 600mA

Im hoping the 3F supercap wont take too much time to charge if i place current limiting resistor to limit in-rush current. Also, i will try not to continously attach the load to the supercap to allow it to be charged by the battery. i will try to time it and pulse the circuit. I guess this time the 600mA pulsed circuit will hurt the battery only circuit and will benefit the battery w/ supercap circuit. waht do u think?

hi,
If you are charging the SCap via a switching or linear controller, what final voltage do you expect the SCap
to be charged to, using 2 * 1.2VBty in series.?

Morning Torben.

 

[picbits]

Supercaps tend to have quite a high leakage current so having one permanently across the battery terminals will cause a higher drain than not having one on the circuit at all.

 

[Boomslang]

If you are charging the super cap from the same batteries that's running your circuit, you will not get any benifit. All that it's doing is moving energy from one storage device to another, with some losses.

 

[Torben]

Morning Eric!

One thing I am wondering about: wouldn't the cap and battery both have to be perfect to even achieve unity here? The circuit's total charge will not be increased; some is just shifted into the caps. And I would think that the faster discharge of the cap would reduce the apparent life of the battery, since a lot of current is used to charge the cap but that charge is put back into the circuit much faster than the battery would have.


Torben

Edit: Ninja'd by Boomslang.

 

[ericgibbs]

Morning Eric!

One thing I am wondering about: wouldn't the cap and battery both have to be perfect to even achieve unity here? The circuit's total charge will not be increased; some is just shifted into the caps. And I would think that the faster discharge of the cap would reduce the apparent life of the battery, since a lot of current is used to charge the cap but that charge is put back into the circuit much faster than the battery would have.
Torben

Edit: Ninja'd by Boomslang.

hi,
I agree the losses would be as boomslang and picasm states.

The formala Q= CV .. where Q = I * t,
so t= [C * V]/I should give the OP the charge time.

BTW, you don't get Ninja'd with a Boomslang,,, they bite and then goodnite.

 

[Torben]

hi,
I agree the losses would be as boomslang and picasm states.

The formala Q= CV .. where Q = I * t,
so t= [C * V]/I should give the OP the charge time.

OK, then I'm not going mad. That makes sense.

BTW, you don't get Ninja'd with a Boomslang,,, they bite and then goodnite.

Ow. I just looked them up. I just thought "Boomslang" was a neat nickname. Nasty little devils. I'm glad the only things we have around here which can hurt you that badly are black widow and hobo spiders and cougars. Cougars you can generally avoid by not being a jogger.


Torben

 

[Boomslang]

BTW, you don't get Ninja'd with a Boomslang,,, they bite and then goodnite.

I only bite in the night.. (if you are a woman, weigh 60kg or less and are less than 30 years old)

 

[Mr RB]

here is what im going to do next:
1. replace supercap with 3F
2. use load which draws about 600mA

Im hoping the 3F supercap wont take too much time to charge if i place current limiting resistor to limit in-rush current. Also, i will try not to continously attach the load to the supercap to allow it to be charged by the battery. i will try to time it and pulse the circuit. I guess this time the 600mA pulsed circuit will hurt the battery only circuit and will benefit the battery w/ supercap circuit. waht do u think?

If I read you right this is a 600mA PULSED load??

You will get worse battery life with the supercap in circuit because the supercap will allow greater current into the load when pulsed, and will then place additional drain on the battery to recharge the cap after each pulse.

With the battery only in circuit, when a pulse occurs the battery's high internal impedance will mean the load recives less current during the pulse as the battery voltage will temporarily "sag".

So with the supercap your load will recieve more total current on average, and the battery will drain faster.

If you want to improve the experiment then use a "constant current" pulsed load that will draw very similar average current regardless of whether the supercap is in circuit.

If your goal is to reduce battery consumption then add a series resistor of the highest value possible that still allows the load to "function" each time, or better still an appropriate inductor if size and costs allow.

 

[tattee]

well, here is what ive read:
1. batteries drain/deteriorate faster with loads that draws high currents.(ex. camera flash)

what im trying to do is replicate this with LEDS in parallel which would equal to about 600mA or higher. When i said pulsed i mean like the camera flash wherein there is a time gap between the next shot.

Il try to charge the supercap during those time gap. and when i connect the LEDs, only the supercap will handle all the current drawn by the LEDs. also, i will place a current limiting resistor when charging the supercap to limit the in rush current.

With the battery alone, the battery will handle all the current draw which i hope will drain it fast.

 

[ericgibbs]

well, here is what ive read:
1. batteries drain/deteriorate faster with loads that draws high currents.(ex. camera flash)

what im trying to do is replicate this with LEDS in parallel which would equal to about 600mA or higher. When i said pulsed i mean like the camera flash wherein there is a time gap between the next shot.

Il try to charge the supercap during those time gap. and when i connect the LEDs, only the supercap will handle all the current drawn by the LEDs. also, i will place a current limiting resistor when charging the supercap to limit the in rush current.

With the battery alone, the battery will handle all the current draw which i hope will drain it fast.

hi,
How are you going to control the duration and current of the discharge pulse.?
What colour are the LED's.?

Did you see from the Q = C*V formula, how long its going to take to charge the SCap 33F, using two AA's.?

 

[tattee]

i havent figured out what i will be using to pulse the circuit.. maybe a 555 timer would do. do you have any suggestion?

Im not using the 33F anymore.. i am planning to replace it with the 3F from nesscap. I think it will be easier and faster to charge.

Btw, thank you for all who replied. i appreciate all your inputs! keep it coming..

 

[ericgibbs]

i havent figured out what i will be using to pulse the circuit.. maybe a 555 timer would do. do you have any suggestion?

Im not using the 33F anymore.. i am planning to replace it with the 3F from nesscap. I think it will be easier and faster to charge.

Btw, thank you for all who replied. i appreciate all your inputs! keep it coming..

hi,
I dont mean the pulse generation ie: 555, I mean the current switching device that connecting the LED's to the SCap, when it gets the pulse from the timer.?

What colour are the LEDs.????

 

[tattee]

ok.. u mean a relay or swtich transistor? il try the relay coz i dont think a transistor cant handle all the current. the LED are all red! just the ordinary kind (20mA i think).

 

[Boncuk]

Hello! I've been doing experiments with supercapacitor. I'm trying to integrate it to help prolong battery runtime.

I tried some scenarios like connecting the Supercap directly in parallel with the battery with no added components/circuit. The thing is it didn't help the battery at all rather it degrades it even more. After reading some articles i've found out that the in-rush current may have been the cause of my problem.

Now, im planning to add some components that would somehow limit the current draw from the battery using only simple components like resistor/transistor/mosfet/etc.. and at the same time charge the supercap fast. Ive seen some chips for supercap chargers but my budget is limited so I opt to use simple components if its possible.

From my experiments, (correct me if im wrong) i have noticed that the higher the current supplied to the supercap the faster it charges. So im planning to use darlington pair so that i can input small amount of current(therefore preserving the battery) and provide an ampified current at the output(thus, charge the Supercap fast).

Is my idea somehow effective or im not making any sense at all? I do appreciate any suggestions you want to share. This is my first time handling supercap and i dont have that much experience with electronics either.

Im using two AA rechargeable batteries(1.2V, 2100mAH) and 1 SuperCap (2.5V, 33F). My Load will take about 200mA-500mA. Im also planning to replace the SUperCAp with smaller value (0.1F, 2.5V) because it's taking too much time to charge this 33F supercap.

Thank you in advance!

You are talking about the wrong caps. These are no supercaps! A supercap starts at 500F and goes up to 600F, 700F is under development.

Also with the 33F cap observe the maximum allowed voltage very closely. They are normally rated 2.5V (not even 2.501V! perhaps BOOM.) To calculate for a current limiting device during charge use the fact that one F is the capacitance which allows 1A of current flow for a time of 1 second, with other words, to charge the 33F within one second you must apply 33A. Since the discharged cap behaves like a dead short it will draw all the current it can get thereby killing your battery.

Boncuk

 

[ecerfoglio]

..........

To calculate for a current limiting device during charge use the fact that one F is the capacitance which allows 1A of current flow for a time of 1 second, with other words, to charge the 33F within one second you must apply 33A. Since the discharged cap behaves like a dead short it will draw all the current it can get thereby killing your battery.

Boncuk

one F is the capacitance which allows 1A of current flow for a time of 1 second per volt of final voltage

with other words, to charge the 33F to 1 V within one second you must apply 33A. - To charge it to 2.5 V at 33 A uses 2.5 seconds, or else you need 82.5A ti charge it in one second. (82.5 = 33 x 2.5)

 

[Hero999]

Charging a capacitor without switching regulator to current limit is very inefficient because half the power is lost in the resistance of the capacitor and wiring.

Connect a resistor in series with a capacitor and calculate the energy dissipated in the resistor when the capacitor is charged and you'll find it's equal to the energy stored in the capacitor.

For effieicncy's sake you need to charge the capacitor using a constant current switching regulator.