# Charging 35V capacitor from 1,5V AA battery.

Discussion in 'The Projects Forum' started by LukeP, Aug 30, 2012.

1. ### LukeP Thread Starter New Member

Aug 29, 2012
4
0
I need to charge a 35V capacitor 1000μF,
starting from 1,5V AA battery.
The capacitor will be discharged into a solenoid, then the cycle restart.
Capacitor charging time must be around 6 seconds.
How i can do it saving board space and maximizing battery life?
Thanks.

2. ### Papabravo Expert

Feb 24, 2006
11,779
2,495
The 1.5 V battery will never charge the capacitor to 35 V.

If you multiply 1000e-6 times the charging resistor eg 1KΩ gives a time constant of 1 second. In this amount of time you will get to 63% of the driving voltage, 1.5V or about 0.945 volts. In 5 or 6 time constants you'll get much closer to 1.5 Volts but you'll never get all the way there. Discharging 1.5 Volts into a solenoid will result in a pretty wimpy response unless the solenoid is designed for that voltage.

3. ### LukeP Thread Starter New Member

Aug 29, 2012
4
0
i was thinking about a boost converter to raise the voltage up to 35V... to charge the capacitor.

4. ### JDT Well-Known Member

Feb 12, 2009
658
87
Bit of a sweeping statement, that! Not directly, of course. But with a flyback switching circuit it can be done. Think of a camera flash for example.

He doesn't say there is any limit to the current that can be drawn from the 1.5V supply. Designing an efficient switcher to work at 1.5V is not easy. But, I repeat, it probably can be done!

5. ### LukeP Thread Starter New Member

Aug 29, 2012
4
0
From various datasheet 1,5V AA batteries can take around 200mAh drawing without reducing battery life. i can double the battery pack, 2x1,5V AA, to have a initial range 2-3V to start with.

Disposable camera flash gain much more V from 1,5V AA (around 300V i think), but they have only 27 shots to do!
i want to do 1000 cycle from that 2x1,5V AA, charging 1000 times the capacitor.
A separate trigger circuit actuate the solenoid.
Then the capacitor restart to charge and so on.
I can also extend charging time from 6seconds to 10-12seconds, to limit excessive battery drain, but less time is better.

So fly back it's a good start?

6. ### JDT Well-Known Member

Feb 12, 2009
658
87
According to my calculations (and maths is not my strong point), to charge 1000μF to 35V requires
Q=C/V = 0.035 coulombs.

To do this in 6 seconds is an average power of 0.035/6 = about 5.8mW.

Sounds do-able.

I would adapt a circuit like this:-
http://ludens.cl/Electron/ledlamp/ledlamp.html

To get more voltage I would add extra turns between the collector and the diode. Experimentation needed!

7. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
Possible but very inefficient at such a low input voltage. Could do it cheap with a free running 555C timer driving an NPN with an inductor to boost up to 35V. The 1000uF cap needs an average current of about 6 mA to charge up to 35V in six seconds. That's roughly 200mW. It's possible.

Here is a similar design I made to generate 13V from 1.5V input.

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8. ### #12 Expert

Nov 30, 2010
18,093
9,683
The 13 volt zener will have to go and probably want a disable command (pin 4 to ground) when 35 volts is achieved.

LukeP likes this.
9. ### LukeP Thread Starter New Member

Aug 29, 2012
4
0
Thanks for imputs!
So, to start, i'll double the initial voltage: 2xAA, so initial range it's 2V-3V.
Yes, i need something to stop charging at desidered level.

10. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
510
http://www.datasheetcatalog.org/datasheets/1150/64052_DS.pdf

The LM3578 will run down to 2V input, it's a very simple and cheap simple boost regulator. You can just set it to regulate at 35V and it will never charge above that.

The 555C/NPN approach I posted will also work, just need a 35V zener to clamp the voltage. It burns more current since it's a shunt regulator. You would need to increase the amount of inductance until you get enough energy to charge the cap in six seconds or less. The inductor shown on the schematic is only 100 uH.

http://www.coilcraft.com/dt1608c.cfm?utm_source=supplyFrame&utm_medium=SEP

Last edited: Aug 31, 2012
11. ### themasterbait New Member

Jul 31, 2017
12
0

Disposable cameras use transformers wired in before the capacitor.
Discharge the battery using a pulsed output circuit and feed the energy into a transformer with more windings in the secondary coil (output coil) and use a diode to polarize the electricity generated by the electromagnet/primary coils pulsed field that induces an electrical potential within the secondary coil (the diode shouldnt be necessary because the electricity source is pulsed dc and not alternating current and the directions of the fields should follow the action described in the right hand rule of physics)

Use the voltage gains to charge the capacitor and use a resistor in the output circuit of the capacitor in order to increase discharge time if you want to have a continuous higher voltage output (which should be wired in parallel to the capacitor as to not increase the capacitor's recharge time).

Did you know that each flash has roughly 30 watts of energy? (315 volts * .1 amp =31.5 watts)

A single double a battery has a power potential during its entire lifespan of less than 10 watts.

(Statement removed by moderator. Overunity may not be discussed in this forum. -dc)

Last edited by a moderator: Jul 31, 2017
12. ### Alec_t AAC Fanatic!

Sep 17, 2013
8,995
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Welcome to AAC, tmb. Note that you have responded to a 5-year old post, so I doubt the thread starter will see this.

13. ### themasterbait New Member

Jul 31, 2017
12
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Its mainly for other stragglers to see. Im figuring use a 1:4 voltage gaining ratio of windings transformer and using ohms law, add a resistor with the corresponding ohm value and wattage requirements at the output.

For example,
10 volts pulsed dc or ac input at like .25 continuous amps for a total of 2.5 watts input into the coil - get like 40 volts output with an amperage of either.1 or .05 amps and use a 10 ohm wire wound resistor to gain 4 additional amps at the output without losing too much voltage because there is only a 10 ohm resistor at the output lol - use pulsed dc to store electricity in a capacitor, the amp gains and voltage level modifications at the output side of the transformer could make the capacitor charge up to the intended voltage level while accounting for charge voltage to stored voltage drops and the extra amperage would make the capacitor charge faster - use a self powered oscillator and a resistor at the capacitor's output to slow down the discharge rate and feed the input of the transformer a fluxuating source signal.

Transformers's coils are like 2 independant circuits and one of them generates an electromagnetic field while the other one acts like the windings in an electrical generator - b=e, the strength of the electromagnetic field determines the electrical potential of the field and varying wire sizes and turns per coil can influence the voltage and amperage of the output coil.

I don't understand why mankind is running short on energy sources sometimes.

14. ### ebeowulf17 Distinguished Member

Aug 12, 2014
2,624
505
Between the suggestion of free energy and the silly user name, I've got to assume this is trolling.

15. ### Bordodynov Well-Known Member

May 20, 2015
2,022
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See

crutschow likes this.
16. ### Colin55 Active Member

Aug 27, 2015
500
47
"Did you know that each flash has roughly 30 watts of energy? (315 volts * .1 amp =31.5 watts)"

This is a false statement.

The flash is only for a millisecond and the current is possibly much higher than that.

When you say "energy" you must use the term watt-second

17. ### themasterbait New Member

Jul 31, 2017
12
0
Watts per second is when you apply the total wattage used per time frame. Im not specifying a time frame, so instead im saying "watts" to apply the measurement to the whole instead of a specific time frame - im low balling the potential gains and high balling the variables that impact the total gains. I do that in order to account for other unaccounted for variables - such as the brand of aa battery.

Have you ever measured a disposable camera's flash circuit's capacitor's values after charging but before discharging?

They may or may not make disposable camera circuits with higher outputted values.

18. ### themasterbait New Member

Jul 31, 2017
12
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Do the math and measuring yourself - instead of assuming, do your own homework and prove me right or wrong.

19. ### themasterbait New Member

Jul 31, 2017
12
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Is that a self-oscillating power-magnifying circuit??

20. ### Colin55 Active Member

Aug 27, 2015
500
47
Watts per second

This is incorrect. It is watt-second.

You don't know your electronics terminology.