# Genius! Please tell me the theory which apply on this

Discussion in 'The Projects Forum' started by ranatungawk, Dec 1, 2012.

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1. ### ranatungawk Thread Starter Senior Member

Oct 30, 2008
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I found this circuit in a cheep rechargeable LED touch. However, as far as I could understand, I saw an issue of this circuit (other than the safety issue). So genius! Please tell me what I think on this, is correct!!

In order to drop the voltage, a Mylar Capacitor is used. Further, the Ni-Cad battery (3.6/700mA) will be charged with 4V/70mA source. However as we know, when the battery-charge go up (after 1-2 hours), the drain current from the source go down (<70mA). However, at the start, the charging current and the voltage was calculated according to V=IR and then a capacitor was selected in order to fulfill the above charging-values. So according to this argument, after 1-2 hours, charging voltage on the battery must be high than starting 4V (Ex; if I=30mA then voltage on the battery should go up to 134V). can this be happened ?? Can it make any harm to the battery?

Please see the attach diagram

Last edited by a moderator: Dec 2, 2012
2. ### ErnieM AAC Fanatic!

Apr 24, 2011
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Don't built this circuit.

If you do build it, don't turn it on. You'll put your eye out.

Seriously, it is very dangerous.

3. ### Audioguru New Member

Dec 20, 2007
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The 230VAC mains is a sine-wave. Its peak voltage is 325V and spikes from appliances cause it to go higher. Then the 250V capacitor will probably blow up or catch on fire.

If the capacitor survives then the battery will continue to be overcharged and might burst. A normal battery charger limits its voltage, detects when the battery is fully charged then shuts off.

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4. ### DickCappels Moderator

Aug 21, 2008
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Remember that since you are using a half-wave rectifier, on average the charging current will only be 70 max x 50%. To get the full 70 ma, use a full wave bridge.

If its a good battery, the voltage one the cell won't get much above 1.4 volts or so.

I agree with Audioguru that the capacitor voltage rating is not high enough. If that's the AC voltage rating of the capacitor, then the voltage rating is in RMS which takes the peak into account. But 250 does not leave sufficient margin -in my town, I have seen the mains run up to 250 V. Its a good idea to use a self-clearing film capacitor such as those sold for X cap and Y capacitor use in EMI filters with a higher voltage rating like 270 VRMS.

The circuit as drawn lacks safety features. As ErnieM said, it might put your eye out. It could use about 100 ohms in series with the AC input to limit inrush current to a few amps, a DPDT switch to completely disconnect the circuit from the mains and a mechanical interlock of some sort so that you can't open the door to the battery compartment unless the switch is in the off position. A little easier but easier to defeat would be to have a mechanical design that made the battery compartment inaccessible when the charger is plugged into a wall outlet. Yes, you should have a batter compartment so that in case the battery explodes, you don't have shrapnel flying all over the place. That's how it could put your eye out!

And be really careful around mains voltage -keep one hand in your back pocket and stay insulated from the floor. Good luck.

Last edited: Dec 2, 2012
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5. ### bertus Administrator

Apr 5, 2008
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Hello,

I am closing this thread as it violates AAC policy and/or safety issues.

Quote:
6. Restricted topics. The following topics are regularly raised however are considered off-topic at all times and will results in Your thread being closed without question:

• Any kind of over-unity devices and systems
• Automotive modifications
• Devices designed to electrocute or shock another person
• LEDs to mains
• Phone jammers
• Rail guns and high-energy projectile devices
• Transformer-less power supplies
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Bertus

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