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Author Topic: Wanted help with Capacitor based power supply  (Read 8791 times)
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user77
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« Reply #25 on: May 10, 2014, 10:27:00 22:27 »

I would suggest you to put a resistor 10-100 ohm in series to circuit so as to limit the current at power on and avoid high peaks and spurious on the power network.
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pushycat
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« Reply #26 on: May 11, 2014, 08:35:01 08:35 »

@ nordiceng
My friend, Instead me writing the whole story again, would you please read the full thread with my posts and all the replies. Even then you have some doubts aprt from this, you may feel free to post....Smiley

@ robotai
Yes i am using this circuit on 220V AC. Connecting a 2.2Mfd with parelllel the output does nothing (I expected that this will boost the voltage) as the ceramic SMD capacitor is not allowing this due to its high internal resistance. The brightness increased when i soldered two capacitors in parallel but using single capacitor of the same value didn't work.
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robotai
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« Reply #27 on: May 13, 2014, 05:32:59 05:32 »

@ robotai
Yes i am using this circuit on 220V AC. Connecting a 2.2Mfd with parelllel the output does nothing (I expected that this will boost the voltage) as the ceramic SMD capacitor is not allowing this due to its high internal resistance. The brightness increased when i soldered two capacitors in parallel but using single capacitor of the same value didn't work.

The capacitor paralleled with output has 2 effect.
1. It helps absorb the voltage spike when turning on/off the circuit.
2. It helps smooth the current flow on the LED series.
Since you are using Triac to control the ON/OFF for this circuit, it should prevent it from voltage spike already. If you didn't use any capacitor in parallel with LEDs, the current on LEDs may vary from 0~50mA. Adding the capacitor should smooth the current more linearly into steady 30mA.

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DreamCat
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« Reply #28 on: May 19, 2014, 06:07:57 06:07 »

Notice from Dreamcat:

If you want to use AC capacitor as step-down element, and if you want chose this kind of capacitor made from China, and use it in AC 220V Grid area. please stop!

reason is very simple, the AC capacitor will be damaged very easy, and other component also will be damaged cause by it.

So, I suggest you had better chose good capacitor as drop-down element, and improve your circuit. for example, use NTC reduce the current within power on state, use TVS or other overvoltage protection device reduce the possibility of capacitor's damaged.

you may know, this kind circuit is not suite to use with Serial Voltage regulator. so, don't use 7805 with it.

btw, this kind circuit can only provide small current. if you want use it in a big current, you can use a SMPS as next stage.

but , why not use a SMPS directly?

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May be I expressed the wrong meaning, sorry for my bad english. Please correct it for me if you can.
user77
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« Reply #29 on: May 19, 2014, 01:53:45 13:53 »

E12 bulb holders it is very little, but may be it's possible to unmount copper contacts from plastic holder, create a little pcb to be mount instead of the plastic holder, you can weld the lamp contacts and components two wire on pcb, replace diode bridge with a simply diode, may be this solution permits you to use the bulb holder
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pushycat
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« Reply #30 on: June 03, 2014, 07:22:01 07:22 »

@ user77
Yes i am using 100E 0.5watt in series to the led's for additional protection.

@DreamCat
SMPS is a bigger option but safer one. The size of this circuit is so small, that it powers 45 Nos of 8mm white led's in series and no SMPS is availble of that capacity in this size. Of course i am keeping the safety issue apart.
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zac
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« Reply #31 on: June 04, 2014, 09:51:28 21:51 »



So, an LED string does not act like a zener diode unless you forward bias the zener, then the zener acts like a forward biased diode. At a given current, an LED  has a voltage drop, sure. But so does a resistor.

The description of an LED as a zener like device was perhaps not the most accurate description, but I understood what he/she meant.  As you can see from the IV plot, each type of LED has a characteristic voltage drop in forward bias.  This characteristic allows LEDs to be used as crude voltage references.  The IV curve is not a straight line so is not like a resistor.   
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Signal
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« Reply #32 on: June 27, 2014, 03:52:18 15:52 »

Can anyone guide me where i am going wrong?
You make common mistake of novices - you miss the fact that real electronic component is not identical to electric parameter it stands for. So you need to read datasheet for _each_ component you use and understand meaning of _each_ parameter it states. Otherwise - endless surprises are guaranteed. If we focus on "illogical" voltage levels while changing capacitors in your case the dielectric material is the key factor. Read this great article for beginning: http://www.maximintegrated.com/en/app-notes/index.mvp/id/5527 .

Second. When you receive wise advises like one that robotai gave the correct way to handle it - to understand what are they about, not to dismiss them just because predicted result seems to you as unmatching. See attached presentation of mentioned factor (LTspice is used for simulation). Note#1: Serial current limiting resistor compensates initial peak through LEDs only partially. Note#2: When you get measured value of voltage or current it is important to know what it represents: peaks, average amplitude (DC component) or RMS.

It's better to ensure that: (a) your LED can live with such short over-current (from datasheet! not from your experience) or that (b) your switch is slow enough or (c) activates only near zero-crossing or (d) you have other hidden current limiter. Or use "capacitive voltage divider" to have a reason to expect long life of your device.

Any other advices seem to me redundant because objective of your device is not defined enough. If the main target is "I like the flicker I've got" then it's done Wink
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