Entire Forum This board This topic Members Entire Site
 Pages: [1] 2  All
 Author Topic: Wanted help with Capacitor based power supply  (Read 6657 times) 0 Members and 1 Guest are viewing this topic.
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « on: February 14, 2014, 01:20:25 13:20 »

As shown in the picture, I am using a simple bridge rectifier in series with an AC capacitor 470K 400v polyester radial lead capacitor. The load is simply some 40 led's in series. where i am getting 136 volts at the rectifiers DC out which glows the led's well.
IF i replace the radial polyester capacitor with a SMD ceramic chip 1210 package with 470K 500V one the output is 121volts.
Now i am using 1Mfd 500volts cermic chip capacitor in the same place but the output is 125 volts. In fact i am trying to get the same voltage (136Volts) what i am getting with local polyester one but with a SMD capacitor. Can anyone guide me where i am going wrong?
 Logged
hate
Hero Member

Offline

Posts: 555

Thank You
-Given: 156

 « Reply #1 on: February 14, 2014, 03:02:50 15:02 »

The output of this design is current limited rather than voltage limited. The impedance of the capacitor (ZC=1/(j*2*pi*f*C)) where f is the frequency of the input sine wave, determines the maximum current you can get as in I=V/ZC (V=input voltage rms). You can't get a fixed output voltage unless you implement some sort of voltage limiter at the output like a zener of desired rating. And beware that the output voltage in your sketch is fluctuating not constant.

So the capacitance of the cap changes the output voltage indirectly by changing the output current. If you will use this only for lighting up LED arrays, I suggest you decide what current you want the LEDs to sink and compute the capacitance of the cap according to that value.
 Logged

Regards...
Unhappy
Senior Member

Offline

Posts: 256

Thank You
-Given: 868

 « Reply #2 on: February 14, 2014, 03:18:52 15:18 »

The output of this design is current limited rather than voltage limited. The impedance of the capacitor (ZC=1/(j*2*pi*f*C)) where f is the frequency of the input sine wave, determines the maximum current you can get as in I=V/ZC (V=input voltage rms). You can't get a fixed output voltage unless you implement some sort of voltage limiter at the output like a zener of desired rating. And beware that the output voltage in your sketch is fluctuating not constant.

So the capacitance of the cap changes the output voltage indirectly by changing the output current. If you will use this only for lighting up LED arrays, I suggest you decide what current you want the LEDs to sink and compute the capacitance of the cap according to that value.

Dear hate, I started to reply and saw your reply to be more idealistic and explaining(if understood properly). I erased my reply before posting and pressed thankyou
 Logged
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #3 on: February 14, 2014, 03:46:58 15:46 »

@hate
Ok.to be precise, i am using 40 red color LED's in series as the load and practically i am looking / comparing at the brightness of the same with those connected with smd components and the through hole components. As per your point, the 1mfd cap should deliver more current thus more brightness than compared to that of 470nf. But here i am getting very small increase in brightness. On the other hand, the same load connected with 470K through hole radial lead capacitor is giving more brightness than the later. My aim is to get the same brightness with smd components what i am getting with through hole ones.
 Logged
zuisti
Senior Member

Offline

Posts: 400

Thank You
-Given: 241

 « Reply #4 on: February 14, 2014, 04:30:55 16:30 »

Hi;
The above is only possible if the SMD ceramic caps have much higher loss (internal) resistor compared to the good old polyester caps. What says the datasheet?
One solution could be: use more SMD caps parallel.
 Logged
hate
Hero Member

Offline

Posts: 555

Thank You
-Given: 156

 « Reply #5 on: February 14, 2014, 05:59:43 17:59 »

@hate
Ok.to be precise, i am using 40 red color LED's in series as the load and practically i am looking / comparing at the brightness of the same with those connected with smd components and the through hole components. As per your point, the 1mfd cap should deliver more current thus more brightness than compared to that of 470nf. But here i am getting very small increase in brightness. On the other hand, the same load connected with 470K through hole radial lead capacitor is giving more brightness than the later. My aim is to get the same brightness with smd components what i am getting with through hole ones.
First of all more current doesn't always mean more brightness. Let's say you're trying to drive single LEDs with 20mA rating, applying 100mA to these type of LEDs won't increase their brightness as in the current ratio. 1uF (I think you meant micro Farad (uF) by mfd) cap will deliver ~70mA rms under 220V, 50Hz and 470nF will deliver ~32mA rms. These two currents won't differ much in the brightness but will decrease the lifetime exponentially. And don't forget these values are rms values.

About the capacitor type, you should be very careful with the type of the capacitor here as it's the only component that stands in between the mains voltage and anything that uses it. There are X2 class capacitors for this job. The idea is they don't get closed circuit if they fail, namely they don't shorten mains voltage to the rest of the circuit. And any cap to be used for this purpose should be rated more then the peak voltage of the mains used (i.e. more than ~310V for 220V rms).
 Logged

Regards...
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #6 on: February 15, 2014, 06:16:07 06:16 »

I guess Zuisti has hit the same point which was in my brain too. The SMD is a "ceramic chip" whereas the through hole is a Polyester one..Hmm..i''ll have to experiment more on this further. Thanks every one.
Regards
Pushy
 Logged
h0nk
Active Member

Offline

Posts: 204

Thank You
-Given: 182

 « Reply #7 on: February 15, 2014, 01:02:02 13:02 »

Hello pushycat,

Multi-layer ceramic capacitors tend to vary their value with the voltage across them.
Take a look in the data sheet and You will see it.

And, i would not use ceramic capacitors for this. Even if they rated 500V...

Polyester with 630V and above or as hate says X2-class would be ok.

Best Regards
 Logged
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #8 on: February 19, 2014, 05:45:10 05:45 »

@h0nk
Now i am using Cermic chip capacitor 1210 package. Just wanted to know are polyester film caps available in same pacakage with 630Volts? A bit more bigger size with SMD will certainly affect my PCB deisgn but i am ready to change my PCB for that.
Regards
Pushy
 Logged
h0nk
Active Member

Offline

Posts: 204

Thank You
-Given: 182

 « Reply #9 on: February 19, 2014, 09:17:15 09:17 »

Hello Pushycat,

take a look here:
http://de.farnell.com/folienkondensatoren/nennspannung/630v/pg/310141112
(Capacitors made for snubber purposes may also fit.)

and here:
http://www.wima.com/EN/article.htm

You should consult the data sheet of the capacitor.
There may be spikes which may be much greater than the regular voltage.
E.g. when someone switches off a big electric motor.
The capacitor should not degrade when this condition arises.

X-Class Capacitors are special made to resist these conditions.

You should also think about some form of fail-safe security.
If You cant use X-Class, put 2 capacitors in series, so one capacitor may fail without doing much harm.

Best Regards
 Logged
solutions
Hero Member

Offline

Posts: 1818

Thank You
-Given: 647

 « Reply #10 on: February 19, 2014, 09:30:38 21:30 »

Your power factor is going to totally suck, but then the power draw is so low, who cares?

I think you want a self healing film cap, not a ceramic,to survive line spikes. The problem is that I know you are Mr Cheap and X-caps are rather expensive.

The two caps in series is a good idea. Short the first one with the first surge, short the second cap with the next surge

ezine - the LEDs do not function as a zener. They are a current controlled voltage drop. And, apart from recombination where you are getting decreasing increments of light per unit power, you will get a monotonic increase in brightness with current until the LED burns out

Your problem with the brightness is related to where you are in the AC waveform. You have a fairly high string voltage, so you will be sensitive in PHASE to when the LEDs actually turn on. Think of it as a TRIAC dimmer, or sinewave PWM if you must.
 Logged
enzine
Junior Member

Offline

Posts: 81

Thank You
-Given: 516

 « Reply #11 on: February 19, 2014, 10:35:30 22:35 »

@solution:
the LEDs do not function as a zener
I apologize for having expressed in a very unhappy but I only meant to simplify the operation of 40 LEDs

Of course, the series of 40 LEDs works like  a zener diode giving a voltage drop of 1.8x40V (the Zener diodes works in reverse breakdown conduction...)

@pushicat
If You wants to realize a commercial product, on mains  is MANDATORY the use of X2 capacitors.

To minimize  risk of electrical shock ,
might be a good idea  to use two X2 capacitors ( 2 x value!) in series of  each live and neutral.

Ciao
 Logged
solutions
Hero Member

Offline

Posts: 1818

Thank You
-Given: 647

 « Reply #12 on: February 19, 2014, 11:13:14 23:13 »

It also works like a battery. I also works like a resistor. It works like many things if you are talking about a device having a voltage drop across it.

In the context of diodes, though, the LED is in forward conduction, a zener is in avalanche breakdown. In avalanche, the Vbr is pretty much constant. in an LED, the IV curve is steep, but you do get a higher Vf as current increases.

ZENER (operates on negative bias - left side of graph):

LED (operates forward boased - right side of graph):

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.

In engineering, we speak precisely. When you say something behaves like something, it should. A zener doesn't. At all. No matter what your simulator says.
 Logged
mexpcb
Active Member

Offline

Posts: 113

Thank You
-Given: 33

 « Reply #13 on: February 20, 2014, 12:48:23 00:48 »

we as engineers sometimes take a look at thinks on the hard way but sometimes there is an easy way to solve problems...

i just look at the original post and see that mainly you wanted to power up the leds but, as some one says on AC the DC block caps is just acting as an impedance device (Reactance) but also is blocking some possible noise coming from the AC source, but i don't really see why you need a cap in series with the bridge and the input, at least you wanted to do something as described before, as solution mention the Power Factor. anyway...

if you see the circuit, you need a current limiter resistor, when you first power on the leds, there will be a high speed change of voltage (F goes high), as you may know caps do like voltage changes, then at very tiny time your cap act like a short circuit and the Xc is not what you calculate, someone mention they it may be already damage and that is what i think..

i suggest to do your calculations again, and just take a look to the LED current needed and calculate a series resistor and if you will use the DC block cap just take the same equations somebody provide you to calculate the Xc...

if this did not make sense i may be misinterpreting things here..

 Logged
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #14 on: February 20, 2014, 05:30:06 05:30 »

@enzine
The cost of X2 capacitors will too count in my case as: If instead using SMD capacitors, i am using through hole polyester capacitors rated 400V in the same place and 1N4007 x 4Nos to lightup same array of led's with good brightness. The only experiment was to reduce the size but naturally at a bit more price of SMD's

@h0nk
Thanks for the links. I procured the capacitors from a a small firm so even he has a bunch of similar caps in a box where no details of any sort are found.

The circuit is widely used in LED's for decorating purposes where a dedicated genset will be used. Where as power consumption is no matter in comparison to those old 6.2volt filament bulbs in series which used to really suck the power.

Another silly point flashed in my brain..that is while we are using a bridge rectifier and the output is an array of led's connected in series, the voltage can be boosted by connecting a capacitor (I used the same SMD 1Mfd what i used in AC side) to the output and the brightness was quite sufficient to lightup 45 white LED's in series. Its kept On since a day now without any heating and failures. So far so good..
 « Last Edit: February 20, 2014, 05:36:22 05:36 by pushycat » Logged
enzine
Junior Member

Offline

Posts: 81

Thank You
-Given: 516

 « Reply #15 on: February 20, 2014, 07:07:04 07:07 »

@pushicat:

@solutions:
 Logged
hate
Hero Member

Offline

Posts: 555

Thank You
-Given: 156

 « Reply #16 on: February 20, 2014, 09:01:00 09:01 »

i just look at the original post and see that mainly you wanted to power up the leds but, as some one says on AC the DC block caps is just acting as an impedance device (Reactance) but also is blocking some possible noise coming from the AC source, but i don't really see why you need a cap in series with the bridge and the input, at least you wanted to do something as described before, as solution mention the Power Factor. anyway...
The cap here is used as a current limiter by taking advantage of its impedance (almost all of it comes from its reactance). The advantage over a series resistor is that the cap spends very little power while limiting the current. If a series resistor would be used to limit current, it would dissipate lots of power for even very low currents. It's a transformerless and practical design for low currents.

if you see the circuit, you need a current limiter resistor, when you first power on the leds, there will be a high speed change of voltage (F goes high), as you may know caps do like voltage changes, then at very tiny time your cap act like a short circuit and the Xc is not what you calculate, someone mention they it may be already damage and that is what i think..
A thermistor (NTC) can be used in series to the cap to surpass 'surge current'. But there is one case to consider if a NTC will be used. It is where the power switches off and on again before the NTC cools down where the NTC will still have low resistance incapable to limit any surge current.
 Logged

Regards...
mexpcb
Active Member

Offline

Posts: 113

Thank You
-Given: 33

 « Reply #17 on: February 20, 2014, 04:48:08 16:48 »

i was trying to explain but when and did a web search just to see if someone already explain it, i find out a good tutorial of a very similar circuit...

just search on youtube "Tutorial: Electrical impedance made easy" and see part 1 and part 2 from (Ben Krasnow)...

if you just want to see what i mean adding an extra resistor, go to first part at 13:27...

then you will see and understand better what i try to write before...

i hope it helps..

Regards
 « Last Edit: February 20, 2014, 05:04:09 17:04 by mexpcb » Logged
hate
Hero Member

Offline

Posts: 555

Thank You
-Given: 156

 « Reply #18 on: February 20, 2014, 10:24:02 22:24 »

mexpcb: I thought you were talking about replacing the cap with a resistor but I now get what you mean. Still a series resistor is not a good design imo as it will dissipate a certain amount power all the time and there is the fact that resistors also have a maximum surge current rating which can also break them when introduced to surge currents above their maximum. An NTC solution would fit better imo.

Last but not the least, diodes also have surge current ratings.
 Logged

Regards...
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #19 on: March 15, 2014, 07:18:51 07:18 »

Yes..i do agree with the reply of "hate" to "mexpcb". I have finalised the above said deisgn and its all working fine on my test desk since a week now. The design is simple and the same one as shown in my first post except i have added a 100E 0.5watt carbon film resistor in series to the Load and the output of the bridge rectifier is connected with 1Mfd/500V non polarised SMD capacitor. Thanks every one who gave their valuable suggession and inputs here.
Regards
Pushy
 Logged
mexpcb
Active Member

Offline

Posts: 113

Thank You
-Given: 33

 « Reply #20 on: April 01, 2014, 10:57:39 22:57 »

well done, if you can just provide the circuit for all the people or images or results we can all see what you did and new people coming here can see what it was at the beginning and what did you did and what is the final result...

regards
 Logged
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #21 on: April 03, 2014, 07:01:32 07:01 »

The Circuit is quite simple with just 3 components and is 99% same as seen in my first post except the capacitor is connected parellel with a 560K resistor for safe discharge and the 100E 0.5 watt is connected on the load side where the DC is fed. rest all are assembled on a double side PCB. Picture attached.
 Logged
nordiceng
Newbie

Muted
Offline

Posts: 29

Thank You
-Given: 1

 « Reply #22 on: April 03, 2014, 07:34:59 07:34 »

why you are using ac capacitor, most of the rectifier ccts does not require this capacitor .please more details on this circuit and what you are trying to do
 Logged
robotai
Junior Member

Offline

Posts: 60

Thank You
-Given: 27

 « Reply #23 on: April 03, 2014, 09:46:41 09:46 »

I suppose you are using 220V AC as the input.
IMO. Add about 5uF capacitor in parallel with your LED series, which should help protect your LED from the spike on turning on/off the power switch.
 Logged
pushycat
Senior Member

Offline

Posts: 448

Thank You
-Given: 198

 « Reply #24 on: April 07, 2014, 07:13:28 07:13 »

@....nordiceng