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jacks
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« Reply #80 on: December 23, 2009, 11:19:03 AM » |
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Bihira yata talaga sir ganyan sa Raon. Try mo sir kung meron sa fan capacitor. Sa Deeco naman sir medyo oblong ang shape base sa spec karaniwan yellow color sya. Or talagang mangangahoy sa mga PSU, dami nun sa Raon.
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« Reply #80 on: December 23, 2009, 11:19:03 AM » |
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insomartin
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« Reply #81 on: December 23, 2009, 11:38:49 AM » |
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hmmm
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jeff0labs
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« Reply #82 on: January 18, 2010, 12:08:59 PM » |
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mas maganda kung may series na capacitor para mababa ang losses. kung magpapa-ilaw ng 20mA LED sa 220V ng resistor lang, masyadong mataas ang power na kailangan.
mas maganda rin kung gagawing series resistor ang computed, madalas kasi hindi given ang rated voltage ng mga resistor kaya safer kung series (half nung computed x 2).
usually 600VDC ang rating ng mga resistors (mof afaik). check niyo datasheet to be sure.
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winn1971
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« Reply #83 on: January 19, 2010, 12:55:15 PM » |
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best p rin talaga kung series capacitor instead na series resistor ang gagamitin. i had my led porch light in operation for few weeks tapos biglang di na lang gumana one day. i thought may bumigay na led (since series connected sila, kahit 1 lang masira, inoperable na yung buong unit) but when i opened it "fried" yung 3 series resistors na ginamit ko to drop the voltage going to the FWB although over-rated pa yung ginamit ko as in 1W each though based the cals ko e pwede na kahit 1/2W lang. todate still can't find a source/supplier for new 250Vac 0.22uF caps kaya tyaga sa pangangahoy from old pc psu, monitors and tv.
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rdpzycho
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« Reply #84 on: January 20, 2010, 11:43:54 PM » |
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^ merong nabibilhan niyan sir somewhere in Raon ('di ko lang alam kung saan banda) kasi gumagamit kami dati niyan sa inverter. IIRC, 630V pa 'yung nabibili ng procurement, though hindi siya rectangular (baka 'yung X caps ang tinutukoy niyo sir, sa RS at Farnell lang may local supply nun), either Mylar or PPE, mostly Mylar.
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zer0w1ng
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« Reply #85 on: January 21, 2010, 07:06:35 AM » |
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Try sourcing from motors/fans component suppliers. These caps are usually used as starting capacitors.
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piona
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« Reply #86 on: January 21, 2010, 01:05:29 PM » |
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@ Winn
dapat siguro 400V or 600V ang rating ng caps mo dahil 220Vac tayo..... hindi 120Vac
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zer0w1ng
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« Reply #87 on: January 21, 2010, 01:13:53 PM » |
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Right. 220V AC is 622V peak to peak.
220ACrms x 2 x 1.414 = 622Vpp
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piona
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« Reply #88 on: January 21, 2010, 01:23:08 PM » |
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^^^thats it! caps are rated in pk-to-pk pagdating sa ac, dapat pala mga 1kv na hehehehe....
@ Winn
try nyo dalawang ceramic na 0.1uF/1kv parallel (0.2uF total), marami nyan sa mga electronics store
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winn1971
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« Reply #89 on: January 21, 2010, 01:28:10 PM » |
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^^^thats it! caps are rated in pk-to-pk pagdating sa ac, dapat pala mga 1kv na hehehehe....
@ Winn
try nyo tatlong ceramic na 0.1uF/1kv parallel (0.3uF total), marami nyan sa mga electronics store
thanks for the suggestion bro. hopefully will also fit sa space. will try & let u guys know.
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winn1971
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« Reply #90 on: January 21, 2010, 01:43:38 PM » |
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Right. 220V AC is 622V peak to peak.
220ACrms x 2 x 1.414 = 622Vpp
Nope. 220Vac rms is 311Vac pp. 240Vrms is 339V pp. No need for the "2" as multiplier. Thus, a 400V cap is good enough but since hard to find yung size na required e i used what is available on hand. not to mention na same rated cap is used sa line side ng old pc psu connected sa 230Vac input plug. it's a 250Vac rated cap. Also, though not shown sa markings nung capacitor, I believe the given Vac rating is the max line voltage application. If i would have used a DC rated cap, then in no way would i go for lower than 400Vdc cap. I could be wrong, just in case, pls correct me. tia.
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marcelino
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« Reply #91 on: January 21, 2010, 02:00:45 PM » |
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zer0w1ng
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« Reply #92 on: January 21, 2010, 02:05:28 PM » |
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Nope. 220Vac rms is 311Vac pp. 240Vrms is 339V pp. No need for the "2" as multiplier. Thus, a 400V cap is good enough but since hard to find yung size na required e i used what is available on hand. not to mention na same rated cap is used sa line side ng old pc psu connected sa 230Vac input plug. it's a 250Vac rated cap. Also, though not shown sa markings nung capacitor, I believe the given Vac rating is the max line voltage application. If i would have used a DC rated cap, then in no way would i go for lower than 400Vdc cap. I could be wrong, just in case, pls correct me. tia.
Please read http://en.wikipedia.org/wiki/Alternating_current . Vpeak is half of the Vpeak-to-peak. Peak to peak is doubled thus the 2 multiplier. Please also note that it is clearly peak-to-peak voltage that I mentioned.
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macky
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« Reply #93 on: January 21, 2010, 02:18:20 PM » |
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ano kaya ginamit ditong cap? bipolar di ba? http://www.discovercircuits.com/PDF-FILES/acwhiteleds.pdfparang pwede na yung mga caps gamit sa CCFL tubes na uso ngayon, dami ko na nakolektang pundidong ccfl, kinakahoy ko ang 13001 xsistor, ganyan din mga caps rated @400VDC ganito:  eto pa: 
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pENdot
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« Reply #94 on: January 21, 2010, 03:28:28 PM » |
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in theory side... and BTW, i have experimented on actual, ganito lang naman.
capacitors have this resistive characteristics when applied to AC signals. we call this reactance or capacitive reactance which we can obtain as
Xc = 1/(2*pi*f*C); where Xc = capacitive reactance, f = frequency and C = capacitance.
As an example, we would like to lit a single LED on a 220Vac, the LED's forward voltage and current are 3V and 20mA, respectively. 3V is negligible to 220V so we ignore the forward resistance of the LED.
so the resistance need is:
Rx = 220/20mA = 11kohms. pwede naman resistor to... pero we would need a bigger power handling resistor for this. the next higher standard value is 12k. I pick next higher to be safe.
Power desipated on resistor = I^2R = 20mA^2 * 12k = 4.8watts... so we will be needing at least 5watts rated resistor. but expect that heat will build up. NOTE: 4.8watts heat is like 20% heat of a 25watts soldering iron.
If we don't want this heat, lets utilize the capacitive reactance. so we want a 11kohms resistance to limit the circuit to 20mA.
from the equation of capacitive reactance, we could calculate the capacitance this way...
C = 1/(2*pi*60*11k)= 241nF. <---- we simply put this in series the LED. we also add a resistor 10% of the resistance we wanted. 11k*.1 = 1.1k (kahit 1/2watts nalang... 1.1k *20mA^2).
****NOTE. this calculation if for supply voltage of 220V ONLY. i would pick the calculation for 240V... pasiguro!
theoritically and practically! go! marce go!. heheheh, from the simple question nagiging complicado but still for the purpose of safety. un lng naman. simple question but w/ the help of techs and engineers here nagiging klaro ung answers. thanks guys!
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winn1971
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« Reply #95 on: January 21, 2010, 03:42:53 PM » |
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Ok, agree on the clarification made, re: Vp and Vpp, thanks. but again, since we are discussing here the component (the cap) in an ac circuit, and the said circuit having a "sine" wave, whether, pure, distorted, etc. then your +Vpeak never coincides with your -Vpeak at any given time. The component will be exposed to only one peak at any given time. Therefore, no point using Vpp, which is 2Vp, in considering the max exposure limit of a cap or any electrical device for that matter. The only time you get to see a Vpp (or 2Vp) is when you attempt to sync two ac power sources at 6 o'clock.
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zer0w1ng
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« Reply #96 on: January 21, 2010, 03:58:42 PM » |
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It will be exposed to 2xVp.
For example, at 90degrees to 270degrees of the sine wave will be -622VDC. And from 270degrees to the next cycle at 90degrees is +622VDC.
If you have doubts on my explanation, you could try this in actual component/circuit.
BTW, I am hesistant like to reply again to this for it may cause flaming but I should correct this wrong notion.
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winn1971
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« Reply #97 on: January 21, 2010, 04:52:26 PM » |
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It will be exposed to 2xVp.
For example, at 90degrees to 270degrees of the sine wave will be -622VDC. And from 270degrees to the next cycle at 90degrees is +622VDC.
If you have doubts on my explanation, you could try this in actual component/circuit.
BTW, I am hesistant like to reply again to this for it may cause flaming but I should correct this wrong notion.
don't worry bro. i'm not the kind na napipikon kung ako ang mali. in fact i appreciate when someone is able to educate me on things that i thought i know but i actually might not. and though we are probably sidetracking from the main topic now, this might also help others here in the forum. so in the spirit of further learning, would like to ask for a simple circuit design that i can make and measure the 2Vp. because if this is so, why are ac components/equipment/devices rated, marked, and, used the way they are? I could be missing something here.
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zer0w1ng
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« Reply #98 on: January 21, 2010, 05:56:33 PM » |
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Marking is a different thing. If the component/capacitor is marked as 220VAC, it means it could handle up to 622VDC. If marked as 400V only (no "AC"), can handle up to 400VDC ONLY and not suitable for 220VAC power line. The "630V" marked is the correct capacitor to use as sir rdpzycho's procurement personnel got. This circuit can measure the voltage across the capacitor: o-------R----+ <----->|-------+-------------+ | 1N4007 | | 220VAC C/DUT C 1uF/630V VOM | | | o------------+ <--------------+-------------+
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TinTopHack
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« Reply #99 on: January 21, 2010, 06:55:05 PM » |
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It will be exposed to 2xVp.
For example, at 90degrees to 270degrees of the sine wave will be -622VDC. And from 270degrees to the next cycle at 90degrees is +622VDC.
If you have doubts on my explanation, you could try this in actual component/circuit.
BTW, I am hesistant like to reply again to this for it may cause flaming but I should correct this wrong notion.
I am a bit confused here. Have we considered the fact that when the sine wave crosses zero, the voltage across the cap reverses? Let us do this in slow motion. Let us assume we have a cap connected directly across a 220VAC line. Starting at 0 deg. the voltage across the cap is 0V. As the sine wave rises towards 90 deg peak the voltage across the cap follows with a little lag considering it is a cap connected directly across the AC line (no resistance except for the negligible wire resistance). At 90 deg the cap voltage will be 220 x 1.414 = +311V. Then beyond 90 the AC starts to drop to zero. The voltage across the cap will also start to drop because it is forced to do so by the source - meaning the cap starts to discharge. This goes on until 180 deg - zero crossing. At this point the AC voltage will be zero and the cap voltage follows. Beyond 180 the voltage reverses and goes down to towards 270 deg peak or -311V. The cap voltage follows also but this time the voltage is reversed (negative). Beyond 270 deg the AC voltage rises back towards zero. The cap voltage follows as well. So if we look at the highest voltage seen by the cap it appears to be just the peak voltage BUT this voltage reverses periodically. From 90 to 270 the cap does not charge to -2VPK because it discharges back to zero at 180 deg. Same thing happens from 270 to 90 deg of the next cycle. -TTH
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The world, as everybody knows, is analog; unless, of course, it's digitized.
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