# Impedance and LED on 120V.

Discussion in 'General Electronics Chat' started by AcousticBruce, Mar 25, 2014.

1. ### AcousticBruceMember

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lets say I have a 120V source, capacitor, 2 LED's rated at 40mA and a resister (or no resister) and I want to run the LEDs on with a low power loss. Questions at bottom.

Circuit #1
+120VAC
+3K Ohm
+2 LEDs ran in parallel.

This runs fine, but power loss on resister is 4.8 Watts!

Circuit #2
+120VAC
+Capacitor 884nF
+2 LEDs ran in parallel.

I used [tex]1/(2pi * f * Xc) = 884nF[/tex]

This just seems like it is not ok.

circuit #3
+120VAC
+300 Ohm resistor
+and I did a 982.6nF cap

The reason I chose that cap is because I needed 3K of impedance in order to get it at 40mA. I took 1/(2pi*(60Hz)(2700 Xc)), well it ended up being over 40mA. I imagine because I needed to do the impedance equation.

Circuit #4
+120VAC
+300 Ohm resistor
+888.6nF cap

I did sqrt(R^2 + Xc^2) = Z. I made Z = to 3k Ohm and derived an equation to solve for Xc. I ended up getting 2985Xc needed.

So for all this I have a few questions.

1) In AC circuits, is it good or bad practice to use a capacitor for Xc instead of using a resistor?

2) Why did the Xc work without the resistor (circuit 2), but when I added a resistor in Circuit 3, it seemed to be of by 5mA?

3) Did I use the formula, Z = sqrt(R^2 + Xc^2) or Z^2 + R^2 = Xc^2, correctly? Is that formula what you need to use a resistor and capacitor at the same time in a circuit?

4) Why is this stuff so damn interesting?

2. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Nice job:

1) As long as you don;t need isolation and use X and/or Y capacitors.
2) You LED's in parallel essentially act as a zener diode at Vf of the other LED. Vf varies by color.
Take a look here http://home.comcast.net/~teachcte/electron/formula1.htm#Z as well.
3) Probably depends of Vf of your model.

Look at this application note: http://www.google.com/url?sa=t&rct=...=Jk5xGlrMb8i20M_m2VRtJQ&bvm=bv.63587204,d.dmQ

a&uact=8&ved=0CCkQFjAA&url=http%3A%2F%2Fwww.kemet.com%2Fkemet%2Fweb%2Fhomepage%2Fkechome.nsf%2Ffile%2FKEMET%2520Kollege%2520Presentations%2F%24file%2FEvoxRifaRFIandSMD.pdf&ei=TR0yU6v1D8aH0AHPg4CYAg&usg=AFQjCNGDyKkyJHDat8vDb0K_8Yfb_m5ppQ&sig2=Q3fTJedjcgXnWRi2TmxICQ&bvm=bv.63587204,d.dmQ

3. ### AcousticBruceMember

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I am loving this transformer-less power supply pdf!!! I have always wondered how my phone gets a 2A 5V supply from a tiny plug with no heavy transformer! Still in the process of reading all this.

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5. ### RatchitWell-Known Member

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"1) In AC circuits, is it good or bad practice to use a capacitor for Xc instead of using a resistor?"

As long as you don't mind the current being out of phase with the voltage by 90°.

"2) Why did the Xc work without the resistor (circuit 2), but when I added a resistor in Circuit 3, it seemed to be of by 5mA?"

Did you measure it or calculate it? Was it high or low? Don't forget, current is existing in each diode only half the time.

"3) Did I use the formula, Z = sqrt(R^2 + Xc^2) or Z^2 + R^2 = Xc^2, correctly? Is that formula what you need to use a resistor and capacitor at the same time in a circuit?"

No, your second equation is wrong.

"4) Why is this stuff so damn interesting? "

You are the only one who can answer that question.

Ratch

6. ### JimBSuper ModeratorMost Helpful Member

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Possible confusion here.

A transformerless power supply, has no transformer, at all.

The light phone charger type power supplies do have a transformer, but it is a small high frequency transformer.
The incoming AC is converted to DC, which in turn is converted to high frequency AC which driver the small light transformer.
These are usually called Switched Mode Power Supplies (SMPS).

A transformerless power supply could be very dangerous when ised as a phone charger.
A correctly designed SMPS is quite safe.

JimB

7. ### NorthGuyWell-Known Member

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It would be the best to only have a capacitor with no resistor because it'll waste less power. But you need a small resistor to protect your diodes from in-rush current in case you connect it to the mains when it's at the peak.

8. ### AcousticBruceMember

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I thought about this very thing! The capacitor starts out at high current flow and if the AC is also at peak, this is just lots of current. So what about an inductor instead. I do not know any math on this right now, I will check it out.

Here is an idea based off the reading I have been doing on inductors... Since the current flows slowly first, what If you make a circuit that starts with an inductor and when that current builds, a transistor switches to a capacitor for reactance. Maybe this is what a smps does? I dont know

9. ### RatchitWell-Known Member

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I don't think you have to have a resistor or worry too much about the inrush of current. Here is why. Let's assume that the voltage is 120 volts at 60 Hz, the capacitor is 1 uF and the total resistance of the everything is 0.1 ohms. Furthermore, let us assume that the cap is energized backwards so that it aids the voltage which turns on at the worse possible time in its sinusoidal cycle. What do we have?

We have a voltage of sqrt(2)*120 volts peak. Double that for the voltage from the cap gives 2*sqrt(2)*120 volts peak. Dividing by the resistance gives (2*sqrt(2)*120)/0.1 = 3394 amps peak into the diode. Ouch!

But look at the time constant, RC = 0.1 usec. How much energy will be dissipated in the diode for, say 5 time constants, after which the inrush will be finished? Since the period of the sine wave is about 17 msec, and 5 time constants of the inrush current is 0.5 usec, we can assume the sinusoid will not change much during that time. Anyway, it turns out that the diode accepts 11 nanojoules of energy during that short period of time. Any worries?

Ratch

Last edited: Mar 26, 2014
10. ### Jony130Active Member

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I can assure you that this inrush limiting resistor greatly improve circuit reliability. Specifically when you use is as a transformerless power supply.
Also don't forget to add resistor in parallel with capacitor for for safety reason (discharge the cap).

http://www.dos4ever.com/flyback/flyback.html

11. ### RatchitWell-Known Member

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In some cases, a current limiting resistor is appropriate. In this particular circuit, it is not necessary due to the extremely low total energy involved and the very short time span of the transient.

Ratch

12. ### tvtechWell-Known MemberMost Helpful Member

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Hi AcousticBruce

Please bear in mind that Transformerless Power Supplies should never be used in any application where there is any sort of input/output coupling to any other device. It's a case of Mains in and that is it. No other sockets/plugs or whatever.

All this must happen in an Enclosure where the user has zero chance of coming into Electrical contact with anything happening inside the power supply.

If you don't heed this advise you are looking for trouble......

Regards,
tvtech

Last edited: Mar 27, 2014
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13. ### NorthGuyWell-Known Member

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I have a bunch of transformerless power supplies, such as phone charger, computer power wall-wart etc. and they all have sockets and plugs. All seems to be Ok and many are UL listed for safe use. Little children charge their phones that way and I haven't heard of them being electrocuted or otherwise hurt by their power supplies.

14. ### tvtechWell-Known MemberMost Helpful Member

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Hi NorthGuy

Have you actually opened any of those you mention above and looked at what is in them??

I guarantee they all have a little Transformer inside and it is a SMPS. Isolated mains output. And therefore safe for input/output connections to other devices.

Check them out

Regards,
tvtech

Last edited: Mar 27, 2014
15. ### NorthGuyWell-Known Member

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No, I didn't check them out. I assumed they're transormerless, which is a bad thing to do.

You're probably right. Most of them have enough room for a small high frequency isolating transformer. But some of them are so small, they don't seem to have enough space even for a capacitor.

After something breaks, I usually open it up to see how it works. So, I have pretty good idea how cofeemakers and toasters are built. But none of these little power supplies broke on me yet

16. ### tvtechWell-Known MemberMost Helpful Member

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Always here to guide Buddy

None of them have broken probably because your Mains supply is way more stable then where I live.

Stay well and keep safe.

Regards,
tvtech

17. ### ronvWell-Known MemberMost Helpful Member

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I think the diode response to the high current messes up your calculation. I never tried to calculate it. Just knew a current limit was needed. Anyway here is a simulation of it. 14 mj.

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18. ### RatchitWell-Known Member

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First of all, I found an arithmetical error in my calculation, and I now get a value of 60 millijoules instead of 11 nanojoules. The diode will not heat up much from that little energy. From looking at your curves, it appears that your time constant is much larger that what I used. I can show you my calculations if you wish.

Ratch

19. ### ronvWell-Known MemberMost Helpful Member

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I don't think the calculations would be to meaningful since most LEDs aren't spec'd there anyway.
The time constant is longer than yours because of the led Vf response to the high current. It is not like a standard diode where the drops stays pretty constant.
Anyway here is an led from Osrom with pulse curves.
http://www.osram-os.com/Graphics/XPic9/00115417_0.pdf/LR T68F - TOPLED.pdf
Pretty conservative, but.

20. ### RatchitWell-Known Member

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I just ran my calculations from 0.1,1.0,10.0,100.0 and 1000 ohms. They all came out to about 60 mjoules. It appears that increasing the resistance raises the time constant, but it also lowers the inrush of current which keeps the energy of the transient at a constant level. I assumed in my calculations that the resistance would be constant for such a short time period, but I have no way of knowing its true resistance level or variation. Interesting exercise.

Ratch

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