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Impedance and LED on 120V.

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Yes it is interesting.
When I model it there is a difference. Mostly in Vf.
For 170 volt step with 1, 10, 100, and 1k... 30ma rated diode 880 nf cap 0 esr
TC .03ms, .1, .3, and 3ms
Vf (peak) 150v, 80v, 17v and 5.4v
Id (peak) 18a, 9, 1.55, .165 amps
mj 11.4, 6.2, 1.6 and .64mj
 
With 1000 ohms the RC contant might be big enough for the voltage to drop away from the peak before the capacitor gets fully charged :)

The time constant for 1000 ohms resistance and 1 uf capacitor is 1 ms. The period of 60 Hz is 17 ms. Therefore, the cap has more than enough time to energize or deenergize before the sinusoidal wave changes significantly. You can see from the current plot below that transient dies out very quickly.

Northguy.JPG


Ratch
 
Yes it is interesting.
When I model it there is a difference. Mostly in Vf.
For 170 volt step with 1, 10, 100, and 1k... 30ma rated diode 880 nf cap 0 esr
TC .03ms, .1, .3, and 3ms
Vf (peak) 150v, 80v, 17v and 5.4v
Id (peak) 18a, 9, 1.55, .165 amps
mj 11.4, 6.2, 1.6 and .64mj

I am wondering about the figures you are putting into Spice. You have to double the 170 voltage because the cap might be left energized at 170 volts when it was last turned off. Then it might sum with the 170 source voltage if the source is turned on at its highest peak. I also don't get the same time constants you do for multiplying R*C. What is Vf, the forward voltage? That is never more than a few volts, is it? How come I see 150 volts, 80 volts, etc? Anyhow, those limits should be looked at after the calculations are made, not during the calculation of the response the of circuit.

Ratch
 
Hi Guys

Let me simplify this discussion with real world working results after a Year of testing...yes and I know this an interesting subject...

1. You need a current limiting Resistor. The resistor serves two roles....
Firstly as a device that limits inrush current into the X2 cap when plugging the PSU into Mains. This prevents too much arcing when plugging the plug in.
Secondly as a safety devise should the X2 Cap fail at any time (as in go shorted).

In my design it is a little 1W wirewound 6.8 Ohm Resistor feeding a 2.2 uf X2 Cap. No issues.

2. Referring to the post above, good practice is to have a discharge Resistor wired across the X2 Cap. I use a 1 Meg 1W for this. No issues.

3. Ensure you never loose the designed load. Probably the most important thing of all.... As I have stated here on another thread your output VOLTAGE from the X2 is going to go sky high if you do. Isolation Transformers don't help here....and you will still get belted if you are fiddling at the wrong time.

And that is all I have to say about Transformerless Power Supplies. They are reliable, interesting, unburstable and just need to be understood.

The Electronics that utilize this reliability are simple....you just need to understand, respect, do. And you have a product that is fail safe.

Regards,
tvtech
 
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Hi Guys

Let me simplify this discussion with real world working results after a Year of testing...yes and I know this an interesting subject...

1. You need a current limiting Resistor. The resistor serves two roles....
Firstly as a device that limits inrush current into the X2 cap when plugging the PSU into Mains. This prevents too much arcing when plugging the plug in.
Secondly as a safety devise should the X2 Cap fail at any time (as in go shorted).

What arcing is that? Expect a maximim transient of 340 volts, and the current to run 1 led plus energize a 1 nf cap in less time than it takes to blink an eye.

In my design it is a little 1W wirewound 6.8 Ohm Resistor feeding a 2.2 uf X2 Cap. No issues.

If the cap shorted, that's 120/6.8 = almost 18 amp. That's enough to take out every component in the circuit.

2. Referring to the post above, good practice is to have a discharge Resistor wired across the X2 Cap. I use a 1 Meg 1W for this. No issues.

Why turn a good cap into a leaky one? The cap energizes and deenergizes in a flash anyway.

3. Ensure you never loose the designed load. Probably the most important thing of all.... As I have stated here on another thread your output from the X2 is going to go sky high if you do. Isolation Transformers don't help here....but you will still get belted if you are fiddling at the wrong time.

Since it is a series circuit, loosing the load will cause the current to go to zero.

And that is all I have to say about Transformerless Power Supplies. They are reliable, interesting, unburstable and just need to be understood.

The Electronics that utilize this reliability are simple....you just need to understand, respect, do. And you have a product that is fail safe.

Regards,
tvtech
 
LOL Ratch

Its OK. Merely stating the experience with my product. So far so good.

Only good and learning can come out this for all. I am allowed to talk about this here but not on some other forums out there.

All the best,
tvtech
 
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Hi Ratch

The thread evolved. Please check out the lot.

Regards,
tvtech
 
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


Would you mind helping me understand how to get that math?
What all variables did you look at? I understand cap charge to max and and mains starting on peak, 2 * sqrt(2) * 120, but where did 100 mili-ohm come from?


also earlier you said Z^2 + R^2 = Xc^2 was wrong. So when you do
Z = sqrt(R^2 + Xc^2) >>>> (Z^2 = R^2 + Xc^2) >>>> (Z^2 - R^2 = Xc^2) How is this wrong?
 
I only see two posts on that thread which I don't understand very well. Not much evolution.

Ratch

It's coming...the evo revo lution is almost here. The whole X2 Transformerless Supply thing will be explained in detail in laymans terms so everybody understands.

And that hopefully will put to rest the myths and dangers for ever.

Regards,
tvtech
 
Would you mind helping me understand how to get that math?
What all variables did you look at? I understand cap charge to max and and mains starting on peak, 2 * sqrt(2) * 120, but where did 100 mili-ohm come from?

I used 0.1 ohm as a guesstimate. 1uf was the value of C.

First we find the current in terms of Laplace expressions.
AcousticBruce.JPG


Then we find the inverse Laplace.

AcousticBruce.JPG


We can plot i for 5 time constants.

AcousticBruce.JPG


We can also plot the power for 5 time constants

AcousticBruce.JPG


Integrating the power with respect to time over 5 time constants we get about 60 millijoules.

AcousticBruce.JPG


Changing the value of R does not change the energy of 60 millijoules much because a higher resistance lowers the current and increases the time constant.

also earlier you said Z^2 + R^2 = Xc^2 was wrong. So when you do
Z = sqrt(R^2 + Xc^2) >>>> (Z^2 = R^2 + Xc^2) >>>> (Z^2 - R^2 = Xc^2) How is this wrong?

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?

Don't you see anything wrong with the second equation?

Ratch
 
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Sure, 340 volts, unless of course you use TVtechs little bleeder. :)
Yes, now in agreement if I use 1 Ufd. And yes the power is about 60mj, but most of it is in the resistor, not the LED.

If you look at Vf curves the Vf may go up .1 volts for each 10 ma of increase in current so the high Vf are not unexpected. Just very painful (for the LED)
 
Sure, 340 volts, unless of course you use TVtechs little bleeder. :)
Yes, now in agreement if I use 1 Ufd. And yes the power is about 60mj, but most of it is in the resistor, not the LED.

If you look at Vf curves the Vf may go up .1 volts for each 10 ma of increase in current so the high Vf are not unexpected. Just very painful (for the LED)

Why use a bleeder resistor to make the capacitor leaky? That would prevent the cap from energizing as fast as possible in order to develop a counter voltage to limit the current. I am using no resistor, so most all the resistance is from the LED. Why is the Vf even considered? The transient is finished in less than a microsecond, and 60 mjoules in not much to worry about. 60 mjoules is the energy, not the power. Looking at the power curve, its peak is over a million watts for an infinitesimal amount of time.

Ratch
 
Why use a bleeder resistor to make the capacitor leaky?
1-Cuts the inrush current in half.
2- Keeps you from getting zapped if it happens to be charged up when you get across it.
Why is the Vf even considered?
Because the high Vf limits the peak current.

The transient is finished in less than a microsecond, and 60 mjoules in not much to worry about. 60 mjoules is the energy, not the power. Looking at the power curve, its peak is over a million watts for an infinitesimal amount of time.
No it's not because the Vf goes way up. Depending on the LED it can be 35 40 usec. But it is still 60 mj. This seems to break down the LEDs around the bond wires. Cree has some write ups.

Ratch[/quote]
 
1-Cuts the inrush current in half.

How does it do that? By crippling the capacitor's ability to energize and develop a back voltage, which will cut the inrush current? Then why have a cap at all? The OP said he wanted the circuit to be efficient. Using a resistor to limit current wastes power. Either let the cap do its job or use an energy wasting resistor in the first place.

2- Keeps you from getting zapped if it happens to be charged up when you get across it.

Are you going to protect yourself from every cap in the circuit? How about the mains terminals? I think that is a specious reason.

Because the high Vf limits the peak current.

Now that really puzzles me. The Vf of a diode is never high. Only a few volts at best. 340 volts is not even going to know it's there. See the Vf curve below.

ronv.JPG



No it's not because the Vf goes way up. Depending on the LED it can be 35 40 usec. But it is still 60 mj.

Vf is measured in volts, not time or joules.

This seems to break down the LEDs around the bond wires. Cree has some write ups.

What about the bond wires? Who is Cree?

Ratch
 
How does it do that? By crippling the capacitor's ability to energize and develop a back voltage, which will cut the inrush current? Then why have a cap at all? The OP said he wanted the circuit to be efficient. Using a resistor to limit current wastes power. Either let the cap do its job or use an energy wasting resistor in the first place.

Are you going to protect yourself from every cap in the circuit? How about the mains terminals? I think that is a specious reason.

As you appear to have extremely limited knowledge of electronics, perhaps you shouldn't be playing with dangerous permanently live circuits?.

1) You have a series resistor, both for surge limiting reasons and to act as a 'fuse' when the capacitor goes S/C (no idea what you even mean by "crippling the capacitor's ability to energize and develop a back voltage", it's just nonsense).

2) The resistor in parallel with the capacitor is also essential (and probably required by legislation in most civilized countries?), to prevent electrical shock across the plug ins when you unplug it.

I can't believe this has gone to 37 posts?, it's a crude, simple (and relatively dangerous) circuit - which is why you perhaps shouldn't be playing with them?.

Series resistor, X2 capacitor, parallel resistor, LED - why 37 posts?.
 
As you appear to have extremely limited knowledge of electronics, perhaps you shouldn't be playing with dangerous permanently live circuits?.

Your assumption and statement above are completely unjustified. Although I have not made a living in electronics, I have studied theory and gained experience with the basics most of my life, which by now is a considerable number of years. In fact, I have probably been exposed to electronics longer than you have, unless you are are older than I am. I consider myself well grounded in the mathematics and the basic theory of electronics, and I believe I can defend my assertions with any member of this forum. Your disagreement with me on this matter should not be interpreted as lack of knowledge, unless you can successfully show that I am wrong in what I say. My discusson of the circuit is about its viability with respect to activating the diode without burning it out, not whether the circut is safe to touch. With that in mind let's peruse your following comments.

1) You have a series resistor, both for surge limiting reasons and to act as a 'fuse' when the capacitor goes S/C (no idea what you even mean by "crippling the capacitor's ability to energize and develop a back voltage", it's just nonsense).

A resistor across the capacitor is not what I call a "series" resistor. That parallel resistor will turn the capacitor into a leaky one. It will also let the surge current bypass the capacitor and be applied directly to the diode. The capacitor by itself will energize quickly and stop the surge in less than a microsecond. I proved that mathemetically in my previous post. Basically, the circuit is using the immitance of the capacitor instead of a resistor to limit the current to the diode. That saves on energy, as the OP originally said he wanted to do. I would appreciate it if you could explain why a resistor is necessary, and why you don't understand that the cap develops a backvoltage when it is energized.

2) The resistor in parallel with the capacitor is also essential (and probably required by legislation in most civilized countries?), to prevent electrical shock across the plug ins when you unplug it.

Oh, you mean a 1 uf cap energized to 60 millijoules, and which will will deenergize in a fraction of a microsecond. I think I will get a bigger shock just by walking across a carpet.

I can't believe this has gone to 37 posts?, it's a crude, simple (and relatively dangerous) circuit - which is why you perhaps shouldn't be playing with them?.
Series resistor, X2 capacitor, parallel resistor, LED - why 37 posts?.

I am not playing with them. I am theorizing for someone else who is experimenting with a LED. This topic has gone on this long due to lack of understanding of what is happening in this circuit. Activating a LED should be simple and crude, not complicated. The danger or safety depends how well the components are shielded from prying paws. I have not addressed that topic.

Ratch
 
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A resistor across the capacitor is not what I call a "series" resistor.

Nor would I, and I didn't - it also wouldn't be much use as a surge limiter or 'fuse' if it was in parallel either.

That parallel resistor will turn the capacitor into a leaky one. It will also let the surge current bypass the capacitor and be applied directly to the diode.

This is one of the reasons we know you have limited electronics knowledge, it's a discharge resistor, while it obviously will leak a little current past the capacitor it certainly won't "let the surge current bypass the capacitor and be applied directly to the diode" with makes no sense at all, the surge is THROUGH the capacitor and caused by it (which is why a limiting resistor in series is always fitted in such designs).

Oh, you mean a 1 uf cap energized to 60 millijoules, and which will will deenergize in a fraction of a microsecond. I think I will get a bigger shock just by walking across a carpet.

Again, showing your lack of knowledge - there's no reason to mention joules anywhere in this design, it's not something which is very often used at all in electronic design. It also certainly won't 'de-energise' (correct term is discharge) is a fraction of a uS through a human body.

The resistor, again universally fitted, discharges the capacitor so a user can't get a shock from the plug pins, without such a resistor it could stay charged for a considerable length of time - while it's not going to be a fatal shock, it could easily be enough to cause you to have an accident, or throw the device and cause damage.
 
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