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
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
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 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.
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
Hi Ratch
Need you to take a look here: https://www.electro-tech-online.com...he-public-domain-at-last.136196/#post-1146578
Regards,
tvtech
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
I only see two posts on that thread which I don't understand very well. Not much evolution.
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?
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?
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)
1-Cuts the inrush current in half.Why use a bleeder resistor to make the capacitor leaky?
Because the high Vf limits the peak current.Why is the Vf even considered?
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.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.
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.
Because the high Vf limits the peak current.
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.
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?.
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.
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.
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