Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Impedance and LED on 120V.

Status
Not open for further replies.
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.

In post #38 of this thread you said,
"1) You have a series resistor, both for surge limiting reasons and to act as a 'fuse' when the capacitor goes S/C....". That appears to me like you said that.

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).

I makes no sense to you because you don't understand what is happening. Current does not pass through a capacitor. If i did, it would be considered defective or "leaky". Energy does pass through a cap, however, and this is what sends the charge of the opposite plate on its way, giving the illusion that the current passed throught the cap. A capacitor does not "charge". No matter how many volts you apply across a capacitor, the net charge will be the same at it was when no voltage was applied. For every coulomb of charge added to one plate, one coulomb of charge is substracted from the opposite place for a net change of zero. The charges become distributed unevenly, but their net amount does not change. This uneven distribution of charge causes a voltage to appear across the plates of the cap. It took energy to create this voltage. A capacitor should not be called charged when it isn't, unless you mean "charged with energy", in which case you might as well say "energized". By putting a resistor across the capacitor, you don't allow it to energize to the max voltage, so the back-voltage does not suppress the surge as effectively as it would if no resistor were there. Now, if you don't agree, you are obligated to supply concise and cogent reasons why this is not so instead of just saying that I don't know what I am talking about.


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.

Certainly there is a reason. A cap with very little stored energy is not a threat, and energy is measured in joules. If that cap deenergized through a body, it would certainly take longer, but the current would be proportionally less. I doubt if anyone could feel it during that infinitesimal time.

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.

As I said above, the energy is so small, I doubt if anyone could even feel it.

Ratch
 
I did burn few diodes with transients, so I believe in using the resistor, however I cannot come up with any scientific method to determine the necessary value. Just from my "feelings" 220 Ohm seems too much. If I were building this, I probably would consider something between 10 and 50 Ohm. I have read the quoted article, and this is the method they use to select the value of the resistor:

The value is chosen using the R=P/I² i.e. 0.25W/(33mA)² = 229W

I don't understand where the 0.25W come from? Just because common resistors are 0.25W?

Why not 30 Ohm or 150 Ohm? With less resistance, power dissipation will be less. Will help Kyoto :)

How do they know that 229 Ohm is enough? Why not to upgrade to 800 Ohm 1W resistor?
 
Ratch please read this link: http://www.marcspages.co.uk/tech/6103.htm

It explains the basics :)

Regards,
tvtech

OK, I studied the single LED section thoroughly. Of interest, but not really pertinent is the fact that his mains are 230 vs my 115, his frequency is 50 vs my 60, and he made a mistake in transcribing the capacitor value as 0.0000001 instead of 0.000001, but his result was correct.

First of all, I have to wonder why he uses a full-wave diode bridge instead of two LEDs back to back like the OP did. Wouldn't that be much simpler? His calculation of the cap to limit the current is correct. The OP did not give me the max current of his diode, but with 1uf, it will need to be 60 ma. If that is too much, the cap will have to be reduced in value.

The 220 resistor has me scratching my head. He says it acts like a fuse in case the cap blows. I think the LED is going to act like a fuse and the resistor will be the last component left standing, right? I don't know what a resistor does for harmonics because it is frequency independent. If he is worried about line harmonics, he should get a filter. And what is a 220 ohm resistor going to do for a transient? Notice he did not do a transient analysis like I did, so he does not mention how quickly transients die out due to short time constants.

In conclusion, he uses components like the bridge and a resistor which I don't think he needs. His circuit uses more power than the OP's circuit does. Other than that, his calculations appear to be correct. By the way, notice he does not put a resistor across the capacitor.

Ratch
 
I did burn few diodes with transients, so I believe in using the resistor, however I cannot come up with any scientific method to determine the necessary value. Just from my "feelings" 220 Ohm seems too much. If I were building this, I probably would consider something between 10 and 50 Ohm. I have read the quoted article, and this is the method they use to select the value of the resistor:

I don't understand where the 0.25W come from? Just because common resistors are 0.25W?

Why not 30 Ohm or 150 Ohm? With less resistance, power dissipation will be less. Will help Kyoto :)

How do they know that 229 Ohm is enough? Why not to upgrade to 800 Ohm 1W resistor?

Why use a resistor at all? Did you see my transient analysis and notice how short the time is and how small is the energy stored? If you had problems, I would be interested in your capacitor values.

Ratch
 
Pity that article tends to use W for ohms though :D

But I agree, he's unlikely to read it, and even less likely to believe it.

It's not perfect. But it does show the way.

Those are the Basics.

So good to have your input Nigel. Only those who have actually walked the road will understand. And you know it well.

People who have been through the Mill......and felt and not theorized...that is the thing. Stay well :)

tvtech
 
Why use a resistor at all? Did you see my transient analysis and notice how short the time is and how small is the energy stored?

I understand your point that only the energy necessary to charge the capacitor will travel through the diode during the transient. You can easily calculate it without much of the analysis - C*V^2/2.

The resistor slows down the process. With 1o Ohm resistor it takes 100 times longer to deliver the energy than with 0.1 Ohm resistor (should really factor in capacitor ESR). Clearly, the speed of energy delivery increases the risk of diode failure. Faster delivery would be more damaging. I've never seen any formulae that would attempt to quantify the damage based on the speed of energy delivery, but it doesn't mean they don't exist.

If you had problems, I would be interested in your capacitor values.

Unfortunately, I cannot recall now.
 
OK, I studied the single LED section thoroughly. Of interest, but not really pertinent is the fact that his mains are 230 vs my 115, his frequency is 50 vs my 60, and he made a mistake in transcribing the capacitor value as 0.0000001 instead of 0.000001, but his result was correct.

First of all, I have to wonder why he uses a full-wave diode bridge instead of two LEDs back to back like the OP did. Wouldn't that be much simpler? His calculation of the cap to limit the current is correct. The OP did not give me the max current of his diode, but with 1uf, it will need to be 60 ma. If that is too much, the cap will have to be reduced in value.

The 220 resistor has me scratching my head. He says it acts like a fuse in case the cap blows. I think the LED is going to act like a fuse and the resistor will be the last component left standing, right? I don't know what a resistor does for harmonics because it is frequency independent. If he is worried about line harmonics, he should get a filter. And what is a 220 ohm resistor going to do for a transient? Notice he did not do a transient analysis like I did, so he does not mention how quickly transients die out due to short time constants.

In conclusion, he uses components like the bridge and a resistor which I don't think he needs. His circuit uses more power than the OP's circuit does. Other than that, his calculations appear to be correct. By the way, notice he does not put a resistor across the capacitor.

Ratch

Somebody help me 0.0000 to 0.000000

Cannot handle this anymore..Stop it Ratch....you are driving me insane :troll::troll::troll::troll::troll::troll::troll:

FFS = For Flipping Sake STOP TRYING TO BE CLEVER. Nobody is paying attention anymore :troll::troll::troll::troll:

Bah :banghead::banghead::banghead::banghead::banghead::banghead:

tvtech
 
I understand your point that only the energy necessary to charge the capacitor will travel through the diode during the transient. You can easily calculate it without much of the analysis - C*V^2/2.

The resistor slows down the process. With 1o Ohm resistor it takes 100 times longer to deliver the energy than with 0.1 Ohm resistor (should really factor in capacitor ESR). Clearly, the speed of energy delivery increases the risk of diode failure. Faster delivery would be more damaging. I've never seen any formulae that would attempt to quantify the damage based on the speed of energy delivery, but it doesn't mean they don't exist.



Unfortunately, I cannot recall now.

In the example I did for the OP, the spike transient current was over 3000 amps, assuming a resistance of 0.1 ohms. But it was for an extremely short time. Increasing the resistance will lower the spike and increase the duration of the transient. The energy dissipation remains the same. So the question is whether those extremely short current spikes damage anything. They certainly don't heat very much. All I can do is calculate the amount of energy dissipated. We need a fabrication specialist to tell us if that kind of current will damage a diode.

Ratch
 
Somebody help me 0.0000 to 0.000000

Cannot handle this anymore..Stop it Ratch....you are driving me insane :troll::troll::troll::troll::troll::troll::troll:

FFS = For Flipping Sake STOP TRYING TO BE CLEVER. Nobody is paying attention anymore :troll::troll::troll::troll:

Bah :banghead::banghead::banghead::banghead::banghead::banghead:

tvtech


I know you are prone to histrionics, but do try to stay cool, calm, and collected. Especially since it was you who asked me to look at the website you designated.

Ratch
 
DSCF0124.JPG DSCF0125.JPG DSCF0126.JPG DSCF0127.JPG ]
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 :(
Sometimes I will open up a good wall-wart for the fun of it. The last one was a Motorola phone charger. I must say that the quality of the PCB and components are very good. I was impressed.
I use some gas or brake fluid to weaken the glue for about 15 - 30 minutes to help open them easier.
 
Last edited:
We need a fabrication specialist to tell us if that kind of current will damage a diode.

That's why I would use a 10 to 50 Ohm resistor if I had to build it right now. It only costs a cent, if not less.

But if I had to build 100 million LEDs for sale, I would study all the details, and do a lot of experiments to find out if I can save half a million dollars by not using the resistor. We can open up few commercial LEDs and see if they have resistors or not. I bet they did their homework.
 
I am learning a lot from this thread. I like the different schools of thought, it helps me see what peoples perspectives are. I will have more questions when I get done studying this thread.

Jony130 said:


This might be the coolest part about electronics that I have seen. I am going to delve deep into this subject.
 
For some reason this whole back and forth on this thread is reminding me of "it's not the volts that kill you, its the amps".

I do want you all to know, that this circuit is NOT complete, it is merely a focused discussion on capacitance reactance and resistance; the difference between the two; using them in different scenarios; As well as how to calculate it properly. I am studying components and making little schematics to learn the math as well as theory.


I absolutely love all the help. I am taking notes and learning with all this.
 
tvtech said:
Hi Guys

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.

So is this X2 cap a film capacitor? Is this the X and Y capacitors that was mentioned earlier?

tvtech said:
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.

I am assuming this is strictly to bleed the cap. Does it serve any other purpose?

tvtech said:
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.

I think I just got it, basically, I just made a transformerless_power-supply circuit. And this is where you are trying to let me know of the dangers.
You mentioned about losing the load and the dangers. Is this the LED or just the resister as load? And wouldn't it just cause an open circuit? Why would it go sky high?
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top