Question on an Electronic Night Lamp

[Aaron.coG]

Hey Guys,

Electronics just fascinates me........

But I am really dumb when it comes to working on it.

I have a night lamp that glows on neon bulbs. I want to replace them with LEDs so that the life will be longer.

The circuit is like the below but to a 230V power supply - from main power supply, this is linked to MPET D105K2E Capacitor and a resistor, and from there, the neon bulbs are connected in a series.

I am not trying to save the power of the lamp. One of the neon bulbs were fused within a matter of days. Not sure what has caused this - may be the voltage supplied is too high for those tiny fellows to handle. As they are in series, the lamp didn't work. I just removed the fused one and connected the others and got the lamp glowing. Again, in a few days, another one went off. So, I thought replacing them with more LEDs, say eight numbers, could get the lamp running at least for a year.

1. What do you think? Would that work?
2. And do you think the LEDs are reliable than the neon bulbs?
3. Is there a way I could connect them in parallel, of course, adding more resistors?
4. Do the LEDs produce more heat?
5. And could heat be the cause for the neon bulbs to go off quickly?

The image below shows the serial connection of the Neon bulbs - the connection starts with a resistor and then the series of bulbs and the end goes to the main power line.

**broken link removed**​

This image shows the main power line coming to the MPET D105K2E Capacitor from right (black wire) and a resistor connected parallel to the capacitor and this line goes to the resistor in the first picture.

**broken link removed**

If needed, I would try to get you the resistor values shortly.

Really very sorry for the poor quality.....

I hope my description of the issue and my question is clear. :smile:

 

[betwixt]

Yes, LEDs will work fine and should give you many years of operation. Your calculation is wrong though, it should be (Vin - Vleds) / Ileds. In other words the voltage dropped from 12V to the total of the voltage dropped by all the LEDs in series, divided by the current through the LEDs. For example, at 10mA if each LED drops 1.5V the total for four in series would be (4 x 1.5) 6V so the resistor drops the remaining 6V at 10mA. Ohms law gives us a resistor or (12-6)/.01 ohms or 600 Ohms. It's power rating should be at least (W=V x I) 6 x 0.01 = 60mW so even a small 0.25W rated resistor would do.

If you want to use 230V instead of 12V things get a bit more complicated. Firstly the LEDs can not withstand more than about 5V in the reverse polarity and being AC, the polarity reverses several times a second. You can fix this by adding a diode in series to prevent the reverse voltage conducting to the LEDs. The resistor becomes (230 - 6) / 0.01 which makes it 22,400 Ohms at 2.24W but it will only conduct 50% of each mains cycle so 1.5 Watt or bigger resistor will be needed.

Brian.

 

[Aaron.coG]

Hi Brian,

Thanks for the reply.

The calculations are really over my heads. To be frank, I don't know much on electronics. This is my first attempt. I would try to understand your reply.

But did you answer this to connect the LEDs in a series or parallel connection? And do you know if the MPET capacitor drops down the voltage? Can you give me ideas on how I can calculate the output voltage of this circuit?

 

[betwixt]

Firstly, forget using the capacitor, there are applications where it's fine to use them but this isn't one of them. The amount of voltage dropped by a capacitor depends on several factors, some beyond your control so I would play safe and just use a resistor.

Ideally, you would use a transformer or "wall wart" to convert the 230V down to a safe low voltage, this also gives you isolation from the mains power which makes the circuit much safer to handle. If you just use a resistor, although it will work electrically, please remember that all parts of the circuit are connected to AC mains and are therefore potentially dangerous. Ensure no metal parts can be touched anywhere in the circuit.

Now the theory - as simple as I can keep it: LEDs only light up when the voltage across them is the right polarity, if you reverse them they are easily damaged. When conducting (lit up) they each try to hold a voltage across themselves of around 1.5V, the exact voltage depends on the LED color and to some degree the manufacturing process used. If you assume 1.5V you won;t be far off though. Each LED holds a slightly different voltage, just from manufacturing differences so if you connected them in parallel (across each other) the ones that held a lower voltage would hog the power at the expense of the ones that needed a bit extra, this would result in them being different brightness and could even result in some being overloaded and damaged.
The solution is to connect them in series (in a chain) just as your diagram shows. Connected this way, the voltages of all the LEDs is added together, one alone would hold 1.5V, two would hold 3V, three would hold 4.5V and so on. So from end to end of the LED chain, the voltage would be 1.5 times the number of LEDs.
The purpose of the resistor is to drop the remaining voltage to make the total 230V. So for example, if you used four LEDs, they would drop 6V and the remainder for the resistor to drop would be 224V (230 minus 6).
Resistors come in all shapes and sizes but are rated in two ways, their value in Ohms which is a measure of how much they 'resist' current and power rating in Watts which is a measure of how much heat they can safely produce before being damaged. The resistance in Ohms is set by the chemical composition of the resistor material and has little relationship to the power rating, you can get tiny and huge resistors over the whole range of resistance values. Generally, the physical size of the resistor determines how much power it can handle, bigger means more surface area for heat to escape from so high power (more Watts) resistors tend to be larger than low power ones.
For the LEDs you need to drop those 224V and you have to decide at what current the LEDs should be operated. The current is a measure of what 'flows' through them as opposed to voltage which is what is put across them to force the current through. For small LEDs, a good choice of current would be 10 milli-amps (mA). The 'milli' is a prefix meaning "thousandths of" so as a fraction 10mA is 10/1000 or 0.01 Amps. The only place you will find the maximum current is the LED manufacturers data sheet but 10mA would be a reasonably safe current which would still give enough brightness.
Now using Ohms law the resistor value can be calculated, the formula is R (resistance in Ohms) = V (voltage to be dropped) / I (current chosen to flow) so with real numbers that's R = 224 / 0.01 which gives 22,400 Ohms.
You can buy resistors in any value you want but there are 'standard' values' which are mass produced and are cheaper than exact values, the nearest standard value to 22,400 is 22,000 or 22K Ohms which would be close enough that you wouldn't notice any difference.
Next you have to take into account that the 224V dropped had to go somewhere and in is actually converted to heat in the resistor. The amount of heat (in Watts) is calculated by multiplying the voltage across it by the current flowing through it (W = V x I) so again with real values W = 224 x 0.01 gives 2.24 Watts so any resistor rated to release (we use the term 'dissipate') more than that should work safely.
There is one extra factor in your circuit, the 230V is AC, this means the voltage is not constant but always moves up and down between a peak a bit higher than +230V and zero then reverses to a peak a bit higher than -230V before repeating again, this is why it's called "alternating", the 'A' in AC. The value 230V is actually quoted using a term called RMS which gives a steady figure 'equivalent' to the effect of the AC being DC so it can be used in calculations. Remember I said LEDs can be damaged by connecting the voltage backwards, well in AC the polarity is always swinging one direction them the other so for half of each mains cycle the polarity is wrong and damage might occur. There are several ways to cope with this but in your application the easiest is to add a diode in series with the resistor, this will have the effect of blocking the current flow on the part of the mains cycle that would cause damage but still letting current through on the half cycles you want. A diode conducts in one direction only so if you make sure the '+' end of the diode (usually marked with a colored band) points the same way as the LEDs it should work The diode should be rated to withstand peak mains voltage and a common part to use would be a 1N4006 which is very inexpensive.
Now a special bonus - you remember the resistor was dissipating 2.24 Watts, well with the diode connected the 2.24W is only for half the mains cycles, the other half isn't passing any current at all because the diode is blocking it. This means the average power is actually only half the 2.24W or 1.12W which means a physically smaller resistor can be used. I would suggest you get a part rated at 2W to give a safety margin. Please remember that even though a 2 Watt rated resistor would be well within it's ratings, it is still producing 2W of heat so it will get quite warm.

Brian.

 

[Aaron.coG]

Wow... Brian...

You are just awesome to explain this so elaborately..... I am just reading it more than once to get the full understanding of this....

I think you have brought me so close to solving this. Give me some time.... let me understand this and get back to you.

 

[Aaron.coG]

**broken link removed**

Hi Brian,

If my previous photos were not clear, this is the circuit that I am having in hand. Instead of the LEDs, the circuit has 5 Neon Bulbs. This is the one I have to modify to adopt the LEDs.

There are two resistors used. I think the values I have specified are correct. I used an online color code calculator to get the values of the resistors.

Hope I could change this circuit to use LEDs instead of the neon bulbs.

 

[betwixt]

Neons and LEDs work in a completely different way. A typical Neon lamp works on about 90 Volts at a current of about 0.5mA whereas an LED works on about 1.5V at 10mA. Neons also light up regardless of the polarity while LEDs are damaged by just a few Volts in the wrong direction.
To adapt your circuit, remove the capacitor and the 160K Ohm resistor completely and replace them with a diode (1N4006 or similar), facing the same way as the LEDs. Also drop the 1.2M Ohm resistor to 22K Ohms.
Incidentally, on the capacitor 'MPET' is probably the manufacturers name , '105' will be the value of 1uF and the '2E' is a manufacturer code, possibly for the maximum voltage it can withstand.
Brian.

 

[Aaron.coG]

Neons and LEDs work in a completely different way. A typical Neon lamp works on about 90 Volts at a current of about 0.5mA whereas an LED works on about 1.5V at 10mA. Neons also light up regardless of the polarity while LEDs are damaged by just a few Volts in the wrong direction.
To adapt your circuit, remove the capacitor and the 160K Ohm resistor completely and replace them with a diode (1N4006 or similar), facing the same way as the LEDs. Also drop the 1.2M Ohm resistor to 22K Ohms.
Incidentally, on the capacitor 'MPET' is probably the manufacturers name , '105' will be the value of 1uF and the '2E' is a manufacturer code, possibly for the maximum voltage it can withstand.
Brian.

Brian, that's great.

I can understand the polarity issue with the LEDs.

Will buy the items tomorrow and try it. But can I connect 6 LEDs instead of the four or do I have to change the resistor value if the count of LEDs is 6?

---------- Post added at 12:27 AM ---------- Previous post was at 12:21 AM ----------

And can I use any of these diodes?

A 1N4001 is rated at 50 Volts.
A 1N4002 is 100V
A 1N4003 is 200V
A 1N4004 is 400V
A 1N4005 is 600V
A 1N4006 is 800V
A 1N4007 is 1000V

As my input voltage is 230V, I thought of using 1N4003.

Or is any one of these is fine?

 

[karesz]

Hi Aron,
If your input AC-voltage is 230V (RMS), its peak values are over 300V!
For have some reserve by variations of nominal value & by transients; I would select minimum a 1N4005...
K.

 

[betwixt]

Agreed with Karesz, either use a 4005, 4006 or a 4007. I partly explained the reason in an earlier message, it's difficult to define a voltage while it keeps moving so we use a term called RMS to give it a value. The voltage is always changing so saying it is 230V is only applicable to one instant in the AC cycle. To make it easier to work with we derive a number of volts which we can consider to be equivalent to the moving voltage, in other words a measure of what an equivalent but not moving voltage would do in a similar circuit. Unfortunately, you can't just use the average voltage because the average of a positive and equal negative is zero ! What we do is a bit of clever math, firstly we square the voltage, this has the effect of making all the measurement positive since squaring a negative number gives a positive result. Then we take the average voltage over one cycle and finally, to undo the squaring, we take the square root of the result. Hence RMS = Root, Mean, Squared. Now, in a 230V sine wave mains, the math says the peak voltage will be the square root of two multiplied by the RMS value so the peak of 230V RMS is 1.4142136 * 230 = about 325 Volts. So the diode needs to be rated to work at 325 volts or greater.that's why we suggest the higher voltage rated parts.

The resistor value should be (230 - total of all LED voltages)/ LED current. Use a value of 1.5V for each LED and a current of 10mA, so multiply the number of LEDs you choose by 1.5 then subtract the result from 230 to see how much the resistor has to drop. Finally, make sure you use a resistor rated at 2W or more.
Brian.

Brian.

 

[karesz]

.. I partly explained the reason in an earlier message...
Brian.

Sorry Brian,
I didnt read all texts before in detail...
Otherwise your explanations are really fine!
Best regards!
K.

 

[pratchaya.aut]

what the benefit do you want to used or design make the usderstand first

 

[karesz]

Hi pratchaya,
What did you wish to tell me; or can you explain it pls?
K.

 

[Aaron.coG]

Hey Brian, Karesz,

Sorry for the delay. This week was a real hectic week.

Here you go... so these are the components I am gonna use.

From 230 V --> IN4007 Diode --> 22K Ohms Resistor --> 6 LEDs serially and the circuit closes.

On the resistor part, Brian, as per your workouts, 230 - (1.5 * 6) = 221 / 0.01 = 22100 Ohms -> 22K Ohms

Do I have to go a bit higher?

This is the final confirmation before I proceed with this.

 

[karesz]

Hi Aaron,
No problem, everybody has it time to time...
Your plus 100 Ohm isnt a real problem, bigger importance has that your LED are not shurly with "only"1.5V to calculate!
Its to see on some concrete datasheet of your types or measure it on a battery/powersupply & some serial resistor that its current will be similar setted as in your application...
Technology & colour versions makes differences from so 1,5V up to more as 3V :-(
Greetings!
K.

 

[betwixt]

Karesz is correct, different LED types have different voltage drops but the overall effect on brightness will be quite small. Be sure that the 22K resistor is rated at 2W or more and expect it to get quite hot!

Remember to keep your fingers out as well, those voltages bite hard!

Brian.

 

[Aaron.coG]

Hi Aaron,
No problem, everybody has it time to time...
Your plus 100 Ohm isnt a real problem, bigger importance has that your LED are not shurly with "only"1.5V to calculate!
Its to see on some concrete datasheet of your types or measure it on a battery/powersupply & some serial resistor that its current will be similar setted as in your application...
Technology & colour versions makes differences from so 1,5V up to more as 3V :-(
Greetings!
K.

Hey Karesz,

Thanks for your support & the information on the LEDs!

Oh mine.... I think I should test on the LEDs. May be I will buy some variants and see what would fit my application.

Thanks again!

---------- Post added at 12:22 PM ---------- Previous post was at 12:05 PM ----------

Karesz is correct, different LED types have different voltage drops but the overall effect on brightness will be quite small. Be sure that the 22K resistor is rated at 2W or more and expect it to get quite hot!

Remember to keep your fingers out as well, those voltages bite hard!

Brian.

Hey Brian,

Thanks again!

I will be playing around with the different LED types and their voltage drops to get an idea.

Oh mine... I didn't count on the brightness factor until now..... I was thinking of using a white color / yellow color LED. I thought this could give more brightness. :-(

Anyway, I would play around and find out what brightness I get. Then I would increase the count of the LEDs and there by the resistor value. Is this suffice?

I would surely take care of the resistor rating... how about 4w? Will this disperse the heat moderately?

And sure I would take caution about the high voltage! I would play safe.

Thanks for your concern guys! :grin:

 

[betwixt]

By all means play with different LEDs, that's the fun of electronics!
There isn't a direct correlation between color and voltage, it depends on the chemistry inside the LED junction. Generally, red is lowest voltage, yellow/orange a bit higher and green/blue are highest but the ranges overlap. It gets even more confusing with so called 'super bright' LEDs which work in a different way. Instead of actually producing the color from the LED, it makes ultra-violet light which makes a fluorescent paint glow. The mix of paint chemicals decides the color of the glow but electrically it's the same LED providing the light source. This is how white LEDs work, the blend of paint produces a mix of red, green and blue which optically look white, there is no 'real' white light source, it's just the paint glowing.

A word of caution, if you measure the voltage across the LEDs you will get a misleading result. The voltage is DC but the 1N400x diode is only letting current flow for half of each mains cycle so it isn't a steady voltage, it falls to zero and rises to a peak once per cycle. It means the LEDs will flicker but probably too fast to be noticeable. When you measure the voltage it will appear lower than you expect because for half the time the current is actually zero. There are ways to steady the voltage and to reduce the flicker but I suggest getting it working as it is for now. The extra components can be added to the existing circuit later.

Yes, a 4W resistor is fine. What confuses most beginners is that a 4W resistor and a 2W resistor generate the same amount of heat. All resistors generate heat, the rating is how much they can safely withstand without being damaged, not how much heat comes out of them.

Brian.

 

[karesz]

Hi,
Yes, both versions of your power resistors will dissipate the same heat volumen, but the 2W type with the possibly 2W dissipation on it; will be on the limit of hes capacity, means it will be very hot, but the 4W type (because has double dissipations capacity) will have considartable lower temperatur_longer life...
Anyway be care with touchings-maybe it will be (to)hot (around 80 Celsius) for your skin too_ & the dangerous line-voltages!
Concrete temperatures are very depending of the exact resistor type.
Have fun with playings!
K.

 

[Aaron.coG]

That's an awesome information on LEDs.

I was just checking this link **broken link removed** and I could see LEDs are identified in MM and Watts.

So, if I think white LEDs (of course, they seem to be white) could be effective in more light, what suits the 10mA value we were talking about?

---------- Post added at 03:41 PM ---------- Previous post was at 03:35 PM ----------

A word of caution, if you measure the voltage across the LEDs you will get a misleading result. The voltage is DC but the 1N400x diode is only letting current flow for half of each mains cycle so it isn't a steady voltage, it falls to zero and rises to a peak once per cycle. It means the LEDs will flicker but probably too fast to be noticeable. When you measure the voltage it will appear lower than you expect because for half the time the current is actually zero. There are ways to steady the voltage and to reduce the flicker but I suggest getting it working as it is for now. The extra components can be added to the existing circuit later.

This should be fine as long as the flicker isn't seen.....

And we could add the extra components once I am successful with this... your suggestion is appreciated.