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# Transformer vs Potential divider Answered

Hi guys!

I got bored and decided to make a study table light using LED's. My setup consists of 4 LED's in an arm and 10 such arms in parallel. I actually designed it in view that the transformer i used converts 230v - 15v. But after it all finished, i found out that all my transformers are used up for other circuits. I was wondering,

Can i use a potential divider for such simple circuits?

I approximated that if i place a 46K and a 3K resistor in series and take the voltage across the 3k resistor, i will have about 15 v.

What are the disadvantages of using such potential divider?

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## Discussions

That it wont work. :)

See, a volatge divider is not a trivial matter it you have a load on them.
Sure: It can be calculated quite easy what voltage you have over either resistor... But now if you add the LEDs or another load, you have that in parallel to the one resistor, which changes the total reistance of this part.
--> It will change considerably. especially since LEDs (and diodes in general) have a very low dynamic resistance. This means: As soon as they start conducting current, small changes in voltage result in big changes in current.
So think about them like something of a "nearly-short".
And a short in parallel to a resistor makes the resistor obsolete.

So a voltage-divider is NOT the way to go. BUT there is another simple way:
A series-resistor.
For simplicitys sake, i say a white LED uses 3V to function.
Also it is propably best if you design the circuitry for every arm individually: You can in theory switch every arm by itself on or off...

So lets get to work!
4 LEDs x 3V each = 12V accross the LEDs.
So you have to "loose" 230V-12V=218V over the resistor.
Only with those datas we cannot do that. We need to know the current needed for the LEDs. Normally thats 20mA (Adjust that value if your LEDs are power-LEDs and need more current!). Thats very important: The current is the thing to take a look at and NOT the voltage. Thats because the voltage is a result of the current in this situation...

So we have a Resistor-value of:
R=U/I = 218V/0.02A=10900 Ohms. --> Make 11kOhms out of that or recalculate for another current...

So the circuitry looks like:
----[11k]---|>--|>--|>--|>-----
where --|>-- is a diode and --[xxx]-- is a reistor of the given value.

But dont forget the power!
A resistor is nothing more than a heating-element! So lets see that we dont fry that! :)

The dissipated power of the resistor is:
P=U*I = 218V*0.02A= 4.36Watt --> Go 5Watt for that resistor.
A normal 0.25W-Reistor will go up in smoke if you take that :)

So we have: around 4.36W waste-heat and "only" (12V*0.02A) 0.24W used energy in the LEDs.

And now you see the big disadvantage of the resistor versus the transformation of the voltage: A big ammount of energy is wasted in the resistor.
The more of the 230V you "use" over the LEDs, the better this proportion becomes.

As an example if you take 50 LEDs in series, that will dropp (if i go for 3.6V each LEDs and 100mA for Creee power-LEDs), that makes a resistor of 1.8kOhm, 18W used in the LEDs and 5W wasted as heat in the resistor.

BUT: Dont forget that LEDs use DC and not AC. so if you plan on hooking them up to the mains, be sure you rectify the AC and smooth it out with a really big capacitor.
As a rule of thumb you should have 1000uF for each ampere you use.
So with 20mA for the LEDs in each arm and 10 Arms that makes 200mA --> 200uF at least. Go for a 470uF if you ask me... :)
If you have 100mA in each arm, that would make 1000uF at least --> Go 2200uF if you ask me...
And since 16V is a common Capacitor-Voltage for electrolyte-Caps, you are in the green also there.

If you have more questions about what i proposed, just ask :)

Thanks for the reply! Your explanation is awesome pal! So i guess i will stick to the transformer circuit for its efficiency.

Also i noticed you mentioned that for 100mA current, use 100uF capacitor. So we have to select the capacitor value depending on the current drawn? say if i want a 500mA load current, i need to use a 500uF capacitor?

As a rule of thumb: Yes. 1A --> 1milliFarad
The exact formula is a bit more complex and also asks about the ripple-voltage allowed and also the moved charge (in Colomb) needs to be taken into consideration.
But in this simple setup we can go for that rule of thumb.

Also the flickering (occures if the capacity is too small) would be at twice the mains-frequency which would result in 100Hz (Given your mains are 50Hz) which is what normal TVs have as refresh-rate. A PC-Monitor often has 60-72Hz and you also dont see flickering... So i think that wouldnt be this bad even if you dont smooth the voltage at all.

Please also keep in mind that you need to know WHERE to put the capacitors.
That would be just after the rectifier. Accross the rectified voltage... As shown here http://forum.allaboutcircuits.com/showthread.php?t=56064
What is shown there with a resistor is for your setup the resistor in series with the 4 LEDs...

And yeah: I just noticed a small error i made in the first post: There i said that 16V is a common voltage. Yeah. It is. But we need 325+V here for our caps... (230V*squareroot(2)). So go for 400V-Rated capacitors. Sorry for the hickup :)
eBay has always such caps on stock:

LV-side is saver, but not this effective. But since those 16V-Caps only cost a few cents, i also would go for a low-side-puffering with maybe a little bit higher capacity.

You can puffer what side you want, but the HV-Side would be "Better" in terms of electronics (Since you only need to puffer one line). If you go the saver way and puffer the low-voltage, you need to puffer every branch or arm separately.
If you arent experienced with mains-voltage i suggest you indeed go for the puffering of the low-voltage-side.

Would need 89 and a half LED (supposed your LEDs are 3.6V white ones).
He
1st: dont want this many
2nd: would have to deal with high voltages just to the very end of ever arm
3rd: would have a 50Hz flickering and
4th: only would have around 25-30% of the time lighted LEDs since the negative half-sine wont drive a diode and as soon as the positive half-sine reaches the Uforward (from 325V downwards) of the Leds would shut off. Same goes for the raising positive halfsine.

--> Makes one crappy lightsource. Expensive, big, dangerous and labor-intensive