LEDs and AC Supply

Why do Engineers always have to re-invent power supplies? <sigh>

Using a transformer is the more energy-efficient way. However, here is one way to do it without one:

• Rectify the 230VAC to 230VDC
• Drop the voltage using a pair of high power resistors
• Bung a voltage regulator across one of the resistors and set it to 9V.
• ...and cook toast on the dissipated heat.

"Robert is your mother's brother".

Of course, such a circuit is not immune to the risk of electric shock from the mains in the way that one that uses a transformer could be......

Er, why not use a commercial lighting power supply unit instead?

(about £8GBP from major home improvement outlets)

Spend a couple of hundred dollars (or thousands INR) and get a SMPS.

BTW: Engineers do not work "Without ..." They work with the most efficient and cost effective piece of equipment available.

Hence I dont want to use transformer is a nonscientific way to look at things unless there is a specific reason (eg a homework the other portions may be imagination ).

I am not too sure. And if I am not wrong this is the second or third question where the OP wants to step up/down the voltage with specific condition no transformer.

Is there an anti-transformer campaign going on?

agreed and GA...isolation from AC mains is important, regardless of the circuit build.

Words of wisom from Jack. Heed them!

As asked, why would you not want to use a transformer?

Why not get a Wall Wart, one of those little black things and plugs in a wall socket and delivers a low DC voltage. You should not have a problem getting one that delivers 9 volts. Assume that your four LEDs will require 1.2 volts DC at a current of 20 miiliamps.

9V - (4*1.2) = 9 - 4.8 = 4.2 volts

You will need to drop 4.2 volts at 20 ma

I=E/R .020 = 4.2/R R = 2100 ohms

If I am not too confused the four LEDs plus a 2100 ohm resister connected in series should work. a 1/4 watt resister should be rugged enough.

I hope that I haven't lead you astray. If you used a 6 volt wall wart you would only have to drop 6-(4*1.2) = 1.2 volts and the current limiting resister could be smaller.

hello Rajesh

easiest solution (possibly) for your problem is buy a cell phone charger for US $1.00 open it, tinker with the zener diode if you will. to obtain the desired low voltage ,delivers about 350 ma though says 500 ma

its low voltage output is Galvanically isolated from the mains and safe ,be carefull about hi voltage on board while experimenting.

thanks

P.K.D

1. visit: http://www.lynklabs.com/ for more knowledge
2. If mains isolation is not required; that is you not going to touch any LED while "ON" then one diode 1N4007 in series with 12Kohm resistor can drive LEDs directly
3. Buy laptop or AC mains adopter or cell phone charger which can power up LEDs