## Introduction: 1 Watt Led Driver

Hello! friends

Welcome to my another 1 watt led driver project. It's simple and easy to build. I just found a 1 watt led driver circuit diagram on internet and i build it because it's helpful for me.

So let's get started.

## Step 1: Circuit Diagram

With the help of this diagram we can build our own 1 watt led driver circuit just follow the steps.

## Step 2: Materials

You'll need

Splitter box

Piece of veroboard

Soldering iron with solder

Some piece of wires

1 watt led

Diodes: 1n4007 x4

Resistors: 1m, 10ohm x 3

Capacitors: 1uf/400v, 4.7uf/250v x 2 and 100uf/50v

## Step 3: Front and Rear View

Cut the piece of veroboard according to the size of splitter box. Place the components on veroboard according to the circuit diagram and then soldering it very carefully.

## Step 4: Finish

Now fit your circuit into the splitter box and now your 1 watt led driver is ready to use.

I hope you guys like it.

Thanks.

you need a constant current for turn on led, and this led is power led 1 watt with 3.2 volt.

so current need is= 1/3.2 =312mA THAT NORMAL CURRENT IS 350ma.

for making 350ma you must sink 350mA from source.

capacitor need is 5uf 400v , but this is very big capacitor and burn after 24h work!

because after work temperate is very hot.

so you must use of 5 *1uf in parallel mode

and use of true resistor for limit current.

don't use of zener diode for limit voltage, because this burn very soon, maybe in start!.

and use of transistor high voltage!.

but in final, i idea is better use of this circuit, because input wattage is very high, and price energy is more.

notice: if any one need circuit similar but with transistor i can help him, and my mail is abozaravar@yahoo.com

have a good day guys.

12v !

when you use of this circuit, often burn your led in start!

because capacitor is empty and for full it , you need a charge time capacitor with this formula:

t=R*C

and resistor can limited current for charging and decharging and is very important in this circuit:

DischargingCharging

V

(

t

)

=

V

0

(

1

−

e

−

t

/

τ

)

{\displaystyle V(t)=V_{0}(1-e^{-t/\tau })}

V

(

t

)

=

V

0

(

e

−

t

/

τ

)

{\displaystyle V(t)=V_{0}(e^{-t/\tau })}

and:

τ

=

R

C

=

1

2

π

f

c

{\displaystyle \tau =RC={\frac {1}{2\pi f_{c}}}}

or, equivalently,

f

c

=

1

2

π

R

C

=

1

2

π

τ

{\displaystyle f_{c}={\frac {1}{2\pi RC}}={\frac {1}{2\pi \tau }}}

similar down picture:

for ever you have a different current in any resistor,because z=R+1/2pfc.

and in final , input wattage is =Vac * I ac

and i test circuit with watt meter and see 15 watt in that device when operate that.

Sir, I think the watt will be 70ma(apx)*12v =8.4 watt. Are u having any clarification?

when you use of capacitor in ac source then capacitor is similar resistor!

impedance capacitor is = 1/jwc so:

z=1/2*p*f*c

z=1/2*3.14*50*0.000001

z=3184 +(R series for limit peak current charge and decharge capacitor)

i=v/z

i=220/3184 =0.069 A = 69 mA

AND MISTAKE IS HERE:

P=V*I

P=220*0.069 =15.18 WATT !

so this circuit is not 1 watt and is 15 watt!

you can check it with watt meter.

for 220v??

Yeah

if you get an Short on the led line your 1 Watt resistor in the ac line and all the other resistors will go very hot. maybe it Start burning ...

Transform the ac Main in the First Step down to a non risk voltage ( under 50 Volt ) then do all the other components.

D'ont forget to put the led in a box as there is no galvanic isolation!