Why there are 2 diodes present around LED. Can anyone tell me why it is done so. Will it remain on in normal condition or remain off.
I want to use led whose dataheet is attached. Which one on ltspice would be best suited.

 

D11 halves the power dissipation, but it's reverse voltage rating is too low for the situation.

The circuit will work fine without D11, but the resistor will get twice as hot.

 

D11 halves the power dissipation, but it's reverse voltage rating is too low for the situation.

The circuit will work fine without D11, but the resistor will get twice as hot.

Isn't it's going to remain same if d12 is removed along with d11? What is the purpose of adding 2 diodes then?

 

Without D11 all positive cycle of AC source would be just waisted on resistor.
D12 clamps the leakage current of D11 to not harm the Led with high reverse voltage.

 

D11 is intended to provide only a positive voltage to the LED. Diode D12 is intended to prevent reverse voltage destruction of the LED. Unfortunately the circuit will fail and probably all three diodes will be destroyed because the supplied voltage exceeds the allowable reverse voltage of the diodes.

 

D11 is intended to provide only a positive voltage to the LED. Diode D12 is intended to prevent reverse voltage destruction of the LED. Unfortunately the circuit will fail and probably all three diodes will be destroyed because the supplied voltage exceeds the allowable reverse voltage of the diodes.

Yeah got your point. But in actually led and diodes of different make will be used. I have only demonstrated it here for an idea. It will be near load and not near supply.
Also will led glow if load is not connected?

 

I calculate the LED current as about 25 mA, a bit above the 20 mA rating. The LED illumination is not related to the load off to the right side of the drawing. That is a parallel connected circuit, they are totally independent.