**TWO LED TROUBLE LIGHT**

**INSTRUCTION SHEET**

Price - Stock No 547 (Aligator Clips)

Price - Stock No 548 (Accessory Plug)

**The 18000 mcd LED**

**Specifications**

Emitted colour:
White

Lens colour: Water Clear

Wave length: 0.315(nm) max **Pd
W: 120mW** **If mA: 30**

If mA (peak): 150

Vf
Min: -**Vf (V) Typ: 3.3Vf Max: 3.6**

Iv Min: 17000

mcd Min:17000

mcd Typ: 18000

Viewing Angle: 15

As with all LED's, the current through the LED is the critical piece of
information - **If mA: 30. **If this is exceeded it is
likely the LED will be destroyed.

It is also interesting to note the normal operating voltage limits - **Vf (V) Typ: 3.3 **and **Vf Max:
3.6.** This indicates that to power the diode without a resistor
the voltage would **Typ**ically be **3.3 V** to a
**Max**imum of **3.6 V**.

** **

**CALCULATIONS**

{1}

Simplify the circuit and look at one branch:

**To calculate the total resistance required to limit the
current to 30 mA,**

we need OHM's Law.

V = 12

I = 30 mA

therefore R = 400 Ohms

This is the total resistance of the unknown resistor value + the inherant resistance of the LED.

**To calculate the resistance of the LED**

If we note that at a maximum voltage of 3.6V the LED needs no resistor - therefore we can assume that the LED will draw 30mA when connected to a 3.6 V source. Again we use ohms law

V = 3.6

I = 0.030 A

Therefore R = 120 Ohms

If we now think of the diode as aResistor in series with the unknown resistor, by deducting the resistance of the LED from the total resistance we learn the unknown resistance to operate the LED without destroying it. So a 280 Ohm resistor should be used with a 12 V source:

400 - 120 = 280 Ohms

You may now ask why a 390 Ohm resistor was used instead of a 280 Ohm resistor - because the voltage on a vehicle varies and can be as high as 14.5 Volts - to eliminate the possiblity of damage to the LED a resistance buffer was included in the value of the chosen resistor.