# Resistor wattage?

I'm new to electronics, what types of resistors should I get to make this circuit?

I have 6 IR LEDs (High output infrared LED 5mm 1.2VDC 100mA 940nm) that will be connected to an Ardunio Uno.

Going to be plugging this into the cigarette lighter which I think goes up to 13.5vdc?

• That circuit you've drawn looks suspiciously like an SCR. If that's the case, you'll be able to turn it on once, but it will then latch on, and you'll have to reset the supply to turn it off. – Thomas O Oct 4 '11 at 6:58
• @Thomas - no it's not an SCR. An SCR is a combination of a PNP and an NPN. The BC337 is a current limiter. – stevenvh Oct 4 '11 at 8:04
• Be warned that the power in a car can spike to in excess of 40V - often up in the 60V region. You should use a properly rated regulator to limit the spikes. – Majenko Oct 4 '11 at 8:53

The 330$\Omega$ will see a very low current, so that can be a standard 1/4W.
Right, that's spoken from experience. Let's do the calculation anyway. You always have to calculate for worst case, and worst case here means the highest voltage drop across the resistor. Therefore let's take 5V in (don't know if Arduino works at 3.3V or 5V) and 0V across the 6.8$\Omega$ resistor. Then

$P = \dfrac{V^2}{R} = \dfrac{(5V - 0.6V)^2}{330\Omega} = 0.059 W$

which confirms what we thought. So a leaded 1/4W resistor is possible, but also an SMT resistor down to 0402 size will do. A 0402 will allow 100mW.

The 6.8$\Omega$ resistor will see slightly more than the collector current $I_C$, which will be limited to 100mA by the current limiting transistor BC337. Then

$P = I^2 \cdot R = (0.1A)^2 \cdot 6.8\Omega = 0.068 W$

So a 1/4W will do here as well.

edit
About the BC337. This works as a current limiter. If the current through the 6.8$\Omega$ resistor is greater than about 100mA the voltage drop will be 0.7V and the BC337 will start to conduct, reducing the base voltage of the BD135. This way the emitter current (and hence the collector current) of the BD135 can't go higher than 100mA.

• So what's the advantage of using this type of circuit (an extra transistor) to limit the current across the LEDs as opposed to a simple resistor? – m.Alin Oct 4 '11 at 11:33
• @m.Alin - A series resistor will drop the voltage difference between your supply and the sum of the LEDs, so the current will depend on the supply voltage. In the given schematic it's independent of that. – stevenvh Oct 4 '11 at 11:36
• Dissipation in the BD135 will be 100 mA * (Vsuppy-Vleds -0.7) or 0.1*(13.5-7.2-0.7) or 560 mW. BD135 junction/ambient is 100 C/W, so the junction will rise 56C. No problem for reasonable ambient temperatures, will never reach the abs. max temperature of 150C. – markrages Oct 4 '11 at 22:34

I can explain the calculation here.

EDIT: (Please note that it is a calculation in which the transistor BC337 is omitted)

Your input voltage Vdc = 13.5V

Voltage drop across a single IR LED Vled = 1.2V

Total voltage across 6 LEDs 6X1.2V = 7.2V

Now the Voltage that must be dropped across the Resistor Vr = 13.5-7.2 V = 6.3V

The current Flows through the LED I = 100 mA

The current Through the resistor is same, I = 100mA

Now the Value of resistor R = V/I = 6.3V/100mA = 63Ohm.

(Puting a standered value, R = 6.8 Ohm)

Now The Power Calculation

The power dissipated in the Resistor P = V X I = 6.3V X 100mA = 0.63W

Now you can't use ordinary 1/4W or 1/2W resistors.

It is better to use a 1W resistor of 68 Ohms.

• You're ignoring the current limiting BC337! – stevenvh Oct 4 '11 at 7:53
• @stevenvh How does the BC337 transistor work in this circuit? – m.Alin Oct 4 '11 at 7:55
• Also, I think you're ignoring the voltage drop across the BD135 transistor. – m.Alin Oct 4 '11 at 7:58
• @m.Alin - I'll add it to my answer. – stevenvh Oct 4 '11 at 8:07
• Saneesh's answer applies if onlt the resistors is used as a current limiter. Here the main voltagedropping is done by the BD135 transistor so Saneesh's answer does not apply in this case. – Russell McMahon Oct 4 '11 at 8:15