# Finding resistor values with a high power LED on an Arduino (Limited experience with circuitry)

I am having difficulty determining resistor values on a 3W LED circuit. It's running off an AT85 Tiny, and being powered by a 9V battery. We want to be able to monitor the battery level, as well as pulse a couple of small LED's using PWM when the battery is low. The 2 LEDs are 3W, 3.5v max at 350mA.

How do we go about calculating the resistor values in this case?

• Please specify the device names of the LED's. Please do read electronics.stackexchange.com/questions/426765/… Mar 26 '19 at 14:38
• A common 9 volt battery will not last long delivering 700 mA (if it can do it at all...) Mar 26 '19 at 15:15

The standard way to determine the correct resistor value (and wattage!) involves the following process:

1) Determine the desired LED current when the LED is on. You can use the data sheet for this. The LED light output is roughly proportional to the current. Don't exceed the recommended maximums. Let's call this $$\I_{targ}\$$.

2) At the target current determined in step 1 ($$\I_{targ}\$$), estimate the LED voltage drop again using the data sheet as a guide. There are usually graphs in the data sheet that tell you what the typical voltage drop will be vs the LED current. Lets call this $$\V_{led}\$$.

3) Subtract the LED voltage drop from the $$\V_{CC}\$$ voltage. This is the voltage the resistor must drop. Let's call this $$\V_r\$$ (voltage across resistor). Assume the voltage drop across the MOSFET switch is minimal. If it isn't then $$\V_r\$$ will be the total non-LED series voltage to drop. $$V_r = V_{CC} - V_{led}$$

4) Use Ohm's law to calculate the resistance as $$\R = V_r / I_{targ}\$$.

5) The power dissipation in the resistor can be calculated as $$\P = V_r \times I_{targ}\$$

Fine points:

• Once you have determined the power dissipation in the series resistor, chose a part that has a wattage rating 2x or more of the actual dissipation. Failure to observe this will result in a very HOT resistor.

• Choose a MOSFET switch with a low $$\R_{DSon}\$$.

• Remember that there could be significant variation in LED voltage drop from unit to unit.

• I added math formatting to your answer; I think it makes it easier to read. You can take a look at my edits to see how it is done. Mar 26 '19 at 15:24
• Thanks Elliot! Looks great. Much appreciated. Mar 27 '19 at 19:47
• It reads well but ignores fundamental errors in question. You cannot possibly choose ANY resistor value to power two 350mA LEDs from a 9V Alk. cell. Mar 28 '19 at 5:10
• Quite true! I figured the questioner would soon discover that a 9V battery would not last long or even be able to supply the required current. My answer was simply limited to answering the more generic question "Given a fixed VCC and LED type, how can I calculate an appropriate resistor value and wattage". Mar 29 '19 at 14:00
• One more thing I just thought of... If the duty cycle was VERY low and a good-sized bulk bypass cap was added across the battery, the circuit could conceivably work. Furthermore, if the duty cycle was << 1, the wattage rating of the resistor could be significantly reduced. Mar 29 '19 at 14:03

3.5V*350mA is 1.3W not 3W which is probably peak pulse power only or large heatsink.

This approach of mismatching load voltage 3V~3.5V LEDs to 8 to 9V battery voltage results in at least 50%~66% power loss and probably thermal neglect to LED heatsink.

Therefore try a Li-Ion solution better matched to LED string voltage and regulate current with a 50mV drop at Imax and choose a FET rated for at least 20x desired current for 50mV drop max and Vgs(th)=1.1 max or 1/3 Vbat (min) whichever is larger.

The 9V Alk. battery has an average of 5~10 Ohms of ESR when new, then rises. ( 1 to 2A short circuit ) With 350mA design current, the battery will heat up and die very quickly.

• I see someone -1 who has no clue to comprehend, so votes in silence and cannot even say something intelligent Mar 28 '19 at 5:01