Introduction: Retrofit an LED Push Light
This project started because I had an LED push light in my closet that wasn't bright enough for me to see well. I thought the batteries were just getting low, but when I replaced them, it didn't get any brighter! I figured I'd pop the light open to see what was going on inside, and whether I could easily add some more LEDs to make it brighter. But of course, colored LEDs make every project more fun, so I decided to add red, green, and blue LEDs instead. This seemed like a perfect opportunity to write something for the Made with Math Contest - if you aren't familiar with Ohm's Law or how to calculate the current-limiting resistor value for an LED, this Instructable will show you how.
If you have any questions, don't hesitate to leave a comment, and I'll try to get back to you!
- Battery-powered push light, also called a puck light or tap light. Available on Amazon or at hardware stores. This project will work with both newer LED lights and older incandescent lights.
- LEDs of your choice
- Assorted resistors - values will depend on how many batteries your light takes and what LEDs you choose (figuring it out is part of this Instructable!)
- Needle nose pliers
- Mini screwdriver set (sometimes needed to take apart the light)
- Soldering iron (recommended)
- Multimeter (recommended)
Step 1: Disassemble Your Light
Start by disassembling your light. You will usually need a mini screwdriver to do this by removing the back cover. Inside this one we can see a simple circuit consisting of the battery terminals, a button, an LED, and a resistor. An incandescent light circuit will look similar, but will not have a resistor - so you will need to add one if you're switching to LEDs.
Step 2: Decide What LEDs You Want, Look Up Specs
There are many different types of LEDs out there, and they come in all different colors. You will need to decide how many and what color LEDs you want to use. The simplest thing to do is probably to add more LEDs of the same color in parallel. If you want to add LEDs in series or mix colors, you'll need to do a little more math - but that's what this Instructable is for! We'll go over each scenario.
Once you've decided on LEDs, you will need to look up their forward voltage drop and forward current rating. This information is usually available on the website where you bought the LEDs or in the datasheet. 5mm LEDs like the ones pictured above are very common, and typically they're rated at a forward current of 20mA. The forward voltage drop ranges from about 2V-4V and depends on the color.
Step 3: Determine Voltages
If you don't have a multimeter, then at at a minimum you will need to determine the voltage of your light's battery compartment. Knowing the voltage of your battery pack is required for calculating the current-limiting resistor. You can do this by just counting the batteries. A single alkaline AA (or AAA) battery provides about 1.5V, a little higher when they're fresh. When you combine batteries in series, the voltages add up. So in this case, with four fresh AA batteries, I should expect just over 6V.
If you have a multimeter handy, it can't hurt to measure the voltages as well. Do this while the LED is on. In this case you can see I'm getting 6.26V from the battery pack, while I have a 3.26V drop over the LED and a 2.98V drop over the resistor. It's interesting to note that I'm dropping almost half my voltage over the resistor - that's a lot of wasted power! More on that later. If you don't feel like reading the color code, you can also measure the resistance of your resistor - do this while the LED is off.
If you don't know how to use a multimeter, there are plenty of good tutorials out there, on Instrucables and elsewhere. I made this one.
Step 4: Calculating the Resistor for a Single LED
Before we go any further, let's explain how to calculate the resistor value for a single LED. A resistor is governed by Ohm's Law, which states that V = IR, or
Equation 1: Voltage [volts] = Current [amps] x Resistance [ohms]
The voltage of your battery and the voltage drop across the LED are (approximately) constant. So in the circuit above, the voltage drop across the resistor is
Equation 2: Vresistor=Vbatt-VLED
Plugging that into Ohm's Law for the resistor gives us:
Equation 3: Vbatt-VLED = IR
If we have a target current for the LED - call it ILED - then we know everything in Equation 3 except for the resistance. We can rearrange that equation to solve for R:
Equation 4: R = (Vbatt-VLED)/ILED
So, for example, say we have a 2xAA battery pack (which supplies 3V), a red LED with a forward voltage drop of 1.8V, and we want 20mA through the LED. Plugging the numbers in equation 4 gives us:
R = (3V-1.8V)/0.02A = 60 Ω
Resistors tend to come in funny values - so odds are you don't have a 60Ω resistor laying around. That's where knowing how to combine resistors in series and parallel comes in handy. However, it's OK if you don't get exactly 20mA through the LED - "close enough" is probably OK for most applications!
Finally, it's a common misconception that the resistor needs to come before the LED in order to drop the voltage or limit the current. That actually isn't true - it turns out the resistor can be placed after the LED instead. Watch a video explaining why that's true here.
Step 5: Adding More LEDs in Parallel
Let's start with the simplest case: you just want to add more LEDs of the same color in parallel to make your light brighter. It might be tempting just to add the LEDs directly in parallel with the existing one, keeping the single resistor. This works, as you can see in the photos above - all three LEDs light up - but there's a problem! The current through the single resistor doesn't actually change when you add more LEDs in parallel. Since the LEDs are in parallel, the current through the single resistor is going to split evenly between them. So before, where I had 20mA through a single LED, now I'm only getting about 20/3 = 6.67mA through each LED - and they won't be as bright!
Instead, if you add an individual resistor in series with each LED, then you will get the full amount of current through each LED. The downside is that this will drain your battery three times as fast. This isn't a big deal in my case because I have the light mounted in a closet, and won't be using it that often.
So if you take this approach, you've avoided doing any math - you just need a few more of the same resistor. In my case, I have a 100Ω resistor and a white LED - I just need two more of each to get a total of three LEDs in parallel. (Unfortunately I forgot to take a picture of that step, before I switched over to colored LEDs).
Step 6: Add More LEDs in Series
What about adding more LEDs in series? This option works if your battery pack voltage is high enough. LEDs won't turn on at all if the battery pack voltage is below their required forward voltage drop. Just like with batteries, when you combine LEDs in series, their voltages add. In my case, my white LED has a voltage drop of 3.4V, so putting two in series would require 6.8V - more than my battery pack is providing. However, for example, you could do it with two (or even three) red LEDs. If each red LED had a voltage drop of 2V and you wanted a current of 20mA, that would give you a resistor value of R = (6 - 4) / 0.02 = 100Ω. Compare that to only putting a single red LED in the circuit: R = (6 - 2) / 0.02 = 200Ω. By adding a second LED in series, you have greatly decreased the size of the resistor - but you're still only drawing 20mA, so you're not draining your battery any faster! You've made the circuit more efficient because you are dissipating less power in the resistor. That brings in another equation - for a resistor, power equals current squared times resistance, or
So 20mA through a 200Ω resistor dissipates 80mW, whereas 20mA through a 100Ω resistor only dissipates 40mW.
(again, apologies that I don't have a picture of this example - I went right to colored LEDs in parallel)
Step 7: Mixing Different Color LEDs
If you want to mix different color LEDs, you have two options:
- Wire them in series - this works as long as the total voltage drop across the LEDs is less than the battery pack voltage (see previous step).
- Wire them in parallel, each with their own resistor - assuming you want all the LEDs to have the same current so they will be the same brightness, you just calculate the resistor value separately for each LED, given its forward voltage drop.
Sometimes it's easier to prototype circuits like this on a breadboard first (if you don't know how to use a breadboard, check out this tutorial). Again, you might not have the exact resistor values you want handy, so you can play around with different combinations in series/parallel to fine-tune your brightness. In this case, I have a 4xAA battery pack providing about 6V, and red, green, and blue LEDs with voltage drops of about 2V, 2V, and 3V respectively. That gives me a resistor value of R = (6-2)/0.02 = 200Ω for the red and green LEDs, and 150Ω for the blue LED. I don't have those exact values available, but I can create a 200Ω resistor by combining two 100Ω resistors in series, and I can get "close enough" to a 150Ω resistor by combining a 100Ω resistor and 47Ω resistor in series.
Step 8: Adding LEDs in Series AND Parallel
Feeling adventurous and want to make this thing REALLY bright? Add LEDs in series AND parallel! Again, remember that in order to add LEDs in series, your battery voltage needs to be higher than the sum of the LED voltage drops; and if you add LEDs in parallel, it will drain the battery faster. Do a separate calculation for the resistor value for each set of LEDs in series.
Step 9: Put It Back Together!
Prototyping on a breadboard, with alligator clips, or by bending leads together with pliers is a good way try out different LED/resistor combinations before your finalize your design. When you're all done, soldering the LEDs and resistors together will help hold them in place when you reassemble the light. Some tape or hot glue can also help prevent them from moving around. With the red-green-blue LEDs, I had to disassemble and reassemble it a few times until I got the LEDs arranged such that they gave an evenly diffused effect for all three colors. The final product here probably won't be very useful in my closet, but it sure looks a lot cooler!
Runner Up in the
Made with Math Contest