Introduction: Doorbell LED
in my previous Instructable, I showed how I went about designing a replacement doorbell button. In the process of putting it together, I broke the bulb that illuminates the button AND the circuit board that held it all together.
In this instructable I will show you how I went about replacing the bulb with white LEDs. I optimized the design for low part count due to space constraints. In the process, I learnt a few more things about these humble doorbell circuits which I'll share with you.
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
Step 1: Measure the Voltage
In my previous instructable, I measured 20VAC with my multimeter across the Red and Black power terminals...somewhat odd for voltage...seemed rather non-standard. I would have expected 24 or 12 or even 5 but 20 just seemed odd. Luckily I have an access to an oscilloscope at home. Not everyone has one, and you don't need one to go through this instuctable, but out of curiosity I just had to understand why anyone would choose 20VAC...
In order to look at the waveform, set the oscilloscope (if you have one) to 'AC coupling' and probe across the power terminals.
In my case, I observed the following:
The Peak to Peak voltage is actually 35.2V (!). Note: Peak to Peak voltage is the difference between the highest and the lowest point of the AC waveform.
The frequency is 60Hz. Note: frequency refers to the cycles per second that a repeating signal (in this case the AC current) that traverses back and forth through the circuit.
Now I realize that this is perhaps a little too technical already for some folks...again, I wanted to ensure that I understood what was going on a a more deeper level while also providing more information for readers with a background in basic electronics. Here is some background for those readers that might need it.
From the measurements so far, 60Hz makes sense: US Household AC has a frequency of 60Hz, so one can deduce that there is a simple transformer between the mains and this doorbell circuit. But what about the waveform with a peak to peak of 35.2? Seems rather odd. It does, until you realize that the 'standard' 24VAC that is used in many household applications such as HVAC, etc, actually refers to the 'RMS' value of the voltage. A relationship between peak to peak that we observed with our oscilloscope and RMS exists and it is: VRMS = VPP/sqrt(2). If we plug that into our measurements: 35.2/sqrt(2) = 24.9V...close to 24VAC. So we can conclude that our power-supply to the doorbell is actually 24VAC...
So why the discrepancy between my multimeter and scope? Seems like another mystery to solve! I believe my scope...but on the other hand, 20VAC that we got from the multimeter is close enough for what we want to do, which is simply to illuminate the button.
Step 2: Find a Suitable Light Source
In my previous instructable, I describe the challenges of finding a suitable (and cheap) replacement bulbs. Small miniature bulbs are hard to come by these days...luckily, recently I had purchased a batch of white LEDs from Ebay, so I started to think about using LEDs instead. It cost $1.76 for 100 pcs! But be warned, that these ship from China so can take several weeks to arrive.
Illuminating LEDs from an AC source has its own set of challenges which I'll describe in the next few steps as well as suitable solution I found for our particular problem.
Step 3: Powering White LEDs From an AC Source
If you know anything about LEDs, besides 'emitting light', they are a special sub-class of electrical devices called diodes. Diodes only allow current in one direction. If we were to send an alternating current through a diode, it only allows current in the 'forward' direction (anode to cathode) while rejecting current flow in the 'reverse' direction (cathode to anode).
The usual means of powering an LED is to convert the AC voltage to DC so that current flows only in the forward direction through the diode. Converting from AC to DC in the classic sense requires a 'diode-bridge'. Following the diode-bridge would typically be some smoothing filter typically a resistor and capacitor and then finally a resistor to reduce the current through the LED below its supported power level. As you can probably tell, the classic means of powering an LED requires a lot of additional componentry to realize; not including the LED, up to 7 components are required.
For this case, I thought optimizing for part-count: Not including the LEDs, only one more component (a resistor) is required! Or if you count the second LED as extra, then two extra components are required...
Like all diodes, LEDs have design parameters that, if exceeded, result in damage. Here are the specifications of this LED
- Emission color: White
- Dimensions (mm): 5 mmLens
- Color: Clear WaterPeak (nm)
- wavelength: N / Anet
- Voltage (V): 3.2 ~ 3.4
- Reverse current (uA):Less than or equal 30
- Intensity light typical Iv (mcd): average 20000
- Note: Life 100,000 hours
- Viewing angle: 20 ~ 25 degrees
- Maximum absolute (Ta = 25 Celsius)
- Max Power Dissipation: 80mW
- Max continuous current: 30mA
- Max peak current: 75mA
- Reverse voltage: 5 ~ 6V
- Lead soldering Temperature: 240 C(<5Sec)
- Operating temperature:-25 Celsius ~ 85 Celsius
- Storage temperature:-30 Celsius ~ 100 Celsius
- Package: 100 x 5 mm white LED
The bold items are of particular interest. The LEDs cannot exceed 30mA current in the forward direction (with a forward voltage drop of 3.2V) and is capable of rejecting up to ~5V in the reverse direction. If we were to connect these LEDs directly to the AC supplying the doorbell, it will face up to ~18V in either direction (~35V/2, see scope-shot), definitely damaging the LED.
To avoid damaging the LED in the forward direction, a resistor is used to limit the forward current at the peak forward voltage of ~18V to below 30mA. This is the classic 'biasing resistor' that is universally used to limit current (and brightness) through a forward biased LED. The resistance value calculation would be, (assuming we want to set forward current to 20ma), using ohm's law:
(18V - 3.2V)/R = 20mA, R = 740ohms; nearest standard resistance value is 750ohms. You can use a value greater than this to decrease the intensity of the LED as desired.
To avoid damaging the LED in the reverse direction, I thought of connecting a second LED in parallel (but reverse) to the first LED that it is forward biased upon the reversing of the alternating current! To clarify, the second LED's Anode must be connected to the first LEDs Cathode and vice versa (see pic). The idea is that the second LED would maintain ~3.2V across the first LED, when the current is in the opposite direction, effectively clamping it to below 5V where damage can occur.
I also attached a picture of the whole thing connected together.
Step 4: Connect It Up
Just connect the blue and red door-bell power wires to the circuit to ensure it lights.
If it is too bright (which it was in my case), you can increase the resistor value.