Mini lamp with white clear LEDs is a lamp that is assembled with 9 white clear LEDs, 2 capacitors, 1 voltage regulator of +9V, and a DC Power Jack; this last component for being connected an  AC/DC Wall Adapter/transformer Single Out +9Volt, 0.3A.  All of components are mounted on round PCB of 30mm diameter. Or you can only use 12 LEDs and a DC Power Jack if you utilize a Rgulated Wall adapter/ transformer.
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Step 1: List of Materials

Picture of List of Materials
DC Power Jack with 2.0mm Center Pin.png
9 White Clear LEDs
1 Round PCB of 30mm diameter
2 Electrolytic Capacitors
1 Voltage Regulator +9V, 1A
1 DC Power Jack for 9V plug in Wall Adapter/Transformer
1 Wall Adapter/Transformer Single Out 9V, 0.3A

Step 2: Mount the components

Mount the LEDs by observing the positive pin (anode pin) and the negative pin (cathode pin). 

Step 3: Complete the project

vricsi941 year ago
What limits the current flowing through the LEDs?
There are 3 LEDs in series - and since the LEDs require about 3V, with 3 in series at 9V, each LED drops 3V across it. Since none of the LEDs are exposed to more voltage than they require, the current doesn't need to be limited as it would in a circuit where there is only 1 LED and 5V across it. Current limiting resistors in cases like this are actually doing 2 things - yes they limit the current, but more importantly, they are dropping the excess voltage that would otherwise burn out the LED.
Technically, each LED in this circuit is using the other 2 that are in series with it as it's current limiters.
I would also like to make clear - if you had a 3V source, and a single LED - the LED will only draw what it needs. You could have a 500amp 3V power supply, and the LED will still only allow 20-40 mA through it. But put a 5V 50mA power supply across an LED without a current limiter, and the LED will likely burn out.
braulio777 (author)  SuperTech-IT1 year ago
Exactly, you're right.
braulio777 (author)  vricsi941 year ago
Interesting question, remember that a power supply provides a determined current (for example: 1Amp) and each device connected to the power supply will take what it needs. For instance, if it needs 0.5 Amp, you will only have 0.5 Amp available to others devices that you want to connect. That is, if you use the correct voltage, the current is not a problem.
I hope you understand my response.
vicvelcro2 years ago
Constructively, a couple of things I think deserve mention:

Electrolytic capacitors are polarity specific. Connecting them in reverse polarity is not a good thing.

The capacitor on the input side does not need to be electrolytic. It can be ceramic, mylar, or tantalum. The value of the input capacitor can be between 1 microfarad and 10 microfarads.

The capacitor on the output side should be electrolytic and should be 10 times the value of the input capacitor.
braulio777 (author)  vicvelcro2 years ago
You are right. But as the Wall adapter/ transfrormer used is unregulated and its voltage is about 17volt, I need to do something to protect my LEDs. What do you think? The critics are welcome.
Your 7809 regulator should be fine accepting an input up to 32 - 35 volts. The higher the input voltage, the more heat has to be dissipated by the regulator, so a heat sink would be a good idea to add. Your LEDs are already protected, even without capacitors. The regulator limits the voltage. The capacitors smooth it out so the light emitted remains steady.

Your circuit is fine, as it is. My previous suggestions regarding the capacitors are only an improvement to what you already had. The difference between a cheap sedan and a luxury coach, basically.

I didn't intend to mislead you into thinking something was 'wrong'. If I did, I apologize.
braulio777 (author)  vicvelcro2 years ago
Really, I appreciate your comments.Thank you very much
braulio777 (author)  vicvelcro2 years ago
Really, the value of the capacitors is not critical, but in this case, I agree with you because one combination of capacitors that you suggest will function well.
You are correct, the values are not critical. However, some values will perform better than others and some values are just overkill. Too low, we may still have ripple in the voltage. Too high is just unnecessary. The input side doesn't need as much capacitance as the output side, for filtering purposes.

Since you are feeding 9 volts into a 9 volt regulator and will be dropping voltage across the regulator, you will be getting between 7 and 7.5 volts on the output side when under load. Because the output is starved, it's a good idea to put more filter on that end. Also, because the input side is being pushed into the regulator, the input capacitor is "stuffed" more fully and won't likely see as much ripple coming in.

There is nothing wrong with using other values in your circuit, though. You simply might notice a slight flickering effect. Then again, maybe it won't be obvious.