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Trying to get back into electronics with a simple lighting project

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TheRabbit2

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It's been a while since I've worked with circuits. Used to work with them while in college as an biomed undergrad but couldn't find work so Im now in the tech support field.

Anyways, the bulb in my lamp burned out, it is a small 12V bulb that produced a lot of heat. I opened up the lamp and found two inductors and a fuse. While experimenting around with replacing the bulb with leds, I blew the fuse out.

So instead of wasting more money and experimenting, I wanted to see if the following schematic will work:

lighting.JPG

I was thinking about using a voltage divider to lower the 12V input to 3.55V to use with 5 LEDs that I have.
The single one on top has a current rating of 20mA and 3.5V-4.0V
The four in parallel have current ratings of 25mA and 3.3 to 3.6V

From what I remember about electronics, for the voltage divider it is Vout = Vin * (R2/(R1+R2)),
therefore I am using 470Ω (R1) and 1120Ω (R2) to get 3.547V.

From there I am using 173Ω (R3) to drop the current down to 20.5 mA
and 33Ω (R4) to drop the current down to 26.8 mA for each bulb.

Now the thing I can't seem to remember is, what is the voltage drop across a resistor leading to a diode connected to ground?

I know that I will be wasting power by using a voltage divider (about 240mW), but I have size constraints so I can't fit a step down coverter into the base of the lighting area. If anyone knows a really small converter setup, Im willing to try it.

Overall, the math I have done checks out, but Im not sure about my logic. Can I simply take the 5V from the divider and connect it to the diodes? Won't the resistors (R3 and R4) create a voltage drop leading to a voltage much less than the 3.55V desired?

Argh, Im beginning to feel so stupid.

Thanks for helping out!
 
LEDs should always be operated in series, and in series with a ballast resistor. Never operate LEDs in parallel without having a series ballast resistor in each parallel mesh.
 
Never ever assume the voltage drop of the actual diodes you use will be the same, because they never will be, that circuit would probably simulate fine and never work in the real world.
 
Hi TheRabbit2,

not only you are wasting power, but the LEDs won't ever light. With a series resistor of 1.12K the LEDs will never get the required current to "glow".

LEDs are current controlled and with the appropriate forward current the forward voltage drop is automatically resulting from current.

You can keep the part count to a minimum using the suggested circuit as attached. (Just three resistors)

Here is the basic calculation for LEDs, single LED: RL=VB-Vf/If - multiple LEDs: RL=VB-(nLED*Vf)/If, where RL is the current limiting resistor (Ω), VB is the supply voltage (V), nLED is the number of LEDs, Vf is the LED forward voltage (V), If is the LED forward current (A)

In your particular case the numbers to be put in are: RL(D1)=12-3.5/0.02 > RL=425Ω. RL(D2,D3)=12-(2*3.3)/0.25 > RL(D2,D3)=216Ω. The same applies to D4 and D5.

The nearest standard resistor value for D1 is 470Ω, for D2 through D5 it is 220Ω.

Boncuk
 

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Thanks for the fast replies!

Argh, I feel really stupid now. I forgot that the supply voltage is the voltage drop across the component. I think I need some sleep. I would have seen this a year ago, need to find my old electronics textbook.

I guess being unemployed really plays with your mind. Well time to sleep, I'll out this together this while I search for jobs tomorrow.
 
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