TheRabbit2
New Member
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 lighting.JPG](https://www.electro-tech-online.com/data/attachments/32/32281-39aaa3fd2845326092176acdaf9a8e63.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!
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 lighting.JPG](https://www.electro-tech-online.com/data/attachments/32/32281-39aaa3fd2845326092176acdaf9a8e63.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!