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5.5 volt power 12 volt device

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sammy004

New Member
Hi guys I just wanted to know if there is any way I can power a 12 volt led rod with 5.5 volt usb by adding a circuit in between to up grade the 5.5 to 12 volts or is this imposible?
 

smallguy

New Member
Some of the following should do what you want
MC34063
LM2585
MAX253

but I would change the leds to a lower voltage or get a new power supply
 

sammy004

New Member
Some of the following should do what you want
MC34063
LM2585
MAX253

but I would change the leds to a lower voltage or get a new power supply
WOW!! I think it might be easier if I just build a new LED Rod with 5 LED powered by 5.5 volts. LED's are 3volt 20ma.
 

sammy004

New Member
I see these LED calculators that tell me to connect a single LED to one resistor like for output 12 I have 10 LED's and each is connected to a 500-550 ohm resistor. I just wanted to know if there is a way of connecting all the led's "neg pos neg" and use just one resistor at the end or at least 5 LED's?
 

solis365

New Member
I see these LED calculators that tell me to connect a single LED to one resistor like for output 12 I have 10 LED's and each is connected to a 500-550 ohm resistor. I just wanted to know if there is a way of connecting all the led's "neg pos neg" and use just one resistor at the end or at least 5 LED's?
yup thats how its done.

However, USB has limited current. Someone should correct me, but its somewhere around 150mA.

Heres how it works: LED has a 2V drop, say. So 3 LEDs connected in series (neg pos neg as you said) will drop 6V. if they are the 20mA kind of LEDs, they will take 20mA and 6V for all three. You could then add another, identical string, in parallel with that one. then you will need 40mA and 6V. However, if you hooked up the two strings end-to-end (series, neg pos neg), then you would need 12V and only 20mA.

Power stays constant. V * A = P, so you can change your parallel or series combination to get the amount of current and voltage. i.e. USB has 5V and 150 mA (I think, double check that), so you need to find the ideal combination of parallel and series strings to maximize that.

Additionally, LEDs rated at 20mA will probably still look about the same brightness at 15mA, so you might run them a little bit under spec to save power/get more LEDs. (simply use a slightly larger resistor on each string. change the current tolerance in those calculators you use to figure it out)
 

MrDEB

Well-Known Member
I recall a circuit that increases the voltage

where they used 1.5 volts to power a couple of leds.
circuit had a DIY step up transformer.
I think if you searched LED flashlight??
 

sammy004

New Member
I was looking at one of these led rods I have that go into vehicles for decorations or stereo accessories and they have 6 led's with one big resistors at the end (Car Voltage is 12-13 volts right?) thats the kind of setup I'm looking for and I will be using 12 volts now not 5.5 USB.
 

MikeMl

Well-Known Member
Most Helpful Member
No the resistors were not included, I have that LED calculator but I was hoping of making a LED with the least amount of resistors I can use like for example 6 LED's to one resistor. Is that possible?
Do an experiment. Assume that the forward drop across a single led is 2V. Calculate out a resistor to get 20mA starting from your car battery. Hook it up. Using a voltmeter to measure the actual voltage across the single LED.

To see how many you can put in series, figure a drop of not less than 3V across the eventual single resistor you will have in your string. With engine stopped your car battery is about 12.6V, so 12.6-3=9.6V. Now divide 9.6 by the forward voltage of the single Led you measured above. Say it was 2.2V, then 9.6/2.2=4.36, which means that you can only run 4 Leds in series.

Knowing that you have 4 2.2V Leds in series, go back to your calculator to figure what single resistor will drop 12.6-(4*2.2) = 3.8V at 20mA (or your desired current). 3.8/0.02=190Ω
 
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