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You are a hero man! I have a bunch of these around and they're exactly what I needed!
You, my friend, are a genius! I just ordered a case of 36 from Dollar Store for a dollar each. Beets paying almost $3 for each input!
if a voltage regulator is generating heat when connected to 12v i suggest to use 2 voltage regulator 7808 that transform 12v to 8v then connect a 7805 voltage regulator to transform it to 5V ;)
im making led turn indicators for my bike...problem is that the leds i want to use burn out if used above 5v....bikes battery is 12v and 4 amps....how should it be done?attaching pic of leds i want to use
I'm doing that too! Not sure what to do just yet. Someone was saying use a MOSFET? But they are all like 60V? I dunno.
A mosfet has a rating of about 60v depending on the one you get. that isnt how much it outputs. usually N-Channel mosfets are used to power something high voltage while controlling it with a low voltage source. eg 12v lights powered by a 12v battery controlled by an arduino.
For the flashing part, use this circuit. I see the example has a resistor in already for the LEDs so they don't blow:
The part doing the flashing/timing is the 555 timer chip.
I have 12 v dc I want to 5 v which ohm resister have to use
And one more question
One ohm resistor oppose to how many volt
Hi guys. I read all your comments about dopping the voltage from 12 to 5. The problem with that, which I have already experienced, is t that using resistors as potential dividers generates a lot of heat for the circuit! The best on would be to use a regulator that dissipates the least heat and so obviate the need for a heatsink.
try a 12v to 5v Usb adapter like this:
could also try buying a bottle holder to clip to the frame and put a battery inside you could have an 11.1V li-po just for your speaker coupled with the 12V to 5V USB adapter stated above...this way no loss of run-time from your e-bikes battery life but a VERY LONG RUN-time on your speakers which would probably be able to outlast the e-bike battery!
try this battery:
put the battery inside the bottle and boom......new power supply!
How do supply 3.7v fom a 7.5v power bank? The voltage divider that I setup for my BlackBerry works only if the BlackBerry battery is less than half-full, else it just reboots. I used equal rsistors of 176 ohms each and that gave me 3.75v. I suspect that these resistors give current that is too little to charge the BlackBerry. My cslculations show that I can get 1Amp from 50 ohm resistors.
open the powerbank and their should be two lithium cells inside there... just swap only ONE of them around so that the battery has got ++ & -- of the batteries joined your voltage will be halved but you amperage will double.....meaning 3.7V but the powerbank will last two times longer when charging your phone as a benefit.
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Posted:May 6, 2009
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