Powering the LED string from the Nano's 5V output means you are limited by the Nano's 5V regulator, which may not provide adequate power. It may work better to power them directly from the main power supply instead (Vin). And the 1K resistor is way too high. Do you have another 220 that you could use, or even two or three 220s in parallel?
from 2v to 1.75. why am i only getting 2v? im getting ~5v from the 5v pin.
i was planning on selling this with the main power supply being from 6 to 12 volts so they don't have to scramble for the exact power adapter.
should i not do this?
Well, 6V probably won't work. The 5 regulator on a Nano clone will require something like 1.5V headroom to regulate properly. And you have the diode in there which will drop another 0.7V. So 9-12V may be a better choice.
So you're only getting about 2mA of current (E=I*R =>I=E/R
2V/1000 ohms =0.002 A
yep.. im going to drop the resister value, i found some 10's. do i put 2 or 3 on there?
If you draw too much current from the 5V pin, it could burn out the regulator. Better to try a 220 first, powering from Vin, and see what happens to the voltage drop.
kk!
And what is the Vin voltage?
from 2v to 1.75. why am i only getting 2v? im getting ~5v from the 5v pin
Across the resistor , right ?
9v now
yes.
you mean "yes, across the resistor" right ?
3.5 to ~1.75
Ok, you start at 9V, then each LED will drop about 1.4V, and the transistor will drop maybe 0.2V, so that leaves 4.6V to drop across the resistor. So a 220R will pass about 20mA of current through the string. A 50R resistor will pass 92mA. But 220R may be low enough because you have three LEDs emitting IR.
ima run a test for this setup then!
You need to use a power source that can source at least 100mA.
1.5V/0.1A=15 ohm
You need to source 100mA and the resistor needs to be between 10 and 15 ohms.
If you connect a 9V battery with the GND to the circuit GND and to POS to the leds
with a 15 ohm resistor you should read about 3V (more or less the same as your are reading now but because the resistor is 15 ohms the 1.5 V = I*R = 0.1A * 15 ohms but without a datasheet for the leds I don't know if that exceeds their rated
operating current
im using a 2 amp power supply
Measure the voltage at each point, starting at Vin, and see if you get what you expect. If the bottom of the resistor is 1.75V, then the transistor is not in full saturation. That could be part of the problem.
im an idiot. i was reading my power brick upside down... its 6v...
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