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Topic: Control when current can flow (Read 5859 times) previous topic - next topic


You may also be able to put 2 LEDs in series with a single resistor with a 5 V source (depends on the LED voltage drop). That would cut your current draw in half.
Making some assumptions:
17mA current flow desired
Vce - 0.5V,voltage drop across the transistor when on
Vled = 2V, voltage drop across the LED when on
Vsource = 5V, as shown on your schematic

Applying Ohms law, V=IR or V/I = R to calculate Rlimit:
(Vsource - Vled - Vled - Vce)/current = Rlimit
(5V -2V - 2V - 0.5)/.017 = 29.4ohm. so use a 30 or 33 ohm resistor
So, look into your specs for the LED and transistor to be used and calculate the current limit resistors needed.
Power rating needed will be small - with 0.5V across the resistor & 17mA, the resistor is only dissipating 8.5mA (P=IV, so 0.017 * 0.5), so 1/8W resistors are fine.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.


Connections into the base still not right.




As far as what "CrossRoads" said earlier. I use the following web site for easy LED diagrams:


It is a wizard that will calculate how many LED's can go in series and what resistor value to use. It also has a single LED version. Great tool.
Just for future knowledge.
Resistance is Futile

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