Go Down

### Topic: Control when current can flow (Read 6677 times)previous topic - next topic

#15
##### Jan 27, 2011, 05:19 am
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.

#### Grumpy_Mike

#16
##### Jan 27, 2011, 05:31 am
Connections into the base still not right.

#### manray

#17
##### Jan 27, 2011, 05:57 pm
So I believe I have it..

#### Grumpy_Mike

#18
##### Jan 27, 2011, 07:04 pm
Yes that looks right.

#### triath5147

#19
##### Jan 27, 2011, 09:38 pm
As far as what "CrossRoads" said earlier. I use the following web site for easy LED diagrams:

http://led.linear1.org/led.wiz

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.

Go Up