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No - the transistor acts as a buffer between the two voltages.
Referring to the diagram in Noom's link:
The transistor sits at the bottom connected to ground.
To turn it "ON", the Netduino needs to apply a voltage of approximately 0.7V to the base.
Since the the Netduino outputs are 3.3V, there is actually too much voltage available, and this could damage the transistor.
Hence the 2K ohm resistor between the Netduino and the transistor to limit the current that can flow into the base.
Above the transistor there is the load and a series resistor to limit the current in the load.
In your case, you want 3.3V at 20mA to flow when the transistor is ON.
If you use the 3.3V rail, you would not need a resistor since the voltage not too large.
However, when the transistor is ON, it is not a perfect switch and there is a small amount of voltage wasted (approx 0.3V).
So the LED will only get about 3V.
My suggestion then is to use a 5V rail with a resistor to limit the LED current:
R = V / I = (5 - 3.0 - 0.3) / 20mA = 1.7 / 0.02 = 85 ohms
Since you can't buy every value of resistor, you could try an 82 ohm resistor.
Hope this helps - Paul
There's quite a bit of literature about this topic, but I'm having trouble understanding all the implications. I'm working on a project which I'd like to have PWM-controlled LEDs.
I've been reading that I need a current-limiting resistor to protect the Netduino's PWM pins from damage. I was wondering if this is always the case.
I have some LEDs which state that they have a 3.3V forward voltage at 20MA. By my read that should not unduly stress out the PWM pin.
Am I misunderstanding the characteristics of the LED?
The Netduino PWM output would turn the transistor on and off repeatedly.
This will send a pulsed current through the LED - by changing the period of the Netduino output, you will acheive the brightness control you are after.
Have fun - Paul
Edit:
To get the most out of your LED, you will want to supply it from the 5V rail, with a series resistor.
If you use the 3.3V rail, you will not be giving it the full 3.3 due to a small voltage drop accross the transistor.
No - the transistor acts as a buffer between the two voltages.
Referring to the diagram in Noom's link:
The transistor sits at the bottom connected to ground.
To turn it "ON", the Netduino needs to apply a voltage of approximately 0.7V to the base.
Since the the Netduino outputs are 3.3V, there is actually too much voltage available, and this could damage the transistor.
Hence the 2K ohm resistor between the Netduino and the transistor to limit the current that can flow into the base.
Above the transistor there is the load and a series resistor to limit the current in the load.
In your case, you want 3.3V at 20mA to flow when the transistor is ON.
If you use the 3.3V rail, you would not need a resistor since the voltage not too large.
However, when the transistor is ON, it is not a perfect switch and there is a small amount of voltage wasted (approx 0.3V).
So the LED will only get about 3V.
My suggestion then is to use a 5V rail with a resistor to limit the LED current:
R = V / I = (5 - 3.0 - 0.3) / 20mA = 1.7 / 0.02 = 85 ohms
Since you can't buy every value of resistor, you could try an 82 ohm resistor.
Hope this helps - Paul
Answered
