Driving LED with NPN transistor from I/O pin
I'm trying to understand a circuit for driving a LED found on a board I purchased.
Below is the circuit driven by an I/O pin (HS2) of a small 3.3V processor.
The HS2 pin is driven by the I/O with these specs:
V(high) = 0.8 · Vsupply = 2.64 V
V(low) = 0.1 · Vsupply = 0.33 V
The I/O pin can be set to drive up to 28 mA, but I am guessing the design of the circuit would rather pull the current from Vsupply rather than the I/O controlling pin (HS2), and that is the reason for using the NPN transistor (S8050) to drive the LED.
The LED is a SMD 3528 white LED for which I do not have specs, but testing shows it lights up nicely at 40 mA, and does fine at 11 mA, with 3.24 V across.
I am wondering if the reason for this design is because this bright white LED is typically looking for a 3 V drop, and if they used a current limiting resistor in series with the LED, then maybe the resistor would drop too much and bring the voltage across the LED to less than 3 V.
How is the 1k/10k voltage divider working to set the current through the flash?
Does this circuit seem like a good one to duplicate if I wanted to create additional LEDs driven by other pins on the processor?
When the transistor is on, is it setting the current, or will the current vary per the gain of the transistor?
I see how this circuit is acting as a switch, but is it also fixing the current through the LED? Or is the current through the LED dependent on the gain of the transistor?
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