Hello,
I’m a student and I’m working on an electronica project that includes this Velleman kit.
Now I don’t know the function of the two transistors behind the op-amp.
Can someone explain this to me please?
Also, what is the function of the coil and diode D5?
The transistors switch the MOSFET (T1) on and off so as to maintain a constant current through the LED(s). Here’s how it works (for simplicity I will use the 350mA configuration, R6 (1.0Ω) only)…
When powered up the 5V regulator’s output is divided by R1 and R2 to provide a 341.6mV reference applied to the comparator’s inverting input, and T2 is forward biased through R4, turning the MOSFET (T1) off–or mostly off. No current flows through the coil, LED and R6, meaning the voltage drop across R6 will be 0.0V. This level is applied to the comparator’s non-inverting input.
When a comparator’s non-inverting input voltage is less than that of the inverting input the comparator’s output will be low. This low output causes T2 to shut down and forward biases T3 driving it to saturation, turning on T1 (and the LED).
Obviously, when the current flowing though R6 (a 1.0Ω resistor) current reaches 341.6mA there will be 341.6mV potential across R6, this applied to the comparator’s non-inverting input. The comparator’s inputs are now equal so it’s output goes high, causing T3 to turn off, and T2 to saturate switching the MOSFET nearly off. Current stops flowing through R6, the comparator output goes low, and the cycle repeats…
This all happens rather quickly, at a frequency governed by the RC filter formed by R5 and C4 (around 1MHz).
The inductor (L1) supplies power, through the Schottky diode, to the LED when the MOSFET is off.
When configured for 750mA operation R7 is installed in parallel with R6, creating an effective 0.5Ω current sensing load–this will of course require twice as much current flow so that the voltage drop will match the 341.6mV reference voltage.
OK now it’s getting clear to me.
Thanks!