It is a "current sensor" circuit. When the PC is getting current the circuit turns on any other device connected to the triac.
What I don't understand is the reason of that 0.82Ω, 2W resistor in parallel with the two BY398 diodes.
Is it not enough to have those diodes in order to turn on the transistor?
Please let me know the theory behind all this: I'm getting crazy.
The 0.82ohm resistor turns on the transistor BC237 when the current is too high (i.e. R*I>0.8V or so), The two diodes BY398 protect the transistor BC237, should the current become really high.
Sorry,
I continue to not understand.
If i have those 2 diodes, when current flows through them, I have a 0.7V that's enough to trigger the transistor, right? Those diodes are 3A, so no problem about current, that is less than 1A.
For me this circuit looks a little bit dodgy, as the forward voltage drop across the top BY398 diode may be to low to allow the transistor to be turned on ..
To provide protection and not to interfere with the transistor, I’d use two diodes each way, and perhaps increase the value of the base resistor from 47ohm to, say 470 ohm ..
I think the resistor is there to lower the voltage drop when the current level is low. when current is low, the diodes are off. lower votlage drop result less power loss