Well Hello and merry xmas! , I got a qeustion of a circuit witch i dont understand (yes its probaly a stupid qeustion but i cant figure it out)
So this is the text from the book
. Because of the difference in resistances and the relatively high turn-on voltage for the LED, the base is forced below its turn-on threshold, and as a result, the transistor will not pass much current. The LED will glow dimly at best. Now what happens when the switch is opened? The base of the transistor loses its negative power supply and has only its positive power supply. It be- comes much more positive, above the turn-on threshold for the transistor, which tells the transistor to lower its resistance and pass more current. The LED now glows brightly. Thus, when the switch is turned off and breaks the con- nection, the LED is turned on.
how doos the voltage from the base effects the voltage to the led?
i dont understand what he means with , the base loses its negative power and it becomes more positive when the switch is open? how? (witch is probaly also qeustion 3)
3.and when the switch is open how comes the voltage increases across the base of Q1? ,
also i looked online for some explanation and i found that to calculate the base(vbb) of transistor you have to do 12 (1k)(11) = 1.09 V thats above 0.7. so shoulnt the transistor be on? bec its above the thresold of 0.7
- whys the D1 Diode there ? in the book it says this about the diode , but how can the current from the relay go up , and why dosent it just go the negative cell of the battery?
One little problem remains: in the new version of the circuit, if the transistor goes off while the relay is still on, current from the relay can flow back up the wire to the emitter of the transistor, where it will try to flow backward through the transistor to the base, which is “more negative,”
sorry for my bad english :/