I’m trying to calculate the voltage needed at the transistor base to “open” the transistor. I know it’s got to be ≥ 0.7v.
Saw the schematic above. I calculate the resistance in parallel as R = 470x1000 / 470+1000 = 320Ω. Therefore current is I = 7v (9v – current drop of LED) 7/320 Ω = 0.02ma. Therefore voltage going through transistor base is V = 0.02x320 = 6.4V.
Is that calculation correct and is 6.4V too high a voltage for the transistor to handle?