As shown in the figure, when the transistor is on, the level of point A is 0; when the transistor is off, how to calculate the level of point A, please advise, thank you.
Rosemarie 發表於 July 3, 2020
(5-2-0.6)/2k=I I*1K=A
Kalie 發表於 July 3, 2020
The theoretical calculation is as follows :(5-2-0.6)/2k=I, then VA=2+1*I, then VA= 3.2v.
In fact, the transistor has a leakage current at cut-off time, so the voltage at VA point <
3.2V. Calculation is as follows: when the cut-off is set, the current of R1 branch is I1 and the leakage current of triode branch is I2, then the following equation is obtained: 5-VA-0.6=1*(I1+I2) (1), VA-2=I1*1(2);
To solve the above equations, VA=3.2-I2/2<
3.2. However, I2 belongs to uA level, so the VA at this point is also very close to the VA calculated in the first step.
Shalom 發表於 July 3, 2020
It’s just the resistor divider, but if it’s less than 2V, it’s 2V.
