In which region transistor with `h_(fe)=80` operates as shown in Fig. – InterviewSolution

InterviewSolution

1.	In which region transistor with `h_(fe)=80` operates as shown in Fig. Also find `R_(B)`.
Answer» Since `V_(CB) lt V_(c c )` and not equal to zero, therefore, Transistor operates in avtive region: Current in the collector circuit is `I_(c)=((9-5)V)/(2kOmega)=2mA` `h_(fe)=I_c/I_b` or `I_b=I_c/H_(fe)=(2mA)/80=2.5xx10^(-5)A` we know that `V_(BE)` in active region is `0.7V`, therfore, `R_(B)=((3.2-0.7)V)/((2.5xx10^(-5))A)=92xx10^(3)Omega=92kOmega`

Discussion

No Comment Found

Related InterviewSolutions

Assuming the ideal diode, draw the output waveform for the circuit given in Fig.Explain the waveform.
Draw the output waveform across the resistor (Fig.)
Calculate the value of output voltage`V_(0)` and I if the Si diode and the Ge diode conduct at 0.7 V and 0.3V respectively, in the circuit given in Fig. If now Ge diode connections are reversed, what will be the new values of `V_(0)` and I.
Why does the width of depletion layer of a p-n junction increase in reverse biasing?
How does the width of the depletion layer of a p-n junction diode change with decrease in reverse bias?
The V-I characteristics of silicon diode is shown in Fig. Calculate the diode resistance in `` forward bias at `V=+0.9V ` reverse bias `V=-3.0V`.
What do the acronyms LASER and LED stands for? Name the factor which determines (i) frequency and (ii) intensity of light emitted by LED.
In a transistor base is made thin and doped with little impurity atoms. Why?
For the circuit shown in Fig., find the current flowing through the `1Omega` resistor. Assume that the two diode are ideal diode.
Why is a photodide operated in reverse bias mode? Fig. shown reverse bias current, under different illuminating intensities `I_(1)I_(2),I_(3)` and `I_(4)` for a given photodiode. Average the intensities `I_(1), I_(2),I_(3)` and `I_(4)` in decreasing order of wavelength.