Calculate VCE in the figure (in volts).
First check to see if Rin (Base) can be neglected.
=150 x kΩ = 150kΩ
Since 150kΩ is more than 10 times R2. The condition is met and Rin(base) can be neglected.
= 6.88 V
Then, VE = VB + VBE
= 6.88 + 0.7
VE = 7.58V
lE = 2.42 mA
From IE we can determine lc and VCE
and VC = ICRC = 2.42 mA x 2.2kΩ
The correct answer is: 2.26
Calculate the minimum value of C2 (in μF) if the amplifier operate over a frequency range of 2kHz to 10kHz?
RE = 560Ω
XC of the bypass capacitor C2 should be
The capacitive value, at the minimum frequency of 2kHz
The correct answer is: 1.42
If the Q-point is 2 Vand 1 mA with the help of a feedback resistor RB. β = 100, then what is the value of RB in? (Take VBE = 0.7V).
The correct answer is: 130
Calculate the value of VCE (in Volts) for the circuit shown below. (β = 100 and VBE =0.7V).
Let us use Thevenin’s theorem
Now, applying KVL to Base circuit
= 9.45 μA
lC = 0.945mA
Now, Vcc = IcRc - VCE
The correct answer is: 6.22
A npn transistor circuit has ∝ = 0.985. If IC = 2mA then the value of lB (in mA) is?
= 0.03 mA
The correct answer is: 0.03
Calculate the collector voltage VC of the transistor circuit should be in the figure :
(given = 0.96, lCBO = 20μA , VBE = 0.3V, RB = 100/kΩ)
The correct answer is: 4.72
In the below figure, the minimum value of lB (in mA) required to saturate this transistor if [Neglect VCE(sat)].
Since,VCE(Sat) is neglected (assume to 0V)
The correct answer is: 50
In a certain oscillator, Av = 50. The attenuation of the feedback circuit must be?
An amplifier of gain 1000 is made into a feedback amplifier by feeding 9.9% of its output voltage in series with the input opposing. If fl = 20Hz and fh = 200kHz for the amplifier without feedback then due to the feedback what is the new frequency (in kHz)?
The correct answer is: 20000
In the transistor below, base current is 35 μA, VBE = 0.7V, what is the value of RB in kΩ?
The correct answer is: 237.1