### Previous GATE Questions on BJT Low Frequency & High Frequency Analysis

1995

4.       An NPN transistor under forward active mode of operation is biased at Ic = 1 mA, and has a total emitter base capacitance CK of 12 pF, and the base transit time τF of 260 psec. Under this condition, the depletion capacitance of the emitter base junction is _____________

1996

19.   A common emitter amplifier with an external capacitors CC connected across the base and the collector of the transistor is shown. Given gm = 5 mA/V, rπ = 20 k, Cπ = 1.5 pF and Cµ = 0.5 pF.

a.       Determine the ac small signal mid band voltage gain, Vo/Vs.
b.      Determine the upper cutoff frequency fH of the amplifier.
Answer:     (a) -33.33 (b) 18.326 KHz

1998

1.       The fτ of a BJT is related to its gm, Cπ and Cµ as follows

1999

4.       An NPN transistor (with C = 0.3 pF) has a unity gain cutoff frequency fτ of 400 MHz at a dc bias current IC = 1 mA. The value of its Cµ (in pF) is approximately (VT = 26 mV)
a.       15
b.      30
c.       50
d.      96

6.       An amplifier is assumed to have a single pole high frequency transfer function. The rise time of its output response to a step function input is 35 nsec. The upper -3 dB frequency (in MHz) for the amplifier to a sinusoidal input is approximately at
a.       4.55
b.      10
c.       20
d.      28.6

2000

4.       The current gain of a bipolar transistor drops at high frequencies because of
a.       Transistor capacitances
b.      High current effects in the base
c.       Parasitic inductance effects
d.      The Early Effect

2001

4.       An NPN BJT has gm = 38 mA/volt, Cµ = 10-14 F, Cπ = 4x10-13 F and DC current gain β = 90. For this transistor fτ and fβ are
a.       fτ =  1.64 x 108 Hz and fβ = 1.47 x 1010 Hz
b.      fτ =  1.47 x 1010 Hz and fβ = 1.64 x 108 Hz
c.       fτ =  1.33 x 1012 Hz and fβ = 1.47 x 1010 Hz
d.      fτ =  1.47 x 1010 Hz and fβ = 1.33 x 1012 Hz

2003

5.       Generally, the gain of a transistor amplifier falls at high frequencies due to the
a.       Internal capacitance of the device
b.      Coupling capacitor at the input
c.       Skin effect
d.      Coupling capacitor at the output

12.   An ideal saw-tooth voltages waveform of frequency of 500 Hz and amplitude 3 volts is generated by charging a capacitor of 2 µF in every cycle. The charging requires
a.       Constant voltage source of 3 volts for 1 ms
b.      Constant voltage source of 3 volts for 2 ms
c.       Constant current source of 1 mA for 1 ms
d.      Constant current source of 3 mA for 2 ms

2010

Common Data Questions:
Consider the common emitter amplifier shown below with the following circuit parameters.
β = 100, gm = 0.3861 A/V, ro = 259 , RS = 1 K, RB = 93 K, RC = 250 , RL = 1 K, C1 = and C2 = 4.7 µF.

5.       The resistance seen by the source VS is
a.       258
b.      1258 Ω
c.       93 KΩ
d.
6.       The lower cutoff frequency due to C2 is
a.       33.9 Hz
b.      27.1 Hz
c.       13.6 Hz
d.      16.9 Hz

2015

3.       At very high frequencies, the peak output voltage Vo (in volts) is ______________