(a)$2.7\mu m$ (c)$2.25 \mu m$
(b)$0.3 \mu m$ (d)$0.75 \mu m$
Answer is ()
EC.16) In an n channel JFET, the gate is
(a)n type
(b)p type
(c)either n or p
(d)partially n & partially p
Answer is (b)
EC.17)Characteristics of a Voltage Follower circuit are(a)high input impedance, low output impedance,unity gain
(b)high input impedance, high output impedance, unity gain
(c)low input impedance, high output impedance,no unity gain
(d)High input impedance, low output impedance, no unity gain
Answer is (a)
Voltage Follower is a circuit in which the output simply follows the input, then what makes the need of such a circuit ?. I read that the High Input impedance in such circuits draws less current from the input.
EC.18) In p-n-p transistor the current Ie has two components viz. Iep due to injection of holes from p-region to n-region and Ien due to injection of electrons from n-region to p-region. Then
(a) Iep and Ien are almost equal
(b) Iep >> Ien
(c)Ien >> Iep
(d)either (a) or (c)
Answer is ()
EC.19) Assertion (A): A p-n junction has high resistance in reverse direction. Reason (R): When a reverse bias is applied to p-n junction, the width of depletion layer increases.
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not a correct explanation of A
(c) A is true but R is false
(d)A is false but R is true
Answer is ()
EC.20) Impurity commonly used for realizing the base region of a silicon n-p-n transistor is
(a) Gallium
(b) Indium
(c) Boron
(d) Phosphorous
Answer is ()
EC.21)For a stable operational amplifier-feedback network, the ideal closed loop gain can be calculated if which of the following conditions is true:(a)A negligible differential voltage applied at input terminals, produces a significant output voltage
(b)The current required at either of the op-amp terminals is negligible
(c)both (a) and (b) are true
(d) only (a) is true
Answer is (c)
The above two conditions are used to describe the amplifier working characteristics.
Consider for the first condition ,
that is the opamp basically amplifies the input voltage
For the non-inverting amplifier
$V_a\bumpeq \frac{Z_1}{Z_1+Z_2} V_o$
considering ideal conditions $V_a = V_i, \therefore \frac{V_o}{V_i} = \frac{Z_1 + Z_2}{Z_1}$
For the second condition , consider inverting amplifier - By Kirchoff's current law $I_a +I_b \bumpeq 0$
$I_a \bumpeq \frac{V_i }{Z_1} ; I_b \bumpeq \frac{V_o}{Z_2} $
Combining above, $\frac{V_o}{V_i} = -\frac{Z_2}{Z_1} $
EC.22) At very high temperatures the extrinsic semiconductors become intrinsic because
(a) drive in diffusion of dopants and carriers
(b) band to band transition dominants over impurity ionization
(c) impurity ionization dominants over band to band transition
(d) band to band transition is balanced by impurity ionization
Answer is ()
EC.23)For the non-inverting integrating amplifier in the figure,the ideal closed-loop gain $\frac{V_0(s)}{V_i(s)}$ equals:
(a) $ \frac {1}{(R1C1s+1)} .\frac{( RCs+1)}{RCs}$
(b) $\frac{1}{(R1C1s+1)}.\frac{RCs}{(RCs + 1)}$
(c) $\frac{1}{(R1C1s)}.\frac{(RCs+1)}{RCs}$
(d) none of above
Answer is (a)
To find the solution, let's go step by step , $\frac{V_o(s)}{V_a(s)} = \frac{R + 1/Cs}{R}$ (Explained above for EC.21)...Eqn 1
Next,the low pass filter at the front , $\frac{V_a(s)}{V_i(s)} = \frac{1}{R1C1s+1}$ ..Eqn(2) Combining Eqns(1)&(2),$\frac{V_o(s)}{V_i(s)} = \frac {1}{(R1C1s+1)} .\frac{( RCs+1)}{RCs}$
EC.24)Identify the op-amp circuit above and define the relationship between the input and output voltage ,given the diode-current voltage :$i_D =I_S(e^{[qv_D/kT]} -1)$.Here, $I_S $is a constant dependent on diode construction.
q is electron charge
k is Boltzmann's constant
T is absolute temperature.
Then,v_i at the input equals
(a)$-RI_S(e^{[qv_o/kT]} -1)$
(b)$-RI_S(e^{[-qv_o/kT]} +1)$
(c)$R/I_S(e^{[qv_o/kT]})$
(d)insufficient data,cannot be determined
Answer is (a)
$i_R = v_i/R, \implies v_i = -Ri_D $
$\therefore v_i = -RI_S(e^{[qv_D/kT]} -1) $
(a) towards positive terminal
(b) towards negative terminal
(c) either towards positive terminal or negative terminal
(d) towards positive terminal for 1 μs and towards negative terminal for next 1 μs
Answer is ()
EC.26) An ideal opamp is an ideal
(a) Voltage controlled current source
(b) Voltage controlled voltage source
(c) Current controlled current source
(d) Current controlled voltage source
Answer is ()
EC.27)The gain bandwidth product for a 741-opamp , given the closed loop gain, A_v=80dB and the cutoff frequency at that gain f_c=80Hz is
(a)800k
(b)80k
(c)8k
(d)none of the above
Answer is ()
EC.28) The number of doped regions in PIN diode is
(a) 1
(b) 2
(c) 3
(d) 1 or 2
Answer is ()
