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# Multiple choice question for engineering

## Set 1

1. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of gm.
a) 12mA/V
b) 24 mA/V
c) 36 mA/V
d) 48 mA/V

2. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of R?p.
a) 625 ohm
b) 1250 ohm
c) 2500 ohm
d) 5000 ohm

3. A transistor with ß = 120 is biased to operate at a dc collector current of 1.2 mA. Find the value of Re.
a) 2.5 ohm
b) 20.6 ohm
c) 25.2 ohm
d) 30.4 ohm

4. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the smallest value found of the resistance looking into the base?
a) 347 ohm
b) 694 ohm
c) 1041 ohm
d) 1388 ohm

5. A transistor operating with nominal gm of 60 mA/V has a ß that ranges from 50 to 200. Also, the bias circuit, being less than ideal, allows a 20% variation in Ic. What is the largest value found of the resistance looking into the base?
a) 1050 ohm
b) 21000 ohm
c) 3150 ohm
d) 4200 ohm

6. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What is the minimum ß he can tolerate for the transistor used?
a) 100
b) 150
c) 200
d) 250

7. A designer wishes to create a BJT amplifier with a gm of 50 mA/V and a base input resistance of 2000 O or more. What emitter bis current should he choose?
a) 1.06 mA
b) 1.16 mA
c) 1.26 mA
d) 1.36 mA

8. Which of the following is true?
a) Ib = ß Ic
b) Ib = ß + 1/ Ic
c) Ib = Ic/ß
d) Ib = Ic/ ß – 1

Answer: c [Reason:] The correct relationship between Ic and Ie is Ib = Ic/ß.

9. The SI units of transconductance is
a) Ampere/ volt
b) Volt/ ampere
c) Ohm
d) Siemens

Answer: a [Reason:] Transcoductance is given by Ic/Vt.

10. Which of the following represents the correct mathematical form of the term denoted by the symbol Rp?
a) ß/gm
b) Vt/Ib
c) All of the mentioned
d) None of the mentioned

Answer: c [Reason:] Both of the expressions are identical.

## Set 2

(Q.1-Q.3) A pseudo-noise (PN) sequence is generated using a feedback shift register of length m = 4. The chip rate is 107 chips per second.

1. The PN sequence length is
a) 10
b) 12
c) 15
d) 18

Answer: c [Reason:] The PN sequence length is N = 2m – 1 = 16 – 1 = 15.

2. The chip duration is
a) 1µs
b) 0.1 µs
c) 0.1 ms
d) 1 ms

Answer: b [Reason:] Tc = 1/(107) or 0.1 µs.

3. The period of PN sequence is
a) 1.5 µs
b) 15 µs
c) 6.67 ns
d) 0.67 ns

Answer: b [Reason:] The period of the PN sequence is T = NTc = 15 x 0.1 = 1.5 s

4. A slow FH/MFSK system has the following parameters.
Number of bits per MFSK symbol = 4
Number of MFSK symbol per hop = 5
The processing gain of the system is
a) 13.4 dB
b) 37.8 dB
c) 6 dB
d) 26 dB

Answer: d [Reason:] PG = Wc/Rs = 5 x 4 = 20 or 26 db.

5. A fast FH/MFSK system has the following parameters.
Number of bits per MFSK symbol = 4
Number of pops per MFSK symbol = 4
The processing gain of the system is
a) 0 dB
b) 7 dB
c) 9 dB
d) 12 dB

Answer: d [Reason:] PG = 4 x 4 = 16 or 12 db.

(Q.6-Q.7) A rate 1/2 convolution code with dfrec = 10 is used to encode a data sequence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz.

6. The coding gain is
a) 7 dB
b) 12 dB
c) 14 dB
d) 24 dB

Answer: a [Reason:] 0.5 x 10 = 5 or 7 db is the coding gain.

7. The processing gain is
a) 14 dB
b) 37 dB
c) 58 dB
d) 104 dB

Answer: b [Reason:] PG = (107)/(2 x 1000) = 5000 or 37 db.

(Q.8-Q.9) An FH binary orthogonal FSK system employs an m 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection.

8. If the hop rate is one per bit, the hopping bandwidth
for this channel is
a) 6.5534 MHz
b) 9.4369 MHz
c) 2.6943 MHz
d) None of the mentioned

Answer: a [Reason:] The length of the shift-register sequence is L = 2m – 1215 = 32767 bits For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate. The hop rate is 100 hops/sec. Since the shift register has L 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz.

9. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is
a) 3.2767 MHz
b) 13.1068 MHz
c) 26.2136 MHz
d) 1.6384 MHz

Answer: b [Reason:] If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum frequency separation for orthogonality 2/T 400 Hz. Since there are N 32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 MHz.

10. In a fast FH spread spectrum system, the information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density 0.5No and an SNR 20 ~13 dB (total SNR over the three hops). The probability of error for this system is
a) 0.013
b) 0.0013
c) 0.049
d) 0.0049

Answer: b [Reason:] The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is

## Set 3

1. The magnitude of the electric charge (e) is given by ____________
a) -1.6*10-19 C
b) 1.6*10-19 C
c) 9.11*10-31 C
d) 1.637*10-37 C

Answer: a [Reason:] The charge of the electron is the magnitude of electric force that an electron exerts on other particles which is equal to -1.6*10-19 C, the negative sign indicates the direction of force.

2. What is the forbidden gap voltage for silicon material?
a) 1.46 V
b) 1.56 V
c) 10 V
d) 1.21 V

Answer: d [Reason:] The forbidden gap voltage of a material is numerically equal to forbidden gap energy of the material which is 1.21 joules for silicon so forbidden gap voltage will be 1.21 V.

3. Which of the following parameters of P-N junction diode increases with temperature.
a) Cut in voltage
b) Reverse saturation current.
c) Ideality factor
d) Resistance

Answer: b [Reason:] Reverse saturation current at temperature T2 is 2[(T2 –T1)/10] times greater than reverse saturation current at temperature T1 where T2 is greater than T1.

4. Which of the following diodes do not exhibits a constant reverse saturation current with the change in reverse saturation voltage.
a) 1N909
b) 1N405
c) 1N207
d) 1N676

Answer: c [Reason:] 1N207 is the germanium diode for which the reverse saturation current is not constant which the change in voltage due to the leakage in the surface of the diode and due to the generation of new current carriers.

5. Which of these P-N junction characteristics are not dependent on temperature.
a) Junction resistance
b) Reverse saturation current
c) Bias current
d) Barrier voltage

Answer: a [Reason:] As the temperature of the P-N junction increases the current increases and the voltage decreases so the barrier voltage, reverse saturation current, bias current changes with temperature but junction resistance is independent of temperature.

6. As the temperature to the P-N junction increases the current increases due to?
a) Leakage in bias region
b) Electron-hole pair
c) Leakage in P region
d) Leakage in N region

Answer: b [Reason:] As the temperature to the P-N junction increases the mobility of charges increases thus increases the electron-hole pair which proportionally increases the current in the P-N junction diode.

7. By what percentage the reverse saturation current increases with 10 C rise in the temperature.
a) 25%
b) 12.5%
c) 50%
d) 7%

Answer: d [Reason:] As the temperature to the P-N junction diode increases the mobility of charges increases thus increasing the current, the reverse saturation current increases by 7% with 10C rise in temperature and doubles with every 100C rise in temperature.

8. What will be the decrease of barrier voltage with the rise in 10C in temperature?
a) 10V
b) 1mV
c) 10mV
d) 2mV

Answer: d [Reason:] As the temperature to the P-N junction diode increases the voltage across the junction decreases and the current increases with every degree rise in temperature the barrier voltage increases by 2mV.

9. What will be the reverse saturation current in the junction when the voltage across the junction is 0?
a) 0.3A
b) 0.7A
c) 0A
d) 1.24A

Answer: c [Reason:] When the voltage across the junction is zero in the sense there will be potential difference between the junctions hence there will be no movement of electrons and holes, hence the current will be 0.

10. The breakdown voltage of the P-N junction diode decreases due to the increase in.
a) Reverse saturation current
b) Reverse leakage current
c) Bias voltage
d) Barrier voltage

Answer: b [Reason:] Breakdown voltage of the diode is inversely proportional to the reverse leakage current so it decreases with the increase in reverse leakage current.

## Set 4

1. If ‘X’ corresponds to a tunnel diode and ‘Y’ to an avalanche diode, then__________
a) X operates in reverse bias and Y operates in forward bias
b) X operates in reverse bias and Y operates in reverse bias
c) X operates in forward bias and Y operates in forward bias
d) X operates in forward bias and Y operates in reverse bias

Answer: d [Reason:] In forward bias, negative resistance helps for tunnel diode to operate. Here, the current decreases with increase in voltage. If they are used in reverse bias, they are called as back diodes. Avalanche diode operates in reverse bias at breakdown region.

2. The range of tunnel diode voltage VD, for which slope of its V-I characteristics is negative would be? (The VP is the peak voltage and VV is the valley voltage).
a) VD > 0
b) 0D < VP
c) VV > VD > VP
d) VD > VV

Answer: c [Reason:] In tunnel diode characteristics, the slope is negative in the region between VV and VP. Here, it offers negative resistance. The characteristics are depicted below:

3. Tunnel diode has a very fast operation in__________
a) gamma frequency region
b) ultraviolet frequency region
c) microwave frequency region

Answer: c [Reason:] Tunnel diode is a type of semiconductor which works on tunneling effect of electrons in microwave region. So, tunnel diode has a very fast operation in microwave region.

4. Which of the following are true about a tunnel diode?
1) it uses negative conductance property
2) it operates at high frequency
3) fermilevel of p side becomes higher than the n side in forward bias
a) 1 only
b) 1 and 2
c) 3 only
d) 2 and 3

Answer: b [Reason:] The negative resistance property helps in the operation of tunnel diode. As the tunnel diode works at high frequency, its applications are mostly in that range. High frequency oscillators are based on the resonant tunneling diode.

5. The depletion layer of tunnel diode is very small beacause______
a) its abrupt and has high dopants
b) uses positive conductance property
c) its used for high frequency ranges
d) tunneling effect

Answer: a [Reason:] When the P and N regions are very highly doped, the depletion layer comes closer. The tunnel diode is also highly doped. Its doping concentration varies within a small scale. So it’s an abrupt diode. For these reasons, the depletion region is small.

6. With interments of reverse bias, the tunnel current also increases because________
a) electrons move from balance band of pside to conduction band of nside
b) fermi level of pside becomes higher than that of nside
c) junction currrent decreases
d) unequality of n and p bandedge

Answer: a [Reason:] When the forward bias is increased, the tunnel current is also increased upto a certain limit. This happens when the electron movement takes place from P to N side.

7. The tunnneling involves_______
a) acceleration of electrons in p side
b) movement of electrons from n side conduction band to p side valance band
c) charge distribution managementin both the bands
d) positive slope characteristics of diode

Answer: b [Reason:] Tunneling means a direct flow of electrons across small depletion region from N side conduction band to P side valance band. The electrons begin to accelerate in the N side of the semiconductor.

8. Tunnel diodes are made up of________
a) Germanium and silicon materials
b) AlGaAs
c) AlGaInP
d) ZnTe

Answer: a [Reason:] Germanium and silicon materials have low band gaps and flexibility. That matches tunnel diode requirements. The remaining materials emits the energy in terms of light or heat.

9. For a tunnel diode, when ‘p’ is probability that carrier crosses the barrier, ’e’ is energy,’w’ is width.
a) p ∝ e(-A*e*w)
b) p ∝ 1/ e(-A*e*w)
c) p ∝ e(A*e*w)
d) p ∝ 1/e(A*e*w)

Answer: a [Reason:] The carrier jump occurs without any loss of energy due to small depletion layer. The probability of the carrier to jump across a barrier depends on the energy and width of the band. This variess exponentially for a given carrier.

10. In the construction of tunnnel diode,why is the pellet soldered to anode contact and a tindot to cathode contact via a mesh screen?
a) for better conduction and reduce inductance respectively
b) for heat dissipation and increase conduction respectively
c) for heat dissipation and reduce induction respectively
d) for better conduction and reduce inductance respectively

Answer: c [Reason:] Anode goes through better heat dissipation. So the pellet is used for the purpose. The tindot via mesh screen resists inductive effects caused at the cathode. Conduction is an independent factor which can’t be controlled.

11. What happens to a tunnel diode when the reverse bias effect goes beyond the valley point?
a) it behaves as a normal diode
b) it attains increased negative slope effects
c) reverse saturation current increases
d) beacomes independent of temperature

Answer: a [Reason:] After the valley point is crossed, the tunnel diode obtains positive slope resistance. That is similar to the characteristics of a normal diode. So it behaves like a normal diode after beyond valley point.

## Set 5

1. [T] = ?

2. [h] = ?

Answer: a [Reason:] V2 = 2I2 + 4I1 and I1 = 0.5I2 + 0.5(V1 – 2V2) => V1 = 4I1 + 1.5V2.

3. [y] = ?

Answer: a [Reason:] I1 = V1(Ya + Yab) – V2(Yab) and I2 = -V1(Yab) + V2(Yb + Yab).

4.The [y] parameter of a two port network is given by

Answer: b [Reason:] The new parameter will be the sum of the previous network and the resistor parameter.

5. The [y] parameter for a 2-port network and the network itself are given below.

The value of Vo/vs is _______
a) 3/32
b) 1/16
c) 2/33
d) 1/17

6. [y] = ?

7. A 2-port resistive network satisfy the condition A = D = 3/2B = 4/3C. The z11 of the network is
a) 4/3
b) 3/4
c) 2/3
d) 3/2

Answer: a [Reason:] z11 = A/C = 4/3.

8. A 2-port network is driven by a source Vs = 100 V in series with 5 ohm, and terminated in a 25 ohm resistor. The impedance parameters are

The Thevenin equivalent circuit presented to the 25 ohm resistor is
a) 80 V, 2.8 ohm
b) 160 V, 6.8 ohm
c) 100 V, 2.4 ohm
d) 120 V, 6.4 ohm