Physics

Semiconductors and Diodes

461 Questions

Semiconductors and diodes form the foundation of modern electronics, covering topics like intrinsic carrier concentration and Zener breakdown. Questions often explore the characteristics of bipolar junction transistors (BJT) and the properties of doped silicon materials. This topic is crucial for physics and electronics engineering competitive exams.

Zener diode breakdownBJT circuit analysisIntrinsic carrier concentrationDoping in semiconductorsEmitter follower circuit

Semiconductors and Diodes Questions

Multiple choice

Two identical NMOS transistors M1 and M2 are connected as shown below. V bias is chosen so that both transistors are in saturation. The equivalent gm of the pair is defined to be $\frac{\partial I_{out}}{\partial V_i}$at constant Vout The equivalent gm of the pair is

  1. the sum of individual gm's of the transistors

  2. the product of individual gm's of the transistors

  3. nearly equal to the gm of M1
  4. nearly equal to $\frac{g_m}{g_o}$of M2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The current in both transistor are equal. Thus $g_m$ is decide by $M_1.$

Multiple choice

For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded.

  1. Ix = Ibias + Is

  2. Ix = Ibias
  3. Ix = Ibias - $\bigg(V_{DD} - \frac{V_{out}}{R_E}\bigg)$

  4. Ix = Ibias - Is
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{By current mirror} \\ I_x = \frac{\Big(\frac{W}{L}\Big)2}{\Big(\frac{W}{L}\Big)_1}I{bias} \\ \text{Since MOSFETs are identical,} \\ \text{Thus} \hspace{1cm} \Big(\frac{W}{L}\Big)2 = \Big(\frac{W}{L}\Big)_2 \\ \text{Hence} \hspace{1cm} I_x = I{bias}$

Multiple choice

Directions: Consider the CMOS circuit shown below, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p −MOSFET is kept constant at 3 V. Assume that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1 mAV-2.

For small increase in VG beyond 1 V, which of the following gives the correct description of the region of operation of each MOSFET?

  1. Both the MOSFETs are in saturation region.

  2. Both the MOSFETs are in triode region.

  3. n -MOSFET is in triode and p −MOSFET is in saturation region.

  4. n -MOSFET is in saturation and p −MOSFET is in triode region.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For small increase in $v_G$ beyond 1V the n-channel MOSFET goes into saturation as $V_{GS} \rightarrow +ive \ and\ p-MOSFET$ is always in active region or triode region.

Multiple choice

Directions: Consider the CMOS circuit shown, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p−MOSFET is kept constant at 3 V. Assume that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1 mA V-2.

Estimate the output voltage V0 for VG = 15 V.
  1. 4 - $\dfrac{1}{\sqrt2}$
  2. 4 + $\dfrac{1}{\sqrt2}$
  3. 4 - $\dfrac{\sqrt3}{2}$
  4. 4 + $\dfrac{\sqrt3}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Vo = 4 + √3/2 = 4 + 0.866 = 4.866 V This voltage is near to supply voltage, while when it increases to more than 15 V, then the output will decrease.

Multiple choice

For a BJT the common base current gain $\alpha$ = 0.98 and the collector base junction reverse bias saturation current ICO 0.6$\mu A$. This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20XA. The collector current IC for this mode of operation is

  1. 0.98mA

  2. 0.99mA

  3. 1.0mA

  4. 1.01mA
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{We have} \hspace{2cm} \alpha = 0.98 \\ \text{Now} \hspace{2.7cm} \beta = \frac{\alpha}{1 - \alpha} = 4.9 \\ \text{In active region, for common emitter amplifier,} \\ I_C = \beta I_B + (1 + \beta) I_{CO} \\ \text{Substituting} I_{CO} = 0.6 \mu A \hspace{0.2cm} and \hspace{0.2cm} I_B = 20 \mu A \text{in above eq we have,} \\ I_c = 1.01 mA$

Multiple choice

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT / q = 25mV. The small signal input vi = Vp cos $(\omega t)$ where Vp = 100 mV.

The bias current IDC through the diodes is

  1. 1 mA

  2. 1.28 mA

  3. 1.5 mA

  4. 2 mA
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The current flows in the circuit if all the diodes are forward biased. In forward biased there will be 0.7 v drop across each diode. Thus $$I_{DC} = \frac{12.7 - 4(0.7)}{9900} = 1 mA$$

Multiple choice

Drift current in semiconductors depends upon

  1. only the electric field

  2. only the carrier concentration gradient

  3. both the electric field and the carrier concentration

  4. both the electric field and the carrier concentration gradient
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Drift current Id = qn$\mu$nE It depends upon electric field E and carrier concentration n.

Multiple choice

The channel resistance of an N-channel JFET shown in the figure below is 600$\Omega$ when the full channel thickness (tch) of 10$\mu$m is available for conduction. The built-in voltage of the gate P+ N junction (Vbi) is - 1 V. When the gate to source voltage (VGS) is 0 V, the channel is depleted by 1$\mu$m on each side due to the built in voltage and hence the thickness available for conduction is only 8$\mu$m

The channel resistance when VGS = - 3 V is
  1. 480$\Omega$
  2. 600$\Omega$
  3. 750$\Omega$
  4. 1000$\Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$At \hspace{0.2cm} V_{GS} = 0V, b = 4 \mu m \\ since \hspace{0.2cm} 2b = 8 \mu m \\ \text{Thus} \hspace{0.2cm} r' = \frac{\rho L}{W a} \times \frac{a}{b} = 600 \times \frac{5}{4} = 750 \Omega$