# BJT, FET and their Biasing Circuits

 Question 1
The cross-section of a metal-oxide-semiconductor structure is shown schematically. Starting from an uncharged condition, a bias of +3V is applied to the gate contact with respect to the body contact. The charge inside the silicon dioxide layer is then measured to be +Q. The total charge contained within the dashed box shown, upon application of bias, expressed as a multiple of Q (absolute value in Coulombs, rounded off to the nearest integer) is __________ .
 A 0 B 1 C -1 D 2
GATE EE 2020   Analog Electronics
Question 1 Explanation:

Overall charge in side the box $q + q - q - q = 0$ charge
 Question 2
The enhancement type MOSFET in the circuit below operates according to the square law. $\mu_nC_{ox}=100\mu A/V^2$, the threshold voltage ($V_T$) is 500 mV. Ignore channel length modulation. The output voltage $V_{out}$ is
 A 100 mV B 500 mV C 600 mV D 2 V
GATE EE 2019   Analog Electronics
Question 2 Explanation:

As, $V_{DS}=V_{GS}$
MOSFET is in saturation,
\begin{aligned} I_D&=\frac{1}{2}\mu _nC_{Ox}\left ( \frac{W}{L} \right )(V_{GS}-V_T)^2 \\ 5 \times 10^{-6}&=\frac{1}{2} \times 100 \times 10^{-6} \times 10 (V_{GS}-0.5)^2\\ V_{GS}&=0.6 \\ V_0&=600mV \end{aligned}
 Question 3
Given, $V_{gs}$ is the gate-source voltage, $V_{ds}$ is the drain source voltage, and $V_{th}$ is the threshold voltage of an enhancement type NMOS transistor, the conditions for transistor to be biased in saturation are
 A $V_{gs} \lt V_{th};V_{ds}\geq V_{gs}-V_{th}$ B $V_{gs} \gt V_{th};V_{ds}\geq V_{gs}-V_{th}$ C $V_{gs} \gt V_{th};V_{ds}\leq V_{gs}-V_{th}$ D $V_{gs} \lt V_{th};V_{ds}\leq V_{gs}-V_{th}$
GATE EE 2019   Analog Electronics
Question 3 Explanation:
For NMOS transistor to be in saturation the condition will be
$V_{gs} \gt V_{th}$
and $V_{ds} \geq V_{gs}-V_{th}$
 Question 4
In the circuit shown in the figure, the bipolar junction transistor (BJT) has a current gain $\beta=100$. The base-emitter voltage drop is a constant, $V_{BE}= 0.7$ V. The value of the The venin equivalent resistance $R_{Th}$ (in $\Omega$) as shown in the figure is ______ (up to 2 decimal places).
 A 70.45 B 85.25 C 90.09 D 105.65
GATE EE 2018   Analog Electronics
Question 4 Explanation:

To calculate $R_{Th}$ D.C. volatage should be short circuited.

$R_{Th}=1k\Omega ||\frac{10k}{1+\beta }$
$\; \; =1k\Omega ||99.0099$
$R_{Th}=90.09\Omega$
 Question 5
For the circuit shown in the figure below, it is given that $V_{CE}=\frac{V_{CC}}{2}$. The transistor has $\beta =29$ and $V_{BE}=0.7V$ when the B-E junction is forward biased.

For this circuit, the value of $\frac{R_B}{R}$ is
 A 43 B 92 C 121 D 129
GATE EE 2017-SET-2   Analog Electronics
Question 5 Explanation:
In the input loop,
$10=(1+\beta )I_B \times 4R+I_B \times R_B+0.7+(1+\beta )I_B \times R$
$10=30I_B \times 4R+I_B \times R_B+0.7+30 \times I_B \times R$
$9.3=150 \times I_B \times R +I_B \times R_B .....(i)$

Ouput loop,
$10=(1+\beta )I_B \times 4R+5V+(1+\beta )I_B \times R$
$5=30I_B \times 4R+30 \times I_B \times R$
$5=150 \times I_B \times R .....(ii)$

using equation (i) and (ii),
$I_BR_B=9.3-5=4.3$
and simultaneously putting value of $I_B R$ from equation (ii) in equation (i),
$9.3=I_BR\left [ 150+\frac{R_B}{R} \right ]$
$9.3=\frac{5}{150}\left [ 150+\frac{R_B}{R} \right ]$
$279=150+\frac{R_B}{R}$
$\frac{R_B}{R}=129$
 Question 6
The circuit shown in the figure uses matched transistors with a thermal voltage $V_T$=25mV. The base currents of the transistors are negligible. The value of the resistance R in k$\Omega$ that is required to provide 1$\mu$A bias current for the differential amplifier block shown is ______.
 A 25.4 B 50.5 C 172.7 D 256.4
GATE EE 2017-SET-1   Analog Electronics
Question 6 Explanation:

$V_{BE1}=V_{BE2}+I_0R$
$I_0R=V_{BE1}-V_{BE2} =V_I ln\left ( \frac{I_R}{I_3} \right )-V_T ln \left ( \frac{I_0}{I_3} \right )$
where, $I_s\rightarrow$ Reverse saturation current
$R=\frac{V_T ln\left ( \frac{I_R}{I_0} \right )}{I_0}=\frac{0.025 ln\left ( \frac{1mA}{1\mu A} \right )}{1\mu A} =\frac{0.025 ln (10^3)}{1\mu A}=172.7 k\Omega$
 Question 7
A transistor circuit is given below. The Zener diode breakdown voltage is 5.3 V as shown. Take base to emitter voltage drop to be 0.6 V. The value of the current gain $\beta$ is _________.
 A 2.7 B 12 C 19 D 28
GATE EE 2016-SET-1   Analog Electronics
Question 7 Explanation:

$V_B=5.3 V$
$V_E=V_B-0.6=4.7V$
$I_E=\frac{V_E}{470\Omega }=10mA$
$I_1=\frac{10-5.3}{4.7k}=1mA$
$I_B=I_1-I_2=0.5mA$
$\frac{I_E}{I_B}=\beta +1=20$
$\beta =19$
 Question 8
When a bipolar junction transistor is operating in the saturation mode, which one of the following statements is TRUE about the state of its collector-base (CB) and the base-emitter (BE) junctions?
 A The CB junction is forward biased and the BE junction is reverse biased. B The CB junction is reverse biased and the BE junction is forward biased. C Both the CB and BE junctions are forward biased. D Both the CB and BE junctions are reverse biased.
GATE EE 2015-SET-2   Analog Electronics
 Question 9
In the following circuit, the transistor is in active mode and $V_{C}$=2 V. To get $V_{C}$=4 V, we replace $R_{C}$ with $R^{'}_{C}$ . Then the ratio $R^{'}_{C}/R_{C}$ is ______.
 A 0.5 B 0.75 C 1 D 1.25
GATE EE 2015-SET-2   Analog Electronics
Question 9 Explanation:
CASE-I:

$V_C=2V$
$i_C=\frac{10-2}{R_C} ......(i)$
$10-i_BR_B-0.7=0$
$i_B=\frac{10-0.7}{R_B} .......(ii)$

CASE-II

when $V_C=4V$
$R_C\rightarrow R'_C$
$i_C=\frac{10-4}{R'_C} .....(iii)$
From above equation (iii) and (i),
$\frac{10-4}{R'_C}= \frac{10-2}{R_C}$
$\Rightarrow \frac{R'_C}{R_C}=\frac{6}{8}=\frac{3}{4}=0.75$
 Question 10
In the given circuit, the silicon transistor has $\beta$=75 and a collector voltage $V_c$=9 V. Then the ratio of $R_{B}$ and $R_{C}$ is ________.
 A 75.25 B 105.13 C 150.36 D 175.45
GATE EE 2015-SET-1   Analog Electronics
Question 10 Explanation:
Consider the circuit shown in figure.

$V_C=9V$
$So,\;\; \frac{15-9}{R_C}=I_E$
$\frac{6}{R_C}=I_E$
$and \;\;\frac{9-0.7}{R_B}=I_B$
$So,\;\;\frac{8.3}{R_B}=I_B$
$So, \;\; \frac{I_E}{I_B}=\frac{6 \times R_B}{R_C \times 8.3}$
$(\beta +1)=\frac{R_B \times 6}{R_C \times 8.3}$
$\frac{R_B}{R_C}=105.13$
There are 10 questions to complete.