A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W , is operated at 15 V . The approximate value of protective resistance in this circuit is $\_\_\_\_$ $\Omega$.
Explanation:
The series protective resistance limits the current flowing through the Zener diode so that it does not exceed the power rating.
The Zener diode maintains a constant voltage $\left(V_z\right)$ across load resistance $R_L$. The excess voltage from the source ( $V_{\text {in }}-V_z$ ) is dropped across the resistor $R_s$.
The Zener breakdown voltage $\mathrm{V}_{\mathrm{Z}}=10 \mathrm{~V}$
Maximum power dissipation $\mathrm{P}_{\text {max }}=0.4 \mathrm{~W}$
Using the power formula $\mathrm{P}=\mathrm{V} \times \mathrm{I}$.
$ I_{\max }=\frac{P_{\max }}{V_z} $
$\Rightarrow $ $ I_{\max }=\frac{0.4 \mathrm{~W}}{10 \mathrm{~V}}=0.04 \mathrm{~A} $
The resistor is connected in series with the Zener diode. According to Kirchhoff's Voltage Law (KVL), the sum of voltage drops must equal the supply voltage.
$V_{i n}=V_{R s}+V_z$
$\Rightarrow $ $V_{R s}=V_{i n}-V_z$
$\Rightarrow $ $ V_{R s}=15 \mathrm{~V}-10 \mathrm{~V}=5 \mathrm{~V} $
To protect the diode, we choose the resistance such that the current flowing through it does not exceed $\mathrm{I}_{\max }$.
$ \mathrm{R}_{\mathrm{s}}=\frac{\mathrm{V}_{\mathrm{Rs}}}{\mathrm{I}_{\mathrm{max}}} $
$\Rightarrow $ $\mathrm{R}_{\mathrm{s}}=\frac{5}{0.04} \Omega=125 \Omega$
A potential divider circuit is connected with a dc source of $20 \mathrm{~V}$, a light emitting diode of glow in voltage $1.8 \mathrm{~V}$ and a zener diode of breakdown voltage of $3.2 \mathrm{~V}$. The length (PR) of the resistive wire is $20 \mathrm{~cm}$. The minimum length of PQ to just glow the LED is _________ $\mathrm{cm}$.
Explanation:
For minimum length of $P Q$,
$\begin{aligned} & \mathrm{V}_{\text {diodes }}=1.8+3.2=5 \mathrm{~V} \\ & \therefore \quad I_{P Q}=\frac{20}{20} \times 5 \\ & =5 \mathrm{~cm} \\ & \end{aligned}$
From the given transfer characteristic of a transistor in $\mathrm{CE}$ configuration, the value of power gain of this configuration is $10^{x}$, for $\mathrm{R}_{\mathrm{B}}=10 ~\mathrm{k} \Omega$, and $\mathrm{R}_{\mathrm{C}}=1 ~\mathrm{k} \Omega$. The value of $x$ is __________.
Explanation:
$ \begin{aligned} & \Rightarrow \mathrm{A}_{\mathrm{v}} \cdot \mathrm{A}_1=\mathrm{B} \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}} \cdot \mathrm{B}=\mathrm{B}^2 \frac{\mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}} \\\\ & =\left(\frac{(20-10) \times 10^{-3}}{(200-100) \times 10^{-6}}\right) \times \frac{1 \times 10^3}{10 \times 10^3}=10^3 \end{aligned} $
Hence $\mathrm{x}=3$
A $8 \mathrm{~V}$ Zener diode along with a series resistance $\mathrm{R}$ is connected across a $20 \mathrm{~V}$ supply (as shown in the figure). If the maximum Zener current is $25 \mathrm{~mA}$, then the minimum value of R will be _______ $\Omega$.
Explanation:
Applying voltage rule in loop ABCD.
$ \begin{aligned} \mathrm{E}-\mathrm{IR}-\mathrm{V}_{\mathrm{Z}} & =0 \\\\ 20-\mathrm{IR}-8 & =0 \\\\ 2.5 \times 10^{-3} \times \mathrm{R} & =12 \end{aligned} $
$\mathrm{R}=\frac{12}{25 \times 10^{-3}}=480 \Omega$
If the potential barrier across a p-n junction is $0.6 \mathrm{~V}$. Then the electric field intensity, in the depletion region having the width of $6 \times 10^{-6} \mathrm{~m}$, will be __________ $\times 10^{5} \mathrm{~N} / \mathrm{C}$.
Explanation:
$E = {V \over d} = {{0.6} \over {6 \times {{10}^{ - 6}}}} = 1 \times {10^5}$
The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 \,k \Omega$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 \,\mathrm{k} \Omega$. The voltage gain of the transistor is ______________.
Explanation:
Vgain = Current gain $\times$ ${{{R_L}} \over {{R_i}}}$
$ = {{\Delta {I_C}} \over {\Delta {I_B}}} \times {{{R_L}} \over {{R_i}}}$
$ = {{5 \times {{10}^{ - 3}}} \over {100 \times {{10}^{ - 6}}}} \times {{2 \times {{10}^3}} \over {0.5 \times {{10}^3}}} = {{10} \over {0.5}} \times 10 = 200$
In the circuit shown below, maximum zener diode current will be _________ $\mathrm{mA}$.
Explanation:
${i_s} = {{60} \over {4 \times {{10}^3}}} = 15 \times {10^{ - 3}} = 15$ mA
${i_L} = {{60} \over {10 \times {{10}^3}}} = 6$ mA
${I_z} = {i_s} - {i_L} = 9$ mA
Two ideal diodes are connected in the network as shown in figure. The equivalent resistance between A and B is __________ $\Omega$.
Explanation:
$R = {{20 \times 20} \over {40}} + 15 = 25\,\Omega $
The energy band gap of semiconducting material to produce violet (wavelength = 4000$\mathop A\limits^o $ ) LED is ______________ $\mathrm{eV}$. (Round off to the nearest integer).
Explanation:
Energy corresponding to wavelength 4000 $\mathop A\limits^o $
$E = {{hc} \over \pi }$
$ = {{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {4000 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}}$ eV
$ = {{12400} \over {4000}}$
$=3.1$ eV
$\approx$ 3 eV
The circuit diagram used to study the characteristic curve of a zener diode is connected to variable power supply (0 $-$ 15 V) as shown in figure. A zener diode with maximum potential Vz = 10 V and maximum power dissipation of 0.4 W is connected across a potential divider arrangement. The value of resistance RP connected in series with the zener diode to protect it from the damage is ________________ $\Omega$.
Explanation:
$ \begin{aligned} & \mathrm{I}_{\mathrm{ZM}}=\frac{P_{z M}}{V_{z}}=\frac{0.4}{10}=0.04 \mathrm{~A} \\\\ & \text { So, } R_{P}=\frac{E_{\max }-V_{z}}{I_{Z M}}=\frac{15-10}{0.04}=\frac{5}{0.04}=125 \Omega \end{aligned} $
A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 $\times$ 105 ms$-$1. The speed with which electron enters the p side will be ${x \over 3} \times {10^5}$ ms$-$1 the value of x is _____________.
(Given mass of electron = 9 $\times$ 10$-$31 kg, charge on electron = 1.6 $\times$ 10$-$19 C.)
Explanation:
Conserving energy,
${1 \over 2}m{v^2} = {1 \over 2}m{(6 \times {10^5})^2} - 0.4\,eV$
$ \Rightarrow v = \sqrt {{{(6 \times {{10}^5})}^2} - {{2 \times 1.6 \times {{10}^{ - 19}} \times 0.4} \over {9 \times {{10}^{ - 31}}}}} $
$ = \sqrt {36 \times {{10}^{10}} - {{1.28} \over 9} \times {{10}^{12}}} $
$ \Rightarrow v = {{14} \over 3} \times {10^5}$ m/s
$ \Rightarrow x = 14$
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 $\mu$A, it brings a change of 10 mA in collector current. If the load resistance is 2 k$\Omega$ and input resistance is 1 k$\Omega$, the value of power gain is x $\times$ 104. The value of x is _____________.
Explanation:
Power gain $ = {\left[ {{{\Delta {i_C}} \over {\Delta {i_B}}}} \right]^2} \times {{{R_o}} \over {{R_i}}}$
$ = {\left[ {{{{{10}^{ - 2}}} \over {{{10}^{ - 4}}}}} \right]^2} \times {2 \over 1}$
$ = 2 \times {10^4}$
$ \Rightarrow x = 2$
A zener of breakdown voltage Vz = 8 V and maximum zener current, IZM = 10 mA is subjectd to an input voltage Vi = 10 V with series resistance R = 100 $\Omega$. In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____________.
Explanation:
Minimum value of RL for which the diode is shorted is ${{{R_L}} \over {{R_L} + 100}} \times 10 = 8 \Rightarrow {R_L} = 400\,\Omega $
For maximum value of RL, current through diode is 10 mA
So ${i_R} = {i_{{R_L}}} + {I_{ZM}}$
${2 \over {100}} = {8 \over {{R_L}}} + 10 \times {10^{ - 3}}$
$10 \times {10^{ - 3}} = {8 \over {{R_L}}}$
${R_L} = 800\,\Omega $
So ${{{R_{L\,\max }}} \over {{R_{L\,\min }}}} = 2$
The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 $\Omega$ is __________ mA.
Explanation:
D1 : conducting
D2 : open circuit
$i = {{1 - 0.6} \over {60 + 40}}A$
$ = {{0.4} \over {100}}A$
$ \Rightarrow i = 4$ mA
As per the given circuit, the value of current through the battery will be ____________ A.
Explanation:
Effective circuit will be
Req = 16 || 16 + 2 = (8 + 2) $\Omega$
= 10 $\Omega$
I0 = 10 / 10
= 1 A
In an experiment of CE configuration of n-p-n transistor, the transfer characteristics are observed as given in figure.
If the input resistance is 200 $\Omega$ and output resistance is 60 $\Omega$, the voltage gain in this experiment will be ____________.
Explanation:
Voltage gain $ = {{{I_C}{R_0}} \over {{I_B}{R_i}}}$
$ = {{(10\,mA)(60\,\Omega )} \over {(200\,\mu A)(200\,\Omega )}}$
$\Rightarrow$ Voltage gain = 15
In the given circuit, the value of current IL will be ____________ mA. (When RL = 1k$\Omega$)
Explanation:
VL = 5 V as VZ = 5 V
$\therefore$ ${I_L} = {{{V_L}} \over {{R_L}}} = {5 \over {{{10}^3}}} = 5$ mA
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10 $\mu$A and the collector current changes by 1.5 mA. The load resistance is 5 k$\Omega$. The voltage gain of the transistor will be _________.
Explanation:
${R_B} = {{10 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$
$ = {10^3}\,\Omega $
$\therefore$ ${A_v} = \left( {{{\Delta {I_C}} \over {\Delta {I_B}}}} \right) \times \left( {{{{R_C}} \over {{R_B}}}} \right)$
$ = {{1.5 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}} \times {{5 \times {{10}^3}} \over {1 \times {{10}^3}}}$
$ = {{1.5 \times 5} \over {10}} \times (1000)$
$ = 750$
Explanation:
and Pzener = 2W
VI = 2
I = ${2 \over {10}}$ = 0.2 A
$\Delta$VRs = I Rs
4 $\times$ 0.2 Rs $\Rightarrow$ Rs = ${40 \over {2}}$ = 20$\Omega$
Explanation:
$\Rightarrow$ Input voltage become zero
$\Rightarrow$ Input current is zero
$\Rightarrow$ Output current is zero
$\Rightarrow$ V0 = 5 volt
Explanation:
$i = {{10V} \over {5k\Omega }}$ = 2mA
$I = {{14V} \over {1k\Omega }}$ = 14 mA
$\therefore$ Iz = 12 mA
$\therefore$ P = IzVz = 120 mW
Explanation:
RC = 1000
$\Delta$V = 0.6
${I_C} = {{0.6} \over {1000}}$
IC = 6 $\times$ 10$-$4
${I_B} = {{{I_C}} \over \beta } = {{6 \times {{10}^{ - 4}}} \over {24}} = 25\mu A$
Explanation:
${n_e} = {{{n_i}^2} \over {{n_h}}} = {{{{(1.5 \times {{10}^{16}})}^2}} \over {4.5 \times {{10}^{22}}}}$
$ = {{1.5 \times 1.5 \times {{10}^{32}}} \over {4.5 \times {{10}^{22}}}}$
$ = 5 \times {10^9}$/m3
Explanation:
For RL $\to$ $\infty$
R = 500$\Omega$
Explanation:
0.5 = 8i
i = ${1 \over {16}}$A
E = 20 = 8 + i Rp
Rp = 12 $\times$ 16 = 192$\Omega$
Explanation:
$ = {{100} \over 4} = 25\Omega $
The estimated current gain from the figure is __________.
Explanation:
$\beta $ = ${{\Delta {I_C}} \over {\Delta {I_B}}} = {{\left( {4 - 2} \right)\,mA} \over {\left( {20 - 10} \right)\,\mu A}}$
= ${2 \over {10}} \times {{{{10}^{ - 3}}} \over {{{10}^{ - 6}}}}$ = 200
Explanation:
Input resistance = 100$\Omega$
Output load resistance = 10K$\Omega$
Power gain = ${\beta^2} \times {{{r_{out}}} \over {{R_{in}}}}$
$ \Rightarrow $ ${10^6} = {\beta ^2} \times {{10 \times {{10}^3}} \over {100}}$
$ \Rightarrow $ $\beta$2 = 104
$ \Rightarrow $ $\beta$ = 100
The value of x, to the nearest integer, is __________.
Explanation:
$i = {7 \over {35}} = {1 \over 5}A$
${i_1} = {{15} \over {90}} = {1 \over 6}A$
${i_2} = i - {i_1}$
${i_2} = {1 \over 5} - {1 \over 6}$
${i_2} = {1 \over {30}}A$
Power across diode; P = V2 i2
$P = 15 \times {1 \over {30}}$
P = 0.5 W
$ \therefore $ P = 5 $\times$ 10$-$1 W
The value of x is ____________.
Explanation:
Explanation:
$I = {{90 - 30} \over 4} = 15mA$
${I_1} = {{30} \over {5K\Omega }} = 6mA$
${I_2} = 15mA - 6mA = 9mA$
Explanation:
Given, forward resistance, R1 = 50 $\Omega$
Reverse resistance, R2 = infinity
Battery voltage = 6V
According to circuit diagram,
In this case, diode D1 is forward biased, whereas diode D2 is reverse biased.
So, D2 will act as open circuit.
$6 - 50I - 130I - 120I = 0$
$ \Rightarrow 6 = 300I$
$ \Rightarrow I = {6 \over {300}} = {1 \over {50}}$
$ = {2 \over {100}} = 0.02$ A = 20 mA
Hence, current through 120 $\Omega$ = 20 mA
Explanation:
In Zener breakdown,
$i = {V \over R} = {5 \over {2 \times {{10}^3}}} = 2.5 \times {10^{ - 3}}$
$\because$ $x \times {10^{ - 4}} = 25 \times {10^{ - 4}}$
$\therefore$ x = 25
When VCE is 10V and IC = 4.0 mA, then value of $\beta $ac is __________.
Explanation:
$\Delta $IB = (30 - 20) = 10 $\mu $A
$\Delta $IC = (4.5 - 3) = 1.5 mA
$ \therefore $ $\beta $ac = ${{1.5 \times {{10}^{ - 3}}} \over {10 \times {{10}^{ - 6}}}}$ = 150
Explanation:
So power dissipited in each diode = VI
= 4 $ \times $ 10-2
= 40 mW
Explanation:
Diode left hand diode is forward biased and right hand diode is reverse biased.