Q: The \$V\$-\$I\$ characteristic of a silicon diode is shown in the figure. Calculate the resistance of the diode at (a) \$I_{D} = 15\ mA\$ and (b) \$V_{D} = -10\ V\$.

We know that,Resistance of diode is given vy \$\cfrac{\triangle V}{\triangle I}\$So, at \$I_V=15 mA\$From \$V-I\$characteristic equation:\$R=\cfrac{\left(0.8-0.7\right)V}{\left(20-10\right)mA}=10\Omega\$And,at \$V_n=-10V\$\$\triangle I=0\$So, \$R\$ is infinite.

Q: (a) Calculate the value of \$V_0\$ and I if the Si diode and the Ge diode start conducting at \$0.7\ V\$ and \$0.3\ V\$ respectively, in the given circuit.(b) If the Ge diode connection be reversed. What will be the new values of \$V_0\$ and I?

As in the given circuit,Voltage,\$V=12\$VResistance,\$R=5\$k\$\Omega\$\$\therefore\$, current,\$I=\dfrac{12-0.3}{5\ k \Omega}=\dfrac{11.7 \ V}{5\times{10}^{3}\Omega}=2.34\$mAOutput voltage \${V}_{0}=RI=\left(5\times{10}^{3}\right)\times\left(2.34\times{10}^{-3}\right)=11.7\$VWhen the connections of Ge diode are reversed, then current will bw through silicon.In this case, current,\$I=\dfrac{\left(12-0.7\right)\ V}{5\times{10}^{-3}\Omega}=2.26\$mA\$\therefore\$, Output voltage,\${V}_{0}=IR\$\$=\left(2.26\times{10}^{-3}\right)\times\left(5\times{10}^{3}\right)=11.3\$V

Q: Calculate the current through the given circuit. (The diodes are ideal)

\${ D }_{ 1 }\$ is reversed biased (No current flow)\$ { D }_{ 2 }\$ is forward biased (current flow)\$ { E }_{ o }\$ total resistance\$={ R }_{ L }+{ R }_{ { D }_{ 2 } }\$\$ { R }_{ total }=5\Omega +10\Omega\$\$ =15\Omega\$ current through the circuit\$ \cfrac { V }{ R } \$Ohm's law\$ =\cfrac { 30V }{ 15\Omega } =2A\$

Q: From the graph between current \$I\$ and voltage \$V\$ shown below,identify the portion corresponding to negative resistance

we know that \$R=\dfrac VI=\dfrac{1}{\text{slope of I-V graph}}\$at CD slope is neagtive, so resistance is negative.Hence option C is correct

Q: If each diode in figure has forward bias resistance of \$ 25 \Omega \$ and infinite resistance in reverse bias , what will be the value of current \$ I_1 , I_2 , I_3 \$ and \$ I_4 \$ ?

\$ I_3 \$ is zero as the diode in that branch is reverse biased. Resistance in the branch AB and EF are each \$ (125 + 25)\Omega = 150 \Omega \$ As AB and EF are identical parallel branches , their effective resistance is \$ \dfrac{150}{2} = 75 \Omega \$ \$ \therefore \, \$ Net resistance in the circuit = \$ (75 + 25) \Omega = 100 \, \Omega \$ \$ \therefore \, \$ Current \$ I_1 = \dfrac{5}{100} = 0.05 \,A \$ AS resistance of AB and EF are equal and \$ I_1 = I_2 + I_3 + I_4 , I_3 = 0 \$ \$ \therefore \, I_2 = I_4 = \dfrac{0.05}{2} = 0.025\,A \$

Q: Two wires A and B of same material and same mass have radius \$2\$r and r. If resistance of wire A is \$34\Omega\$, then resistance of B will be?

The radius has a ratio of \$2:1\$, therefore the cross-sectional area is of ratio \$4:1.\$ The resistance per unit length is therefore \$ 1:4.\$ Because we are told the mass is the same, wire \$B\$, the smaller section wire will have to be \$4\$ times as long to have the same mass, neglecting the effects of any insulation on the mass. So the final resistance of \$ B\$ will be \$16\$ times the resistance of \$A\$ or \$544\$ Ohms.

Q: The \$V-i\$ graph for a conductor makes an angle \$\theta\$ with \$V-axis\$. Here \$V\$ denotes the voltage and \$i\$ denotes current. The resistance of conductor is given by

From ohm's law \$R=\dfrac{V}{i}\$also the slope of V-i Graph \$S=\dfrac{V}{i}\$angle with y axis is given by \$\tan \theta =\dfrac{I}{V}\$so \$R=S=\dfrac{1}{\tan\theta}=\cot\theta\$

Question 1

The $V$-$I$ characteristic of a silicon diode is shown in the figure. Calculate the resistance of the diode at (a) $I_{D} = 15\ mA$ and (b) $V_{D} = -10\ V$.

Accepted Answer

We know that,Resistance of diode is given vy $\cfrac{\triangle V}{\triangle I}$So, at $I_V=15 mA$From $V-I$characteristic equation:$R=\cfrac{\left(0.8-0.7\right)V}{\left(20-10\right)mA}=10\Omega$And,at $V_n=-10V$$\triangle I=0$So, $R$ is infinite.

Question 2

In a p-n junction diode, the current I can be expressed as $I = I_{0} e x p (\frac{e V}{2 K _{B} T} - 1)$ , where $I_{0}$ is called the reverse saturation current, $V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, $k_{B}$ is the Boltzmann constant $(8.6 \times 1 0^{- 5} e V / K)$ and $T$ is the absolute temperature. If for a given diode $I_{o} = 5 \times 1 0^{- 12}$ $A$ and $T = 300 K$ , then
(a) What will be the forward current at a forward voltage of $0.6 V$ ?(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 V$ ?(c) What is the dynamic resistance?(d) What will be the current if reverse bias voltage changes from $1 V$ to $2 V$ ?

Accepted Answer

(a)In a p-n junction diode, the expression for current is $I=I_0exp[\dfrac{eV}{2k_BT}-1]$where $I_0$ is the reverse saturation current = $5\times 10^{-12}A$$T=$Absolute temperature = $300\ K$$V$ is the voltage across diode.For forward voltage $V = 0.6\ V$ $I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times 10^{-23}\times 300}-1] = 0.0256\ A$&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;(b)&#10;&#10;In&#10;a p-n junction diode, the expression for current is&#10;$I=I_0exp[\dfrac{eV}{2k_BT}-1]$&#10;&#10;where&#10;$I_0$ is the reverse saturation current = $5\times 10^{-12}A$&#10;&#10;$T=$Absolute&#10;temperature=$300\ K$&#10;&#10;$V$ is the voltage across diode.&#10;&#10;For&#10;forward voltage, $V = 0.7\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times&#10;10^{-23}\times 300}-1]=1.257\ A$&#10;&#10;Hence, increase in current = $1.257\ A-0.0256\ A=1.23\ A$&#10;&#10;(c)&#10;&#10;In&#10;a p-n junction diode, the expression for current is&#10;$I=I_0exp[\dfrac{eV}{2k_BT}-1]$&#10;&#10;where&#10;$I_0$ is the reverse saturation current = $5\times 10^{-12}A$&#10;&#10;$T=$Absolute&#10;temperature=$300\ K$&#10;&#10;$V$ is the voltage across diode.&#10;&#10;For&#10;forward voltage $V = 0.6\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.6}{1.376\times&#10;10^{-23}\times 300}-1]=0.0256\ A$&#10;&#10;For&#10;forward voltage $V = 0.7\ V$ &#10;&#10;$I=5\times&#10;10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times&#10;10^{-23}\times 300}-1]=1.257\ A$&#10;&#10;Hence&#10;increase in current=$1.257\ A-0.0256\ A = 1.23\ A$&#10;&#10;Dynamic&#10;resistance = Change in voltage /Change in current&#10;&#10;$=\dfrac{0.7-0.6}{1.23}=0.081\Omega$&#10;&#10;(d)&#10;&#10;If&#10;reverse bias voltage changes from $1\ V$ to $2\ V$, the current will be&#10;almost constant, because dynamic resistance in the reverse bias is&#10;infinite.

Question 3

(a) Calculate the value of $V_0$ and I if the Si diode and the Ge diode start conducting at $0.7\ V$ and $0.3\ V$ respectively, in the given circuit.(b) If the Ge diode connection be reversed. What will be the new values of $V_0$ and I?

Accepted Answer

As in the given circuit,Voltage,$V=12$VResistance,$R=5$k$\Omega$$\therefore$, current,$I=\dfrac{12-0.3}{5\ k \Omega}=\dfrac{11.7 \ V}{5\times{10}^{3}\Omega}=2.34$mAOutput voltage ${V}_{0}=RI=\left(5\times{10}^{3}\right)\times\left(2.34\times{10}^{-3}\right)=11.7$VWhen the connections of Ge diode are reversed, then current will bw through silicon.In this case, current,$I=\dfrac{\left(12-0.7\right)\ V}{5\times{10}^{-3}\Omega}=2.26$mA$\therefore$, Output voltage,${V}_{0}=IR$$=\left(2.26\times{10}^{-3}\right)\times\left(5\times{10}^{3}\right)=11.3$V

Question 4

Calculate the current through the given circuit. (The diodes are ideal)

Accepted Answer

${ D }_{ 1 }$ is reversed biased (No current flow)$ { D }_{ 2 }$ is forward biased (current flow)$ { E  }_{ o  }$ total resistance$={ R }_{ L }+{ R }_{ { D }_{ 2 } }$$ { R }_{ total }=5\Omega +10\Omega$$ =15\Omega$ current through the circuit$ \cfrac { V }{ R }  $Ohm's law$ =\cfrac { 30V }{ 15\Omega  } =2A$

Question 5

From the graph between current $I$ and voltage $V$ shown below,identify the portion corresponding to negative resistance

Accepted Answer

we know that  $R=\dfrac VI=\dfrac{1}{\text{slope of I-V graph}}$at CD slope is neagtive, so resistance is negative.Hence option C is correct

Question 6

Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode (i) in forward bias and (ii) in reverse bias. Draw the typical V-I characteristics of a silicon diode.
Describe briefly the following terms :
(i) " minority carrier injection " in forward bias
(ii) " Breakdown voltage " in reverse bias.

Accepted Answer

(Refer Fig. 1) Two processes occurring the formation of a p-n junction are diffusion and drift. Due to the concentration gradient across p and n sides of the junction , holes diffuse from p-side to n-side ( p $ \rightarrow $n)and electrons diffuse from n-side to p-side (n$\rightarrow$p) . This movement of charge carriers leaves behind ionised acceptors (negative charge $ \phi $-immobile ) on the p-side and donors (positive charge immobile) on the n-side of the junction. This space charge region on either side of the junction together is known as depletion region. (b) (Refer Fig. 2) The circuit for studying the V-I characteristics of a diode are shown in Fig.(a) and Fig.(b) . For different values of voltages the value of current is noted . A graph between C and I is obtained as in Fig. (c).From the V-I characteristics of a junction diode it is clear that it allows current to pass only when it is biased. So if an alternating voltage is applied across a diode the current flows only in part of the cycle when the diode is forward biased. This property is used to rectify alternating voltages. (i) Minority Carrier Injection : Due to the applied voltage , electrons from n-side cross the depletion region and reach p-side (where they are minority carriers) . Similarly , holes from p-side cross this junction and reach the n-side *where they are minority carriers). This process under forward bias is known as minority carrier injection. (ii) Breakdown voltage : It is a critical reverse bias voltage at which current is independent of applied voltage.

Question 7

If each diode in figure has forward bias resistance of $ 25 \Omega $ and infinite resistance in reverse bias , what will be the value of current $ I_1 , I_2 , I_3 $ and $ I_4 $ ?

Accepted Answer

$ I_3 $ is zero as the diode in that branch is reverse biased. Resistance in the branch AB and EF are each $ (125 + 25)\Omega = 150 \Omega $ As AB and EF are identical parallel branches , their effective resistance is $ \dfrac{150}{2} = 75 \Omega $ $ \therefore \, $ Net resistance in the circuit = $ (75 + 25) \Omega = 100 \, \Omega $ $ \therefore \, $ Current $ I_1 = \dfrac{5}{100} = 0.05 \,A $ AS resistance of AB and EF are equal and $ I_1 = I_2 + I_3 + I_4 , I_3 = 0 $ $ \therefore \, I_2 = I_4 = \dfrac{0.05}{2} = 0.025\,A $

Question 8

The reading of the ammeter for a silicon diode in the given circuit is :-

Accepted Answer

Silicon diode is in forward bias.Hence across diode potential barrier$\Delta V = 0.7\ volts$$I= \dfrac{V-\Delta V}{R}=\dfrac{3-0.7}{200}$$=\dfrac{2.3}{200}= 11.5 mA$

Question 9

Two wires A and B of same material and same mass have radius $2$r and r. If resistance of wire A is $34\Omega$, then resistance of B will be?

Accepted Answer

The radius has a ratio of $2:1$, therefore the cross-sectional area is of ratio $4:1.$&#10;&#10;The resistance per unit length is therefore $ 1:4.$&#10;&#10;Because we are told the mass is the same, wire $B$, the smaller section wire will have to be $4$ times as long to have the same mass, neglecting the effects of any insulation on the mass. &#10;&#10;So the final resistance of $ B$ will be $16$ times the resistance of $A$ or $544$ Ohms. &#10;&#10;

Question 10

The $V-i$ graph for a conductor makes an angle $\theta$ with $V-axis$. Here $V$ denotes the voltage and $i$ denotes current. The resistance of conductor is given by

Accepted Answer

From ohm's law $R=\dfrac{V}{i}$also the slope of V-i Graph $S=\dfrac{V}{i}$angle with y axis is given by $\tan \theta =\dfrac{I}{V}$so $R=S=\dfrac{1}{\tan\theta}=\cot\theta$

Revise with Concepts

V - I Characteristics of P - N Junction Diode

Quick Summary With Stories

V - I Characteristics of P-N Junction Diode

Problems on V - I Characteristics

The $V$ - $I$ characteristic of a silicon diode is shown in the figure. Calculate the resistance of the diode at
(a) $I_{D} = 15 m A$ and
(b) $V_{D} = - 10 V$ .

(a) Calculate the value of $V_{0}$ and I if the Si diode and the Ge diode start conducting at $0.7 V$ and $0.3 V$ respectively, in the given circuit.
(b) If the Ge diode connection be reversed. What will be the new values of $V_{0}$ and I?

Calculate the current through the given circuit. (The diodes are ideal)

From the graph between current $I$ and voltage $V$ shown below,identify the portion corresponding to negative resistance

If each diode in figure has forward bias resistance of $25 Ω$ and infinite resistance in reverse bias , what will be the value of current $I_{1}, I_{2}, I_{3}$ and $I_{4}$ ?

The reading of the ammeter for a silicon diode in the given circuit is :-

Two wires A and B of same material and same mass have radius $2$ r and r. If resistance of wire A is $34 Ω$ , then resistance of B will be?

The $V - i$ graph for a conductor makes an angle $θ$ with $V - a x i s$ . Here $V$ denotes the voltage and $i$ denotes current. The resistance of conductor is given by

Revise with Concepts

V - I Characteristics of P - N Junction Diode

Quick Summary With Stories

V - I Characteristics of P-N Junction Diode

Problems on V - I Characteristics

The V-I characteristic of a silicon diode is shown in the figure. Calculate the resistance of the diode at (a) ID​=15 mA and (b) VD​=−10 V.

(a) Calculate the value of V0​ and I if the Si diode and the Ge diode start conducting at 0.7 V and 0.3 V respectively, in the given circuit. (b) If the Ge diode connection be reversed. What will be the new values of V0​ and I?

Calculate the current through the given circuit. (The diodes are ideal)

From the graph between current I and voltage V shown below,identify the portion corresponding to negative resistance

If each diode in figure has forward bias resistance of 25Ω and infinite resistance in reverse bias , what will be the value of current I1​,I2​,I3​ and I4​ ?

The reading of the ammeter for a silicon diode in the given circuit is :-

Two wires A and B of same material and same mass have radius 2r and r. If resistance of wire A is 34Ω, then resistance of B will be?

The V−i graph for a conductor makes an angle θ with V−axis. Here V denotes the voltage and i denotes current. The resistance of conductor is given by

The $V$ - $I$ characteristic of a silicon diode is shown in the figure. Calculate the resistance of the diode at
(a) $I_{D} = 15 m A$ and
(b) $V_{D} = - 10 V$ .

(a) Calculate the value of $V_{0}$ and I if the Si diode and the Ge diode start conducting at $0.7 V$ and $0.3 V$ respectively, in the given circuit.
(b) If the Ge diode connection be reversed. What will be the new values of $V_{0}$ and I?

From the graph between current $I$ and voltage $V$ shown below,identify the portion corresponding to negative resistance

If each diode in figure has forward bias resistance of $25 Ω$ and infinite resistance in reverse bias , what will be the value of current $I_{1}, I_{2}, I_{3}$ and $I_{4}$ ?

Two wires A and B of same material and same mass have radius $2$ r and r. If resistance of wire A is $34 Ω$ , then resistance of B will be?

The $V - i$ graph for a conductor makes an angle $θ$ with $V - a x i s$ . Here $V$ denotes the voltage and $i$ denotes current. The resistance of conductor is given by