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.

In a p-n junction diode, the current I can be expressed as \\(I = I_0\,exp\, \left( \dfrac{eV}{2K_BT} - 1 \right)\\), 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 10^{-5} eV/K)\\) and \\(T\\) is the absolute temperature. If for a given diode \\(I_o = 5 \times 10^{-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\\)?

(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\\) (b) 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 = 0.7\ V\\) \\(I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257\ A\\) Hence, increase in current = \\(1.257\ A-0.0256\ A=1.23\ A\\) (c) 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\\) For forward voltage \\(V = 0.7\ V\\) \\(I=5\times 10^{-12}\times exp[\dfrac{1.6\times 10^{-19}\times 0.7}{1.376\times 10^{-23}\times 300}-1]=1.257\ A\\) Hence increase in current=\\(1.257\ A-0.0256\ A = 1.23\ A\\) Dynamic resistance = Change in voltage /Change in current \\(=\dfrac{0.7-0.6}{1.23}=0.081\Omega\\) (d) If reverse bias voltage changes from \\(1\ V\\) to \\(2\ V\\), the current will be almost constant, because dynamic resistance in the reverse bias is infinite.

(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

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\\)

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

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.

(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.

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 \\)

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

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\\)

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.

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\\)

V - I Characteristics of P - N Junction Diode | Definition, Examples, Diagrams

diagram

Current-Voltage characteristics of a diode

Current v/s voltage characteristics are attached in the figure.

definition

Threshold voltage

Threshold Voltage :
The minimum value of forward voltage after which a diode conducts is called the threshold voltage of the diode.

V_{t, S i} \sim 0.7 V

V_{t, G e} \sim 0.3 V

definition

Reverse saturation current

A diode has very small value of current in the reverse bias due to the drift of the charge carriers known as the Reverse saturation current. It is in the order of

μ A

definition

Avalanche Breakdown

Avalanche breakdown is a phenomenon observed in a p-n junction diode in reverse bias. When a large reverse bias voltage is applied across a p-n junction diode, thermal current starts increasing to a large value. This results in the heating of the diode. Some avalanche diodes are used as voltage regulators similar to zener diodes.

definition

Dynamic resistance

Dynamic resistance is used to quantify the resistance of non-ohmic materials. It is defined as the ratio of differential change in voltage to a differential change in current.

r = \frac{d V}{d I}

Note:
Dynamic resistance is itself a function of the current (or voltage) through the material for a non-ohmic conductor.

V - I Characteristics of P - N Junction Diode

diagram

Current-Voltage characteristics of a diode