Semiconductor Diode

Semiconductor Electronics · CBSE

Theory

The device and its symbol

A semiconductor diode is basically a p-n junction with metallic contacts provided at the two ends for the application of an external voltage. It is a two-terminal device.
Circuit symbol in words: a horizontal line broken by a solid triangle whose apex touches a short bar drawn perpendicular to the line. The triangle side is the p-region (anode), the bar is the n-region (cathode). The direction of the arrowhead is the conventional direction of current when the diode is forward biased.
With no external voltage the junction sits at equilibrium: a depletion region of immobile ions and a barrier potential $V_0$ across it, with zero net current. Applying an external voltage $V$ alters that barrier, and that is the whole behaviour of the device.

Forward bias

Connection and what happens to the barrier

The diode is forward biased when the p-side is connected to the positive terminal of the battery and the n-side to the negative terminal.
Almost the whole applied voltage drops across the depletion region, because that region has no free charges and therefore a very high resistance compared with the bulk p- and n-material. The drop across the p-side and n-side themselves is negligible.
The applied voltage $V$ is directed opposite to the built-in potential $V_0$ . As a result the depletion layer narrows and the barrier height is reduced to an effective value

$V_{\text{eff}} = V_0 - V$

Why the current grows so steeply

At small applied voltage the barrier is only slightly reduced, so only the few carriers already in the uppermost energy levels have enough energy to cross. The current is very small.
Increase the applied voltage significantly and the barrier falls a lot, so far more carriers have the required energy — and the current increases sharply.

Minority carrier injection (a named 2- or 5-mark item)

Under forward bias, electrons from the n-side cross the depletion region and reach the p-side, where they are minority carriers; similarly holes from the p-side cross into the n-side, where they are minority carriers. This process is called minority carrier injection.
At the junction boundary on each side, the minority carrier concentration therefore rises well above its equilibrium value — much higher than at points far from the junction.
Because of this new concentration gradient, the injected electrons diffuse from the junction edge of the p-side towards the far end of the p-side, and the injected holes diffuse from the junction edge of the n-side towards the far end of the n-side. This motion of carriers on either side is what constitutes the current.
The total forward current is the sum of the hole diffusion current and the conventional current due to electron diffusion. Its magnitude is usually in milliamperes.

Reverse bias

The diode is reverse biased when the n-side is connected to the positive terminal and the p-side to the negative terminal.
Again most of the applied voltage drops across the depletion region, but now the applied voltage is in the same direction as the barrier potential. The barrier height increases and the depletion region widens:

$V_{\text{eff}} = V_0 + V$

This suppresses the flow of electrons from n to p and of holes from p to n, so the diffusion current falls enormously compared with the forward-biased case.
A small current still flows. The junction field is such that any minority carrier — an electron on the p-side or a hole on the n-side — that wanders close to the junction in its random motion is swept across to its own majority zone. This drift of minority carriers is the reverse current, and it is of the order of microamperes. (The same drift current exists under forward bias too, but there it is negligible beside the milliampere injection current.)
The reverse current hardly depends on the applied voltage. Even a small reverse voltage is enough to sweep essentially all the available minority carriers across the junction, so what limits the current is the concentration of minority carriers on the two sides, not the magnitude of the voltage. That is why the reverse characteristic is almost flat.

Breakdown

The reverse current stays essentially voltage-independent up to a critical reverse voltage called the breakdown voltage $V_{br}$ (also written $V_z$ on a graph).
At $V = V_{br}$ the reverse current increases sharply: even a slight further increase in the bias produces a large change in current. Physically, the junction field has become strong enough to break covalent bonds directly, or existing carriers gain enough energy between collisions to knock more carriers loose in an avalanche. Either way an enormous number of extra carriers is released at once.
If the reverse current is not limited by an external circuit below the manufacturer's rated value, the junction is destroyed by overheating. The same is true in forward bias if the rated forward current is exceeded. General-purpose diodes are not used beyond the reverse saturation region.
Definition to quote: the breakdown voltage is the minimum reverse voltage that must be applied across the p-n junction diode for the reverse current to increase abruptly.

The V-I characteristic and how it is obtained

Circuit for forward bias

Connect the diode so that the battery's positive terminal feeds the p-side and its negative terminal the n-side. Put the battery through a potentiometer or rheostat so that the voltage applied to the diode can be varied continuously.
Put a milliammeter in series with the diode (the forward current is large, of the order of mA) and a voltmeter across the diode.
Increase the forward voltage step by step from a low value, noting the current at each step, and plot current against voltage.

Circuit for reverse bias

The same arrangement with the battery's polarity reversed, so the n-side is driven positive with respect to the p-side, again through the rheostat.
Replace the milliammeter with a microammeter, since the reverse current is only of the order of $\mu$ A. Keep the voltmeter across the diode.
Increase the reverse voltage step by step and note the current.

The curve, described in words

Draw one set of axes: voltage along the horizontal, current along the vertical. The forward branch occupies the first quadrant, the reverse branch the third quadrant, and the two use different scales — volts and mA for the forward part, volts and $\mu$ A for the reverse part.
Forward branch: the current first increases very slowly, almost negligibly, while the applied voltage climbs. Once the voltage crosses a certain value the current increases significantly, in fact exponentially, for a very small further increase in voltage — the curve turns a sharp knee and rises almost vertically. Extending the steeply rising part back to the voltage axis cuts it very close to the barrier voltage; that is how the threshold voltage is read off the graph.
That voltage is called the threshold voltage or cut-in voltage: about $0.2$ V to $0.3$ V for a germanium diode and about $0.7$ V for a silicon diode.
Reverse branch: a very small current (of order $\mu$ A) that stays almost constant as the reverse voltage is increased — the reverse saturation current — until the breakdown voltage is reached, where the curve drops almost vertically as the current suddenly becomes large.
The characteristic is therefore strongly asymmetric: the diode conducts appreciably in one direction only. Forward resistance is low and reverse resistance is very high, and this property is what is used for the rectification of ac voltages.

Threshold voltage versus breakdown voltage

Feature	Threshold (cut-in) voltage	Breakdown voltage
Bias in which it appears	forward	reverse
What happens at it	the current stops being negligible and begins to rise steeply	the small reverse current suddenly increases abruptly
Typical size	about $0.2$ to $0.3$ V for Ge, $0.7$ V for Si	tens of volts, set by the manufacture of the diode
Is the diode damaged?	no, this is the normal working region	yes, unless the current is limited externally

Resistance of a diode

$r_d = \dfrac{\Delta V}{\Delta I} \qquad\qquad R_{\text{static}} = \dfrac{V}{I}$

Dynamic resistance $r_d$ is defined as the ratio of a small change in voltage to the corresponding small change in current. Use it on the steep forward part of the characteristic, where a single ratio $V/I$ would be meaningless.
Static resistance $V/I$ is used at a point on the flat reverse branch, where the characteristic is essentially a horizontal line and a single reading is all you have.

Worked patterns

Forward, from a graph (NCERT Example 14.4): treating the characteristic as a straight line between $I = 10$ mA and $I = 20$ mA, at $I = 20$ mA, $V = 0.8$ V and at $I = 10$ mA, $V = 0.7$ V, so $\Delta V = 0.1$ V, $\Delta I = 10$ mA, and $r_{fb} = 0.1/(10\times10^{-3}) = 10\,\Omega$ .
Forward, from the same silicon curve read between $I = 15$ mA and $I = 20$ mA (the version CBSE has set): at $I = 20$ mA, $V = 0.8$ V and at $I = 15$ mA, $V = 0.7$ V, so $\Delta I = 5\times10^{-3}$ A, $\Delta V = 0.1$ V, and

$R = \dfrac{\Delta V}{\Delta I} = \dfrac{0.1}{5\times10^{-3}} = 20\,\Omega$

The two answers differ only because the two readings are taken between different pairs of points on the curve. Always state the two points you used.
Reverse: at $V = -10$ V the curve reads $I = -1\,\mu$ A $= -1\times10^{-6}$ A, so

$R = \dfrac{V}{I} = \dfrac{10}{1\times10^{-6}} = 1.0\times10^{7}\,\Omega$

Dynamic resistance from a bare pair of readings: when the drop across a diode rises from $0.70$ V to $0.71$ V the current changes by $10$ mA, so $r_d = (0.71-0.70)/(10\times10^{-3}) = 0.01/0.01 = 1\,\Omega$ .

Circuit questions CBSE sets

Is this diode forward or reverse biased? Compare the potentials of the two terminals. The diode is forward biased only if the p-side is at the higher potential. Example: p-side grounded at $0$ V and n-side connected through a resistor to a $+5$ V supply — the p-side is at the lower potential, so the diode is reverse biased.
Diode below threshold: a silicon diode of threshold $0.7$ V in series with a $100\,\Omega$ resistor across a $0.5$ V battery, connected in the forward direction. The arrangement is forward biasing, but the applied $0.5$ V is less than the threshold $0.7$ V, so the diode does not conduct appreciably and the current is $0$ A. Name both parts of the answer: forward biasing, zero current.
Unbiased junction: the n-side is at the higher potential relative to the p-side, since the n-region has lost electrons to the p-region. (One CBSE key words the same fill-in-the-blank the other way round; write the physics and justify it with the direction of the barrier field, from n to p.)
Doping and depletion width: increasing the doping concentration decreases the depletion-layer width.
Net current at equilibrium: zero, because the diffusion and drift currents are equal and opposite — not because carriers stop crossing.

Forward bias versus reverse bias

Feature	Forward bias	Reverse bias
Battery connection	p to $+$ , n to $-$	n to $+$ , p to $-$
Effective barrier	$V_0 - V$ (lowered)	$V_0 + V$ (raised)
Depletion width	decreases	increases
Main carriers of the current	injected majority carriers diffusing across	minority carriers drifting across
Order of current	mA	$\mu$ A
Dependence on applied voltage	exponential rise past the threshold	almost independent, up to breakdown
Junction resistance	low (a few $\Omega$ to tens of $\Omega$ )	very high (about $10^{6}$ to $10^{7}\,\Omega$ )
Meter used in the experiment	milliammeter	microammeter

Standard results

Effective barrier: $V_0 - V$ forward, $V_0 + V$ reverse.
Threshold voltage: about $0.2$ to $0.3$ V for Ge, about $0.7$ V for Si.
Forward current in mA, reverse saturation current in $\mu$ A.
$r_d = \Delta V/\Delta I$ ; static reverse resistance $V/I$ , of order $10^{7}\,\Omega$ .
Reverse current is limited by the minority-carrier concentration, not by the applied voltage — hence the flat reverse characteristic up to $V_{br}$ .
At breakdown, covalent bonds break on a large scale (or an avalanche occurs) and the reverse current rises abruptly.
Forward resistance is low and reverse resistance is very high; this one-way conduction is what makes the diode a rectifier.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
In an unbiased p-n junction diode, the p-side of the junction is at ______ potential as compared to that on the n-side of the junction.
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 1 ↓
2
How does an increase in doping concentration affect the width of depletion layer of a p-n junction diode?
1 markCBSE Class 12 Physics — Very Short Answer · 2020, Ap · Section A, 1 markAnswer 2 ↓
3
Explain the formation of the barrier potential in a p-n junction.
2 marksCBSE Class 12 Physics — SA · Term II 2021-22, U · Section B, 2 marksAnswer 3 ↓
4
With the help of the circuit diagram, explain the working of a silicon p-n junction diode in forward biasing and draw its I-V characteristics.
2 marksCBSE Class 12 Physics — SA · 2020C · Section B, 2 marksAnswer 4 ↓
5
Explain the term 'depletion layer' and 'potential barrier' in a p-n junction diode. How are the (a) width of depletion layer, and (b) value of potential barrier affected when the p-n junction is forward biased?
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 5 ↓
6
Draw V-I characteristics of a p-n junction diode. Explain, why the current under reverse bias is almost independent of the applied voltage up to the critical voltage.
2 marksCBSE Class 12 Physics — SA · 2020 · Section B, 2 marksAnswer 6 ↓
7
The V-I characteristic of a silicon diode is as shown in the figure (a curve for silicon that, in the forward direction, passes through approximately $V=0.7$ V at $I=15$ mA and $V=0.8$ V at $I=20$ mA; and in the reverse direction shows a nearly flat current of about $I=-1\,\mu$ A for $V=-10$ V). Calculate the resistance of the diode at (i) $I=15$ mA, and (ii) $V=-10$ V.
2 marksCBSE Class 12 Physics — SA · Foreign 2015, An · Section B, 2 marksAnswer 7 ↓
8
Draw V-I characteristics of a p-n junction diode. Answer the following giving reasons: (a) Why is the reverse bias current almost independent of applied voltage up to the breakdown voltage? (b) Why does the reverse current show a sudden increase at breakdown voltage?
3 marksCBSE Class 12 Physics — SA · 2022C, U · Section C, 3 marksAnswer 8 ↓
9
(a) Why is an intrinsic semiconductor deliberately converted into an extrinsic semiconductor by adding impurity atoms? (b) Explain briefly the two processes that occur in the p-n junction region to create a potential barrier.
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 9 ↓
10
Explain the formation of potential barrier and depletion region in a p-n junction diode. What is the effect of applying forward bias on the width of depletion region?
3 marksCBSE Class 12 Physics — SA · 2020 · Section C, 3 marksAnswer 10 ↓
11
In the following diagram, is the junction diode forward biased or reverse biased? (The circuit shows a p-n junction diode connected between ground on the p-side and, through a resistor, a $+5$ V supply on the n-side.)
3 marksCBSE Class 12 Physics — SA · 1/3, AI 2017, R · Section C, 3 marksAnswer 11 ↓
12
Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode in (i) forward biasing and (ii) reverse biasing. Draw the typical V-I characteristics of a silicon diode. Describe briefly the following terms: (i) minority carrier injection in forward biasing, and (ii) breakdown voltage in reverse biasing.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 12 ↓
13
Explain briefly with the help of necessary diagrams, the forward and the reverse biasing of a p-n junction diode. Also draw their characteristic curves in the two cases. OR Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in (i) Forward biasing, and (ii) Reverse biasing. (Delhi 2014) OR Draw the circuit arrangement for studying the V-I characteristics of a p-n junction diode in (i) forward and (ii) reverse bias. Briefly explain how the typical V-I characteristics of a diode are obtained and draw these characteristics. (AI 2014C)
5 marksCBSE Class 12 Physics — Long Answer · Delhi 2017, Delhi 2014, AI 2014C, An · Section E, 5 marksAnswer 13 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

14
The threshold voltage for a p-n junction diode used in the circuit is 0.7 V. The type of biasing and current in the circuit are (the circuit has a diode $D$ in series with a $100\,\Omega$ resistor, connected across a $0.5$ V battery, with the diode oriented so that the battery's polarity drives it in its forward direction):
- (a)
  Forward biasing, 0 A
- (b)
  Reverse biasing, 0 A
- (c)
  Forward biasing, 5 mA
- (d)
  Reverse biasing, 2 mA
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 14 ↓
15
At equilibrium, in a p-n junction diode the net current is
- (a)
  due to diffusion of majority charge carriers.
- (b)
  due to drift of minority charge carriers.
- (c)
  zero as diffusion and drift currents are equal and opposite.
- (d)
  zero as no charge carriers cross the junction.
1 markCBSE Class 12 Physics — Multiple Choice · 2020, R · Section A, 1 markAnswer 15 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The p-side of the junction is at HIGHER potential compared to the n-side. (This is because, as electrons diffuse away from the n-side and accumulate on the p-side across the depletion region, the n-side is left positively charged relative to the p-side — equivalently, the internal barrier field points from n to p, meaning the p-side sits at the higher potential of the two.)
2.↑ back to the question
Increasing the doping concentration DECREASES the width of the depletion layer. A higher doping concentration means more donor/acceptor ions (charge carriers) are available near the junction; this larger concentration of immobile ions builds up a stronger internal electric field (and a larger potential difference) over a SHORTER distance to achieve the same barrier — so equilibrium is reached with a narrower depletion region.
3.↑ back to the question
When a p-n junction diode is formed, two processes occur simultaneously: diffusion and drift. Diffusion: the p-side has more holes than the n-side, and the n-side has more free electrons than the p-side. Due to this concentration gradient, holes diffuse from the p-side to the n-side, and electrons diffuse from the n-side to the p-side. As electrons diffuse from the n-side to the p-side, they leave behind ionised donor atoms (positively charged) on the n-side near the junction. Likewise, as holes diffuse from the p-side to the n-side, they leave behind ionised acceptor atoms (negatively charged) on the p-side near the junction. These layers of opposite immobile charge on either side of the junction create an electric field pointing from the n-side to the p-side (i.e. from the positive ions to the negative ions), within the region now called the depletion region. Drift: minority charge carriers on either side are swept across the junction by this field, into their respective majority zones. At equilibrium, the drift current and the diffusion current become equal (and opposite). The potential difference that has now developed across the depletion region, due to this layer of opposite charges, is called the BARRIER POTENTIAL — this is how the barrier potential is formed in a p-n junction.
4.↑ back to the question
Circuit: a p-n junction diode is connected through a milliammeter (mA) in series with a rheostat (variable resistor) and a battery, such that the battery's positive terminal connects (through the rheostat) to the p-side and its negative terminal to the n-side (forward bias); a voltmeter is connected across the diode to read the voltage across it. Starting from a low value, the forward bias voltage is increased step by step (read on the voltmeter), and the corresponding forward current is noted (on the milliammeter). A graph is then plotted between voltage and current — this is the forward I-V characteristic of the diode. At the start, when the applied voltage is low, the current through the diode is almost zero, because the applied voltage has to first overcome the internal potential barrier (about 0.7 V for silicon). Until the applied voltage exceeds this barrier, the current increases only very slowly with voltage (a nearly-flat initial portion of the curve). Once the applied voltage exceeds the potential barrier (the threshold or cut-in voltage, about 0.7 V for silicon), the current increases very rapidly with further small increases in voltage — the curve rises almost vertically, and in this region the diode behaves almost like a conductor. If this rapidly-rising part of the curve is extended back (extrapolated) to the voltage axis, it cuts the axis very close to the potential barrier voltage — this is why the threshold/cut-in voltage is also read off the graph this way.
5.↑ back to the question
Depletion layer: the small region in the immediate vicinity of the junction, which is depleted of free (mobile) charge carriers and contains only immobile (fixed) ions, is called the depletion layer. Potential barrier: due to the accumulation of negative charge (ionised acceptors) on the p-side and positive charge (ionised donors) on the n-side of the depletion layer, a potential difference $V_B$ is set up across the junction. This is called the potential barrier, and it opposes the further diffusion of electrons and holes across the junction. (a) In forward biasing, the applied external field is directed opposite to the internal barrier field and is able to partly overcome it, so the width of the depletion layer DECREASES. (b) In forward biasing, the forward voltage directly opposes the potential barrier, so the effective (net) barrier value DECREASES.
6.↑ back to the question
V-I characteristic: for FORWARD bias, the current stays nearly zero until the applied voltage crosses the threshold (about 0.2 V for Ge, 0.65–0.7 V for Si — "ON voltage"), after which the current rises very steeply (nearly vertically) with further small increases in voltage. For REVERSE bias, only a very small, almost constant current flows (of order tens of $\mu$ A) as the reverse voltage is increased, until a certain critical (breakdown/zener) voltage $V_z$ is reached, at which point the reverse current rises abruptly (a near-vertical drop in the graph beyond $V_z$ ). The reverse current is due to minority charge carriers (a few thermally-generated electrons on the p-side and holes on the n-side). Even a small applied reverse voltage is already sufficient to sweep essentially ALL of these minority carriers across the junction, from one side to the other. So the reverse current here is not limited by the MAGNITUDE of the applied voltage (there simply aren't more minority carriers available to be swept across, no matter how much harder you push, until breakdown) — it is limited instead by the concentration of minority carriers available on either side of the junction. This is why the reverse current stays almost constant (independent of voltage) up to the critical (breakdown) voltage.
7.↑ back to the question
(i) From the given curve, at $I=20$ mA, $V=0.8$ V; and at $I=15$ mA, $V=0.7$ V. $\Delta I=(20-15)\text{ mA}=5\times10^{-3}$ A, $\;\Delta V=(0.8-0.7)\text{ V}=0.1$ V. Resistance, $R=\dfrac{\Delta V}{\Delta I}=\dfrac{0.1}{5\times10^{-3}}=20\,\Omega$ . (ii) For $V=-10$ V, we have $I=-1\,\mu\text{A}=-1\times10^{-6}$ A. $R=\dfrac{V}{I}=\dfrac{10}{1\times10^{-6}}=1.0\times10^{7}\,\Omega$ .
8.↑ back to the question
V-I characteristics of a p-n junction: for germanium and for silicon, the forward-bias current stays negligible until the respective threshold voltage (about 0.3 V for Ge, 0.7 V for Si) and then rises steeply; the reverse-bias current stays small (of order a few $\mu$ A) and nearly flat over a wide range of reverse voltage, until breakdown. (a) The current in reverse bias (of order a few $\mu$ A) is due to the drifting of minority charge carriers from one region to the other across the junction. Even a small amount of applied reverse voltage is already sufficient to sweep essentially all the available minority charge carriers through the junction — so increasing the reverse voltage further does not create more current, since it is not the voltage but the (fixed, small) concentration of minority carriers that limits the current. Hence the reverse current stays almost independent of the applied voltage, up to breakdown. (b) At the breakdown voltage, the electric field across the junction becomes strong enough to break an enormous number of covalent bonds directly (or, in avalanche breakdown, existing carriers gain enough energy to knock loose many more by collision). As a result, a very large number of additional charge carriers is suddenly released, and the current increases abruptly at the breakdown voltage.
9.↑ back to the question
(a) Intrinsic semiconductors have very small conductivity at room temperature (only the thermally-generated electron-hole pairs are available as carriers). When a small, controlled amount of a suitable impurity ( $\sim1$ part per million) is added, the electrical conductivity of the doped (extrinsic) semiconductor increases to a great extent, making it far more useful in practical electronic applications — this is why intrinsic semiconductors are deliberately converted into extrinsic ones. (b) The two processes are diffusion and drift. Due to the difference in carrier concentration on the two sides, holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side (diffusion). This leaves behind immobile ionised donor (positive) ions on the n-side and ionised acceptor (negative) ions on the p-side near the junction, forming the depletion region and setting up an internal field. This field then drives minority carriers across the junction in the opposite sense (drift). The potential difference $V_0$ set up across the depletion region by this layer of immobile charges is the potential barrier.
10.↑ back to the question
The two processes that take place in the formation of a p-n junction are diffusion and drift. When a p-n junction is formed, free electrons from the n-side diffuse over to the p-side (filling holes there), and holes from the p-side diffuse over to the n-side. This leaves a layer of uncompensated positive charge (ionised donors) on the n-side of the junction and a layer of uncompensated negative charge (ionised acceptors) on the p-side. This thin region of immobile ions is called the depletion layer — it grows within a very short time of the junction being formed, preventing any further net movement of charge carriers across the junction. This layer of opposite charges sets up a potential difference $V_0$ (of order 0.1 to 0.3 V) across the junction, called the potential barrier $V_B$ , which opposes the further diffusion of electrons and holes across the junction. In forward biasing, the applied forward voltage OPPOSES the potential barrier and adds enough energy for the external field to overcome the strong internal electric field of the depletion layer — as a result, the width of the depletion region DECREASES (and the effective barrier height is reduced).
11.↑ back to the question
In this circuit, the p-side of the diode is connected (through the ground) to $0$ V, while the n-side is connected (through the resistor) toward the $+5$ V supply — so the voltage at the p-side is LESS than the voltage at the n-side of the diode. Since the p-side must be at a HIGHER potential than the n-side for forward biasing, and here the opposite is true, the junction diode is REVERSE BIASED.
12.↑ back to the question
Forward-bias circuit: the diode (p-n) is connected in series with a milliammeter (mA), with a voltmeter across the diode, and this combination connected (through a rheostat, to vary the voltage) to a battery whose positive terminal drives current into the p-side — so the diode is forward biased. The forward bias voltage is increased step by step and the corresponding current noted, to plot the forward characteristic (current initially near zero, rising rapidly once past the threshold voltage, about 0.7 V for silicon). Reverse-bias circuit: the same diode, now with the battery's polarity reversed (so the n-side is driven positive relative to the p-side, through the rheostat), with a microammeter ( $\mu$ A, since reverse current is very small) in series and a voltmeter across the diode. The reverse voltage is increased step by step; a small, nearly constant reverse current flows until the breakdown voltage is reached, beyond which the current rises abruptly. (i) Minority carrier injection in forward biasing: in a p-type semiconductor, electrons are minority carriers. When the p-n junction is forward biased, electrons from the n-side (where they are majority carriers) are pushed across the junction into the p-type material, where they become minority carriers there in larger-than-equilibrium numbers — this process of majority carriers from one side becoming injected minority carriers on the other side is called minority carrier injection. (ii) Breakdown voltage in reverse biasing: it is the minimum reverse voltage that must be applied across the p-n junction diode for the reverse current to increase abruptly (suddenly) — beyond this critical voltage, covalent bonds break down on a large scale (or carriers gain enough energy to cause an avalanche of further ionisation), releasing a very large number of additional charge carriers.
13.↑ back to the question
Forward biasing: the p-n junction diode is connected through a milliammeter and a rheostat to a battery, oriented so the p-side is driven positive relative to the n-side. Starting from a low value, the forward voltage is increased step by step (read by a voltmeter across the diode) and the corresponding current is noted. At the start, current is almost zero (opposed by the potential barrier); once the applied voltage exceeds the threshold voltage, current rises very rapidly with further small increases in voltage — a graph of this gives the forward characteristic (a curve that stays flat near zero, then bends sharply upward past the threshold voltage — about 0.2 V for Ge, 0.7 V for Si). Reverse biasing: the same diode is now connected with the battery's polarity reversed (n-side driven positive relative to the p-side), with a microammeter in place of the milliammeter (since the reverse current is very small). As the reverse voltage is increased, a small, almost constant reverse current (due to minority carriers) flows — nearly independent of the applied voltage — until the breakdown voltage is reached, after which the current increases abruptly. A graph of this gives the reverse characteristic (nearly flat near zero, dropping sharply once breakdown is reached).
14.Answer: (A)↑ back to the question
The diode is connected in the circuit's forward direction (the battery's positive terminal faces the diode's p-side), so this IS a forward-biasing arrangement. However, the applied voltage of the battery ( $0.5$ V) is LESS than the diode's threshold (cut-in) voltage of $0.7$ V. Since the applied voltage has not yet reached the threshold voltage, the diode does not conduct any appreciable current — so the current in the circuit is $0$ A. Option (a): Forward biasing, 0 A.
15.Answer: (C)↑ back to the question
At equilibrium, the diffusion current (majority carriers moving down their concentration gradient) and the drift current (minority carriers swept across by the internal barrier field) are equal in magnitude but opposite in direction — so charge carriers DO keep crossing the junction in both directions, but the NET current is zero because the two currents exactly cancel. Option (c).