201. A forward-biased silicon diode has \(0.75\,V\) across it and carries \(15\,mA\). Its static resistance at this operating point is:
ⓐ. \(20\,\Omega\)
ⓑ. \(75\,\Omega\)
ⓒ. \(200\,\Omega\)
ⓓ. \(50\,\Omega\)
Correct Answer: \(50\,\Omega\)
Explanation: \( \textbf{Given voltage:} \) \(V=0.75\,V\).
\( \textbf{Given current:} \) \(I=15\,mA=15\times10^{-3}\,A\).
\( \textbf{Required quantity:} \) Static resistance at the operating point.
\( \textbf{Static resistance relation:} \)
\[
R=\frac{V}{I}
\]
\( \textbf{Why this relation applies:} \) Static resistance uses the ratio of total voltage to total current at a chosen point on the \(V\)-\(I\) curve.
\( \textbf{Substitution:} \)
\[
R=\frac{0.75}{15\times10^{-3}}
\]
\( \textbf{Simplification:} \)
\[
R=\frac{0.75}{0.015}
\]
\( \textbf{Calculation:} \)
\[
R=50\,\Omega
\]
\( \textbf{Final answer:} \) The static resistance at the operating point is \(50\,\Omega\).
202. A diode is reverse biased when:
ⓐ. \(p\)-side is connected to the positive terminal and \(n\)-side to the negative terminal
ⓑ. the junction is connected with zero external voltage only
ⓒ. both terminals are shorted together with no potential difference
ⓓ. \(p\)-side is connected to the negative terminal and \(n\)-side to the positive terminal
Correct Answer: \(p\)-side is connected to the negative terminal and \(n\)-side to the positive terminal
Explanation: Reverse bias is the opposite polarity of forward bias. In reverse bias, the \(p\)-side is connected to the negative terminal of the battery and the \(n\)-side to the positive terminal. This polarity pulls majority carriers away from the junction. As a result, the depletion layer becomes wider and the effective barrier increases. The connection blocks majority-carrier current under ordinary reverse-bias conditions.
203. Under reverse bias, the depletion region of a \(p\)-\(n\) junction:
ⓐ. widens as majority carriers are pulled from the junction
ⓑ. narrows because majority carriers are pushed into the junction
ⓒ. disappears because fixed ions become neutral everywhere
ⓓ. becomes a metallic conductor instantly
Correct Answer: widens as majority carriers are pulled from the junction
Explanation: In reverse bias, the \(p\)-side is connected to the negative terminal and the \(n\)-side to the positive terminal. This pulls holes in the \(p\)-region and electrons in the \(n\)-region away from the junction. More uncovered fixed ions are exposed near the junction, so the depletion region becomes wider. The increased depletion width raises the effective barrier against majority carriers. Reverse bias therefore strengthens the blocking condition of the diode.
204. A reverse-biased diode allows only a small current before breakdown mainly because this current is due to:
ⓐ. majority carriers crossing freely
ⓑ. fixed ions drifting through the entire crystal
ⓒ. protons emitted from the \(p\)-side
ⓓ. minority carriers
Correct Answer: minority carriers
Explanation: Reverse bias blocks the motion of majority carriers across the junction by increasing the barrier and widening the depletion region. However, minority carriers are still present due to thermal generation. The internal field can sweep these minority carriers across the junction, producing a small reverse current. This current is called reverse saturation current over a range of reverse voltage. Its size is strongly affected by temperature because minority carrier concentration depends on thermal generation.
205. In reverse bias, the effective barrier potential of a \(p\)-\(n\) junction:
ⓐ. decreases
ⓑ. becomes exactly zero
ⓒ. increases
ⓓ. changes into resistance measured in \(\Omega\,m\)
Correct Answer: increases
Explanation: Reverse bias applies the external voltage in the same sense as the built-in barrier for majority carriers. This increases the effective potential barrier across the junction. Majority carriers are therefore pulled away from the junction and find it harder to cross. The depletion layer widens at the same time. The barrier potential remains a voltage-related junction idea, not a resistivity measured in \(\Omega\,m\).
206. A diode has \(p\)-side at \(0\,V\) and \(n\)-side at \(+6\,V\). The diode is:
ⓐ. reverse biased
ⓑ. forward biased
ⓒ. unbiased
ⓓ. short-circuited in forward conduction
Correct Answer: reverse biased
Explanation: The \(p\)-side is at a lower potential, while the \(n\)-side is at a higher potential. This means the \(p\)-side is effectively connected to the negative side relative to the \(n\)-side. That is the reverse-bias condition for a \(p\)-\(n\) junction diode. Reverse bias increases the barrier and widens the depletion region. The diode therefore blocks majority-carrier current except for a small minority-carrier reverse current.
207. Study the comparison table for diode biasing.
| Row | Bias | Barrier | Depletion width | Main carrier effect |
|---|
| P | Forward | Reduced | Narrower | Majority carriers cross more easily |
| Q | Reverse | Increased | Wider | Majority carriers are blocked |
| R | Forward | Increased | Wider | Majority carriers are blocked |
| S | Reverse | Increased | Wider | Small minority-carrier current may flow |
The row that is not suitable is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: Row R gives the reverse-bias behaviour but labels it as forward bias. In forward bias, the barrier is reduced and the depletion width becomes narrower. Majority carriers can cross the junction more easily, producing large current after the knee region. Rows Q and S correctly describe reverse bias, where the barrier increases and the depletion layer widens. The main contrast is that forward bias assists majority carriers, while reverse bias blocks them.
208. Consider the following statements about reverse bias.
I. The \(p\)-side is connected to the negative terminal.
II. The depletion region widens.
III. The reverse current before breakdown is mainly due to majority carriers.
The valid statements are:
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is valid because reverse bias connects the \(p\)-side to the negative terminal and the \(n\)-side to the positive terminal. Statement II is valid because this polarity pulls majority carriers away from the junction and widens the depletion layer. Statement III is not valid because majority carriers are blocked in reverse bias. The small reverse current before breakdown is mainly due to minority carriers. This is why reverse current is sensitive to temperature.
209. A reverse-biased diode has a small current even when the reverse voltage is increased moderately. This small current is called reverse saturation current because it:
ⓐ. increases linearly like an ohmic resistor for all reverse voltages
ⓑ. is produced mainly by fixed donor ions drifting through the lattice
ⓒ. stays nearly constant before breakdown
ⓓ. becomes zero whenever temperature is raised
Correct Answer: stays nearly constant before breakdown
Explanation: In reverse bias, majority carriers are pulled away from the junction and are blocked by the widened depletion region. A small current still flows because thermally generated minority carriers can be swept across the junction by the electric field. Once these available minority carriers are collected, increasing reverse voltage moderately does not increase the current very much. This gives an approximately saturated reverse current before breakdown. The word saturation here refers to limited minority-carrier supply, not to absence of carrier motion.
210. The reverse saturation current of a diode increases strongly with temperature mainly because:
ⓐ. more minority carriers are thermally generated
ⓑ. fixed ions become mobile majority carriers
ⓒ. the diode symbol changes direction
ⓓ. resistance of every semiconductor becomes infinite
Correct Answer: more minority carriers are thermally generated
Explanation: Reverse current before breakdown is mainly due to minority carriers. Minority carriers are produced by thermal generation of electron-hole pairs. When temperature increases, the number of thermally generated minority carriers rises. The reverse-bias electric field can then sweep more of these carriers across the junction. This is why reverse saturation current is much more temperature-sensitive than an ideal fixed leakage current.
211. A silicon diode is reverse biased by \(5\,V\), and its reverse saturation current is \(2\,\mu A\). If the reverse voltage is changed to \(8\,V\) while staying far from breakdown, the current is most likely closest to:
ⓐ. \(0\,\mu A\)
ⓑ. \(2\,\mu A\)
ⓒ. \(5\,\mu A\)
ⓓ. \(8\,\mu A\)
Correct Answer: \(2\,\mu A\)
Explanation: \( \textbf{Given condition:} \) The diode is reverse biased and remains far from breakdown.
\( \textbf{Initial reverse current:} \) \(I_s=2\,\mu A\).
\( \textbf{Voltage change:} \) Reverse voltage changes from \(5\,V\) to \(8\,V\).
\( \textbf{Relevant behaviour:} \) Reverse saturation current is nearly constant over a moderate reverse-voltage range.
\( \textbf{Reason:} \) The current is limited mainly by the supply of thermally generated minority carriers.
\( \textbf{Not an ohmic relation:} \) The current is not expected to become \(\frac{8}{5}\) times larger simply because reverse voltage is increased.
\( \textbf{Breakdown condition:} \) The question states that the diode is far from breakdown, so sudden current rise is not considered.
\( \textbf{Final answer:} \) The reverse current remains closest to \(2\,\mu A\).
212. A graph is described for a reverse-biased diode.
The horizontal axis is reverse voltage magnitude, and the vertical axis is reverse current magnitude. The curve remains almost flat at a small current for some range, then rises sharply at a much larger reverse voltage.
The nearly flat part of the graph represents:
ⓐ. forward current due to majority carriers
ⓑ. minority-carrier reverse saturation current
ⓒ. ohmic conduction through a metal wire
ⓓ. complete absence of thermally generated carriers
Correct Answer: minority-carrier reverse saturation current
Explanation: In reverse bias, majority carriers are blocked by the widened depletion region. The small current that remains is mainly due to minority carriers swept across the junction. Since the supply of minority carriers is limited, the current changes only slightly with moderate reverse voltage. This creates the nearly flat reverse-current part of the graph. The later sharp rise belongs to breakdown, not ordinary reverse saturation.
213. A diode under reverse bias suddenly carries a very large current when the reverse voltage is made high enough. The most suitable description is:
ⓐ. ordinary forward conduction has occurred
ⓑ. the diode has become unbiased
ⓒ. breakdown has occurred
ⓓ. fixed ions have become the only mobile carriers
Correct Answer: breakdown has occurred
Explanation: In ordinary reverse bias, a diode allows only a small reverse saturation current. When the reverse voltage becomes very large, the junction may enter breakdown. In breakdown, reverse current rises sharply because new carrier-generation processes become significant. This does not mean the diode is forward biased. A current-limiting arrangement is usually needed because the reverse current can otherwise become damaging.
214. Assertion: A reverse-biased diode has only a small current before breakdown.
Reason: Reverse bias widens the depletion region and blocks majority carriers, while the remaining current is mainly due to minority carriers.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is true because a reverse-biased diode normally carries only a small current before breakdown. The Reason is also true because reverse bias pulls majority carriers away from the junction and widens the depletion region. This makes majority-carrier crossing very difficult. The small current that remains is produced mainly by minority carriers swept across by the junction field. The Reason directly connects the reverse-bias condition with the small-current behaviour.
215. A claim says: “Reverse bias blocks a diode because no carriers of any kind exist in the semiconductor.” The best evaluation is:
ⓐ. The claim is suitable because reverse bias destroys all electron-hole pairs permanently
ⓑ. The claim is suitable only for germanium and never for silicon
ⓒ. The claim is unsuitable because reverse bias always gives large majority-carrier current
ⓓ. The claim is unsuitable; minority carriers cause a small reverse current.
Correct Answer: The claim is unsuitable; minority carriers cause a small reverse current.
Explanation: Reverse bias blocks the flow of majority carriers across the junction. It does not remove every mobile carrier from the semiconductor. Thermally generated minority carriers still exist in both \(p\)-type and \(n\)-type regions. The reverse-bias field can sweep these minority carriers across the junction, producing a small reverse current. Blocking in reverse bias therefore means strong suppression of majority-carrier current, not complete absence of all carriers.
216. Study the reverse-bias observations.
| Row | Observation | Interpretation |
|---|
| P | \(p\)-side connected to negative terminal | Reverse bias |
| Q | Depletion region becomes wider | Majority carriers are pulled away from junction |
| R | Small current before breakdown | Minority-carrier current |
| S | Small current before breakdown | Large majority-carrier injection |
The row that is not suitable is:
ⓐ. Row P
ⓑ. Row S
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: Row S is not suitable because large majority-carrier injection is a forward-bias effect. In reverse bias, majority carriers are pulled away from the junction and are blocked by the increased barrier. Rows P, Q, and R describe reverse bias properly. The small reverse current before breakdown is mainly due to minority carriers. The table separates the cause of reverse leakage from the cause of large forward current.
217. A diode symbol is used in a circuit diagram. The terminal connected to the \(p\)-type side is called the:
ⓐ. anode
ⓑ. cathode
ⓒ. collector
ⓓ. base
Correct Answer: anode
Explanation: In a \(p\)-\(n\) junction diode, the \(p\)-type side is called the anode. The \(n\)-type side is called the cathode. These terminal names are used when deciding forward or reverse bias in a circuit. Collector and base are transistor terminals, not diode terminals. Remembering anode as the \(p\)-side helps identify diode polarity from a circuit diagram.
218. In a \(p\)-\(n\) junction diode, the cathode is connected to:
ⓐ. the \(p\)-type side
ⓑ. the \(n\)-type side
ⓒ. the base region
ⓓ. the collector region
Correct Answer: the \(n\)-type side
Explanation: The cathode of a diode is the terminal connected to the \(n\)-type side. The anode is connected to the \(p\)-type side. These names are not the same as transistor terminal names. In forward conduction, conventional current is taken from anode to cathode. The cathode label therefore helps decide whether the \(n\)-side is connected to the proper polarity for forward or reverse bias.
219. A circuit has the anode of a diode connected to \(+3\,V\) and the cathode connected to \(0\,V\). The diode is:
ⓐ. reverse biased
ⓑ. forward biased
ⓒ. unbiased
ⓓ. in reverse breakdown necessarily
Correct Answer: forward biased
Explanation: The anode of a diode is the \(p\)-side, and the cathode is the \(n\)-side. Here, the anode is at a higher potential than the cathode. This is the forward-bias polarity for a \(p\)-\(n\) junction diode. Forward bias reduces the barrier and narrows the depletion region. Breakdown is not implied simply because a diode is connected in forward bias.
220. A diode has its cathode at \(+10\,V\) and its anode at \(+2\,V\). The bias condition is:
ⓐ. forward bias
ⓑ. zero bias
ⓒ. impossible to decide because both voltages are positive
ⓓ. reverse bias
Correct Answer: reverse bias
Explanation: Bias depends on the relative potentials of the anode and cathode, not on whether both values are positive relative to ground. The anode is at \(+2\,V\), while the cathode is at \(+10\,V\). Thus, the cathode is at a higher potential than the anode. Since the \(p\)-side is at a lower potential than the \(n\)-side, the diode is reverse biased. The potential difference across the diode is what decides the bias state.
