Diode Circuits and Applications Questions and Answers

Answer: A

Series connection divides the reverse voltage among the devices so that none exceeds its rating, which is how rectifiers for tens of kilovolts are built. Forward drops add rather than reduce, current capacity is unchanged, and recovery time is not improved. Devices in series share voltage and devices in parallel share current, and knowing which problem is being solved settles which connection to use. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: B

Because leakage currents differ from device to device, equalising resistors across each diode are needed or the one with least leakage takes most of the voltage. Series capacitors would block the direct current entirely, inductors do nothing for static sharing, and a single series resistor limits current without equalising anything. Any series string of nominally identical devices needs help to share, because the least leaky device otherwise takes the largest share and fails first.

Answer: C

Parallel connection shares the forward current, allowing a higher total than one device could carry. Voltage sharing needs a series connection, the forward drop is unchanged by paralleling, and the breakdown voltage is that of a single device. The complication is that a diode's forward drop falls as it heats, so the hottest device takes more current - a positive feedback that small series resistors are used to break. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: D

The device with the lowest drop takes the most current, heats, drops less still, and runs away until it fails. Leakage and recovery time are not the concern, and the group's forward drop is that of the lowest device rather than higher. Any device whose conduction improves as it heats needs a ballast element in series when paralleled, which is why the same precaution appears with transistors and with light emitting diodes. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: A

The resistor takes the surplus voltage at the wanted current:

\[ R = \frac{5-2}{0.02} = 150 \ \Omega \]

so 150 ohms, dissipating sixty milliwatts. 250 ohms would follow from a 2 V surplus and 100 ohms from taking the whole five volts across the resistor at a higher current. Subtracting the device's own drop before dividing is the one step this calculation turns on, and forgetting it is the commonest error.

Answer: B

Three drops must be subtracted from the supply:

\[ R = \frac{12 - 6}{0.02} = 300 \ \Omega \]

giving 300 ohms. The wrong answers count one drop or none. Driving several such devices in series from one resistor is far better than giving each its own, because they then carry identical currents and match in brightness - one of the small practical points that separates a working design from a sketch.

Answer: C

Dividing the power rating by the operating voltage:

\[ I = \frac{0.5}{10} = 50 \ \mathrm{mA} \]

so 50 mA, and a prudent design would run well below it since the rating assumes a stated ambient temperature. The other options misplace a decade. A power rating converted into a current limit at the actual operating voltage is the form the designer needs, and doing that conversion first avoids a great deal of confusion later.

Answer: D

The resistor must pass the full load current plus the keep-alive current:

\[ R \le \frac{20-10}{0.035} = 286 \ \Omega \]

so 286 ohms at most. 333 ohms would pass only thirty milliamps and starve the diode at full load. Two hundred ohms works but dissipates more, and four hundred is far too large. Sizing the resistor from the worst-case corner rather than the nominal condition is what makes a regulator work at both extremes.

Answer: A

All the resistor's current now passes through the diode:

\[ \begin{aligned} I &= \tfrac{10}{286} = 35 \ \mathrm{mA} \\ P &= 10 \times 0.035 = 350 \ \mathrm{mW} \end{aligned} \]

so about 350 milliwatts, which sets the diode's minimum rating. This is the second worst-case corner, and it is the one that decides the device rather than the resistor - which is why both corners have to be checked before anything is ordered.

Answer: B

The arrangement draws roughly constant current whatever the load, so at light load almost all of it is wasted in the diode. The resistor's drop is roughly constant rather than proportional to load, output impedance is low and does not rise that way, and the temperature effect is minor. A regulator that discards surplus current rather than declining to draw it is inherently wasteful, which is why series and switching regulators displaced it wherever efficiency matters.