Diode Circuits and Applications Questions and Answers

Answer: A

The output voltage is 3 V.

The transducer is a potentiometer: 5 V is spread uniformly along the resistance element between A and B, and the wiper picks off a fraction of it in proportion to its position. The output is taken between B and the wiper (via C). With the wiper 40 mm from A on a 100 mm stroke, it is 60 mm from B:

\[ V_{0} = \frac{60}{100} \times 5\ \text{V} = 3\ \text{V} \]

The total resistance of 10 kΩ does not enter as long as the output is not loaded, because only the ratio of the two parts matters.

2 V is the voltage between A and the wiper (40% of 5 V), measured from the wrong end. 2.5 V is the mid-stroke value, and 1.5 V has no basis.

Answer: B

The line regulation is 45%.

Line regulation measures how much the load voltage moves when the line voltage swings from its low to its high value. With the definition used in standard electronics texts, the change in load voltage is expressed as a percentage of the low-line load voltage:

\[ \begin{aligned} \text{LR} &= \frac{V_{HL} - V_{LL}}{V_{LL}} \times 100\% \\ &= \frac{12.8 - 8.8}{8.8} \times 100\% \approx 45\% \end{aligned} \]

Such a large figure is typical of an unregulated rectifier, whose output simply follows the line.

22.5% is half the value, and 4.5% and 2.25% are ten times smaller; none of them follows from the 4 V change on an 8.8 V base.

Answer: B

The DC output voltage is 16 V.

The transformer steps 240 V rms down by 20 : 1, and the capacitor charges to the peak of the secondary voltage less one diode drop:

\[ \begin{aligned} V_{2,\text{rms}} &= 240/20 = 12\ \text{V} \\ V_{2,\text{peak}} &= 12\sqrt{2} \approx 16.97\ \text{V} \\ V_{dc} &\approx 16.97 - 0.7 \approx 16.3\ \text{V} \end{aligned} \]

With 500 mF across 1 kΩ the ripple is negligible (the time constant is 500 s against a 20 ms cycle), so the output sits at about 16 V.

17 V is the ideal-diode peak, which ignores the diode drop. 12 V is the rms value, and 18 V exceeds the peak of the secondary altogether.

Answer: A

The barrier potential of a silicon diode is about 0.7 V.

The built-in potential across the depletion region of a silicon PN junction at room temperature is about 0.6-0.7 V. So a silicon diode conducts noticeably only once the forward bias reaches roughly that cut-in voltage.

0.3 V is the figure for germanium. 1.1 V (more exactly 1.12 eV) is the band-gap energy of silicon, not its barrier potential. 0.1 V is too low for any common semiconductor junction.

Answer: B

The reverse saturation current increases with temperature.

The reverse saturation current comes from minority carriers generated thermally near the junction. Their number rises steeply with temperature, so the current roughly doubles for every 10 degrees Celsius rise in silicon and germanium diodes. This is why diodes and transistors are prone to thermal drift.

It therefore does not decrease with temperature and is not independent of it. By definition it is the current under reverse bias, so it does not depend on the forward voltage.

Answer: A

The matching is a-i, b-iii, c-iv, d-ii. A diode is used for rectification. A tunnel diode shows negative resistance and is used in oscillators. A Zener diode keeps a steady voltage in reverse breakdown, so it serves as a voltage regulator. A PIN diode has a wide intrinsic region and works as a switch at microwave frequencies. Option (B) gives the tunnel diode as a regulator and the Zener as an oscillator. Option (C) gives the Zener diode as the oscillator. Option (D) gives the tunnel diode as a regulator, the Zener diode as a microwave switch and the PIN diode as an oscillator. Only (A) is right for all four.