Two Questions on Zener Diode Voltage Regulators

“It is unwise to be too sure of one’s own wisdom. It is healthy to be reminded that the strongest might weaken and the wisest might err”

– Mahatma Gandhi

Zener diodes, as you know, are widely used as reference voltage elements in a variety of voltage regulator circuits. Today I give you a couple of multiple choice questions involving the use of zener diodes in simple voltage regulators:

(1) In the circuit shown, D₁ is a silicon diode which has a voltage drop of 0.7 V while in full conduction under forward bias. The zener diode D₂ has a breakdown voltage of 6.8 V. What is the current through the 450 Ω resistor?

(a) 1 mA

(b) 10 mA

(c) 20 mA

(d) 100 mA

(e) 0 mA

You can use the voltage drop across a forward biased ordinary silicon diode as a reference voltage in voltage regulator circuits. In the circuit shown, a reverse biased zener diode (of breakdown voltage 6.8 V) and a forward biased silicon diode (of voltage drop 0.7 V) in series make a reference voltage of 7.5 V. The output voltage of the circuit is thus 7.5 V.

Since the input voltage to the regulator circuit is 12 V, the voltage drop across the 450 Ω resistor is 12 V – 7.5 V = 4.5 V.

The current through the 450 Ω resistor is (4.5 V)/(450 Ω) = 0.01 A = 10 mA.

(2) The circuit shown in the adjoining figure is the simplest shunt voltage regulator using a zener diode of breakdown voltage 6 V. What is the power dissipated in the zener diode?

(a) 100 mW

(b) 120 mW

(c) 240 mW

(d) 360 mW

(e) 480 mW

Since the input voltage to the regulator circuit is 10 V and the regulated output voltage is 6 V, the voltage drop across the 40 Ω resistor is 10 V – 6 V = 4 V.

Therefore, the current through the 40 Ω resistor (current limiting resistor) is (4 V)/ 40 Ω = 0.1 A = 100 mA. This is the total current flowing into the parallel combination of the zener diode and the100 Ω load resistor.

The current through the load resistor of 100 Ω is (6 V)/(100 Ω) = 0.06 A = 60 mA.

Therefore, the current flowing through the zener diode is 100 mA – 60 mA = 40 mA.

Power dissipated in the zener diode is 6 V×40 mA = 240 mW.

Kerala Medical Entrance 2008 (KEAM 2008) Questions on Communication Systems

The following multiple choice questions on communication systems appeared in the Kerala Medical Entrance 2008 (KEAM-2008) examination question paper:

(1) A 1000 kHz carrier wave is modulated by an audio signal of frequency range 100-5000 Hz. Then the width of the channel in kHz is

(a)10

(b) 20

(c) 30

(d) 40

(e) 50

Let us assume that the system uses amplitude modulation of the usual double sideband type. (It should have been mentioned in the question)

The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.

[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].

(2) If the critical frequency for sky wave propagation is 12 MHz, then the maximum electron density in the ionosphere is

(a) 1.78×10¹²/m³

(b) 0.178×10¹⁰/m³

(c) 1.12×10¹²/m³

(d) 0.56×10¹²/m³

(e) 0.148×10¹²/m³

The critical frequency f_c for reflection by the ionosphere is given by

f_c = 9 N^1/2 where N is the maximum electron number density.

Therefore, N = f_c²/81 = (12×10⁶)² /81 = 1.78×10¹²/m³

Communication Systems -Two Questions Involving Line of Sight Space Wave Communication

Questions on communication systems at Class 12 level are often simple but occasionally you may get questions which are a little bit confusing and time consuming. I give below a couple of practice questions for you:

(1) A television transmitting antenna is mounted at a height of 100 m. For satisfactory line of sight communication at a distance of 40 km, what should be the minimum height of the receiving antenna? (Radius of the earth = 6400 km).

(a) 20 m

(b) 25 m

(c) 30 m

(d) 35 m

(e) 40 m

If h_t and h_r represent the heights of the transmitting antenna and the receiving antenna respectively, the maximum separation (d) between them for satisfactory line of sight communication is given by

d = √(2Rh_t) + √(2Rh_r) where R is the radius of the earth.

Therefore, we have

40 = √(2×6400×0.1) + √(2×6400×h_r)

[Note that we have expressed all distances in kilometre and hence we will obtain the height of the receiving antenna in km].

Squaring, 1600 = 2×6400 (0.1 + h_r)

Therefore, (0.1 + h_r) = 1/8 from which

h_r = 0.025 km = 25 m.

(2) The minimum service area covered by a TV transmitter antenna mounted at a height h is (radius of the earth = R)

(a) πR²h

(b) 2πR²h

(c) πh²R

(d) πRh

(e) 2πRh

The coverage area will be minimum when the receiving antenna is at the ground level. In other words, the height of the receiving antenna is zero.

The distance (d) up to which line of sight communication is possible is therefore given by

d = √(2Rh_t) where h_t is the height of the transmitting antenna.

The minimum coverage area is πR²d =π[√(2Rh_t)]² = 2πRh_t = 2πRh

You can access all posts on communication systems on this site by clicking on the label, ’communication system’ below this post.

Apply for Entrance Examination for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2011 (KEAM 2011), Kerala

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2011-12.

(a) Engineering: B.Tech, B.Tech. (Agricultural Engg.) and B.Tech. (Dairy Sc. & Tech.)

(b) Architecture: B.Arch.

(c) Medical: (i) MBBS (ii) BDS (iii) BAMS (iv) BSMS (v) BHMS

(d) Agriculture and Allied Courses: (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)

(e) Veterinary: BVSc. & AH

Dates of Exam:

Engineering Entrance Examination

18.04.2011 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

19.04.2011 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)

20.04.2011 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

20.04.2011 2 P.M. to 4.30 P.M. Paper-II: Biology.

Application form and Prospectus will be distributed from 19.1.2011(Wednesday) to 14.02.2011 (Monday) through selected Post Offices.

Last Date for receipt of Application by CEE: 14-02-2011 (Monday) - 5 PM

You will find complete details and information updates at http://www.cee-kerala.org/

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To access earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the search button.

Publishing now on Custom Domain

Some of you might have noted that when you search for a page in ‘physicsplus’ at the URL with domain name physicsplus.blogspot.com, you get the required page at physicsplus.in. Please note that I have started publishing using custom domain name physicsplus.in.

So the blog you find is the same and no need to worry!

Two Questions (MCQ) on Photo Diodes

“He who joyfully marches in rank and file has already earned my contempt. He has been given a large brain by mistake, since for him the spinal cord would suffice.”

– Albert Einstein

As you know, photo diodes are special purpose p-n junction diodes. They are very popular photo detectors used for detecting optical signals in communication systems.

Today we will discuss two multiple choice questions on photo diodes:

(1) In using a photo diode as a photo detector, it is invariably reverse biased. Why?

(a) The power consumption is much reduced compared to reverse biased condition

(b) Electron hole pairs can be produced by the incident photons only if the photo diode is reverse biased

(c) Light variations can be converted into current variations only if the photo diode is reverse biased

(d) When photons are incident on the diode, the fractional change in the reverse current is much greater than the fractional change in the forward current

(e) The photo diode will be spoilt if it is operated under forward biased condition

Whether the photo diode is reverse biased or forward biased, the number of electron hole pairs produced by the incident photons is the same. In other words, the change in the diode current is the same in both cases. But in the reverse biased condition the current drawn by the diode in the absence of the photons is extremely small, of the order of nanoamperes or microamperes where as in the forward biased condition this current is significant, of the order of tens of milliamperes. The fractional change in the current because of the incident photons is therefore large and easily measurable if the photo diode is reverse biased. The correct option is (d).

(2) The maximum wave length of photons that can be detected by a photo diode made of a semiconductor of band gap 2 eV is about

(a) 620 nm

(b) 700 nm

(c) 740 nm

(d) 860 nm

(e)1240 nm

The wave length λ (in Angstrom unit) of a photon of energy E (in electron volt) is given by

λE = 12400, very nearly.

Therefore, λ = 12400/E

[The above expression can be easily obtained by remembering that a photon of energy 1 eV has wave length 12400 Ǻ and the energy is inversely proportional to the wave length].

Since E = 2 eV we have λ = 12400/2 = 6200 Ǻ = 620 nanometre.

Photons with wave length greater than 640 nm will have energy less than 2 eV so that they will be unable to produce electron hole pairs in the semiconductor of band gap 2 eV. So the correct option is (a).

“Cheers to a New Year and another chance for us to get it right”

- Oprah Winfrey

“Happy New Year 2011”

Kerala Engineering Entrance Questions (MCQ) on Communication Systems

Questions on communication systems are generally simple and interesting and so you can easily score marks by attempting them. The following three multiple choice questions were included in Kerala Engineering Entrance 2005 question paper and will be able to answer them in less than three minutes:

(1) If a radio receiver is tuned to 855 kHz radio wave, the frequency of local oscillator in kHz is

(a) 1510

(b) 455

(c) 1310

(d) 1500

(e) 855

Since the tuned frequency is 855 kHz, the receiver is an AM (amplitude modulation) receiver. Modern AM receivers are super heterodyne receivers employing a higher frequency local oscillator. On mixing the local oscillator output with the incoming amplitude modulated carrier, an amplitude modulated wave at intermediate frequency (IF) of 455 kHz (by convention) is produced. Since the intermediate frequency is 455 kHz, it follows that the frequency of local oscillator is 855 kHz + 455 kHz = 1310 kHz.

(2) If n₁ and n₂ are the refractive indices of the core and the cladding respectively of an optical fibre,

(a) n₁ = n₂

(b) n₁ < n₂

(c) n₂ < n₁

(d) n₂ = 2n₁

(e) n₂ = √(2n₁)

Since the optical fibre confines the light signal within the fibre by total internal reflection, the refractive index of the cladding should be less than that of the core. Therefore, n₂ < n₁ [Option (c)]

(3) A TV tower has a height of 100 m. What is the maximum distance up to which TV transmission can be received? (Radius of the earth, R = 8×10⁶ m)

(a) 34.77 km

(b) 32.7 km

(c) 40 km

(d) 40.7 km

(e) 42.75 km

We have maximum distance, d ≈ √(2Rh) where h is the height of the antenna.

Substituting given values, d ≈ √(2×8×10⁶ ×100) = 40×10³ m = 40 km.

[The mean radius of the earth is nearly 6400 km. The value is given as 8000 km in the problem to make your calculation simple.

You should remember that the height of the transmitting antenna (or receiving antenna) is the height with respect to the ground level. If an antenna is mounted on a mast of height h₁ and the mast is erected on a hill or building of height h₂, the height of the antenna will be h = h₁ + h₂]

Multiple Choice Questions (MCQ) on Alternating Currents

Most of you might have noted that electric power stations generate alternating current rather than direct current even though the majority of electrical and electronic appliances require direct current. At the power generation stage, alternating current is preferred since the current can be controlled, without power loss, using reactive elements (inductors and capacitors). Transmission of electric power over long distances without appreciable losses is possible by using transformers. The idea, as most of you should know, is to transmit electrical energy at low current and high voltage so that the Joule heating (which is proportional to the square of the current) in the transmission lines is minimized.

You will find some earlier posts on alternating currents on this site. You can access all the posts by trying a search for ‘alternating current’ using the search box provided on this page. Or, you may click on the label ‘alternating current’ below this post.

Let us discuss a few more questions on alternating currents:

(1) A short circuited coil is placed in a time varying magnetic field. Electric power is dissipated due to the current induced in the coil. If the number of turns were quadrupled and the wire radius halved, the electric power dissipated would be

(a) halved

(b) same

(c) doubled

(d) quadrupled

The above question appeared in IIT 2002 screening test paper.

Electric power dissipation is given by P = V²/R. where V is the voltage induced in the coil and R is the resistance of the coil. When the number of turns is quadrupled, the induced voltage V is quadrupled so that V² becomes 16 times. But the resistance of the coil also becomes 16 times since resistance R = ρL /A where ρ is the resistivity (specific resistance), L is the length and A is the area of cross section of the wire. The length becomes 4 times when the number of turns is quadrupled and the area of cross section becomes one-fourth when the radius is halved. The resistance therefore becomes 16 times.
The power dissipated is therefore unchanged [Option (b)].

(2) An electric heater consumes 1000 watts power when connected across a 100 volt D.C. supply. If this heater is to be used with 200V, 50 Hz A.C.supply, the value of the inductance to be connected in series with it is

(a) 5.5 H

(b) 0.55 H

(c) 0.055 H

(d)1.1 H

(e)11 H

The current drawn by the heater is 1000 W/100V = 10 A . When the heater is used with A.C. supply, it will draw 10 A itself. (Note that the current and the voltage values are R.M.S. values when you deal with electric power). If ‘L’ is the inductance required, the expression for the current I is

I = V/√(R² + L²ω²)] where V, R and ω are respectively the alternating voltage, the resistance of the heater and the angular frequency of the A.C.

Substituting, 10 = 200/√[10² + L² (100π)²], since ω = 2πf where f’ is the frequency of the A.C.

Squaring and rearranging, L² = 300/100π)² from which L = √3/10π = 0.055 H.
(3) An alternating emf V = 6 cos1000t is applied across a series LR circuit of 3 mH inductance and 4 Ω resistance. The amplitude of the current is

(a) 0.6 A

(b) 1.2 A

(c) 1.4 A

(d) 1.8 A

Amplitude (maximum value or peak value) of current (I_max) is given by

I_max = V_max/√(R²+L² ω²) = 6/√[4²+(3×10^-3)²×1000²] = 1.2 A

[Note that the values of V_max and ω are obtained from the expression for the emf V which is in the form, V = V_max cos ωt].

Here is a very simple question which you should answer carefully:

(4) The voltage V applied across an A.C. circuit and the current I flowing in it are given by

V = 12 cos ωt volt and I = 20 sin ωt milliampere respectively.

The power dissipated in the circuit is

(a) 120 watt

(b) 120 milliwatt

(c) 240 watt

(d) 249 milliwatt

(e) zero

In alternating current circuits the power is given by P = V_rms I_rms cosΦ where Φ is the phase difference between the applied voltage and the resulting current. Since the voltage is a cosine function and the current is a sine function, the phase difference Φ is π/2. [Note that the voltage can be written as V = 12 sin(ωt + π/2) volt]. Therefore cosΦ (which is called the power factor) is zero. The correct option is (e).

All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) Postponed

The Central Board of Secondary Education has rescheduled the All India Engineering/Architecture Entrance Examination (online/pen & paper) 2011 (AIEEE 2011) from 24-4-2011 to 1-5-2011 in consideration of the Easter festival. The sale of Information Bulletins from all selling centres will start from 22.12.2010 instead of 15.12.2010. Rest of the things/schedule will remain unchanged.

Visit the web sites www.cbse.nic.in and www.aieee.nic.in for details.

Electronics - Multiple Choice Questions on Transistor Amplifiers

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“How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though sometimes sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our own happiness depends.”

– Albert Einstein

Questions on transistor amplifiers at the class 12 (plus two or higher secondary) level are generally simple and interesting even though some among you may have unclear ideas. Today we will discuss a few questions on common emitter transistor amplifiers:

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(1) The adjoining figure shows a common emitter transistor amplifier which uses a silicon transistor. If the quiescent emitter current is 1 mA what is the base biasing voltage?

(a) 4.7 V

(b) 3.7 V

(c) 2.7 V

(d) 1.7 V

(e) 0 V

Because of the emitter current the voltage drop across the 1 KΩ resistor connected to the emitter is 1 V.

[1 mA×1 KΩ = (1/1000) A×1000 Ω = 1 V].

The voltage drop across the base-emitter junction of the silicon transistor is 0.7 V. Therefore, the base voltage under no signal (quiescent) condition is 1 V + 0.7 V = 1.7 V.

(2) In the amplifier circuit shown in Question No.1 what is the function of the capacitor C₁ connected across the 1 KΩ emitter resistor?

(a) To produce positive feed back.

(b) To produce negative feed back.

(c) To pass the excess signal to the ground.

(d) To act as filter capacitor for the transistor supply voltage.

(e) To bypass the signal current so that it will not flow through the emitter resistor.

The capacitor C₁ provides an easy path (bypass) for the signal component of the emitter current. If C₁ is absent the signal component of the emitter current will produce signal voltage drop across the emitter resistor, thereby reducing the signal output at the collector.

The correct option is (e).

(3) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, the quiescent base current of the transistor is very nearly equal to

(a) 1 mA

(b) 1 μA

(c) 2 μA

(d) 4 μA

(e) 5 μA

In the common emitter mode, the current amplification factor (current gain) under no signal condition (β_dc) is given by

β_dc = I_C/I_B where I_C is the collector current and I_B is the base current (both under no signal conditions).

Since the collector current is almost equal to the emitter current I_E (because of large value of β_dc), we have

β_dc ≈ I_E/I_B

Therefore I_B ≈ I_E/β_dc = 1 mA/200 = 0.005 mA = 5 μA.

(4) If the common emitter current gain β_dc of the transistor used in the amplifier circuit shown in Question No.1 is 200, what is the voltage drop across the base biasing resistor R under quiescent conditions?

(a) 12 V

(b) 11 V

(c) 10.3 V

(d) 5.4 V

(e) 4.7 V

The quiescent base current is 5 μA as shown in answering question no.3 above. The base biasing voltage is 1.7 V as shown in answering question no.1. The power supply voltage is 12 V. Therefore, the voltage drop across the base biasing resistor R under quiescent conditions is 12 V – 1.7 V = 10.3 V.

(5) The base biasing resistor in the circuit shown in Question No.1 is

(a) 1 KΩ

(b) 4.7 KΩ

(c) 1.03 MΩ

(d) 1.87 MΩ

(e) 2.06 MΩ

The quiescent base current of the transistor is very nearly equal to 5 μA as shown in answering Question No.3. The voltage drop across the base biasing resistor R under quiescent conditions is 10.3 V as shown in answering Question No.4. Therefore, the base biasing resistor is given by

R = (10.3)V/(5 μA) = 2.06 MΩ

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You can access all questions (with solution) on electronics posted on this site by clicking on the label ‘electronics’ below this post or by trying a search for ‘electronics’ using the search box provided on this page

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