Sample Exam 2

The capacitance of a varactor diode is dependant on the voltage applied to the diode. Therefore it acts like a variable capacitor.

When you connect a low resistance (few µΩ) parallel with ( as shunt) a micro ammeter or milli ammeter, it will be converted to an ameter. More than 99% of current bypass through the shunt. If you connect a large resistor (several kΩ ) in series it will be converted to a Voltmeter.

V = I x R, I = 10mA = 0.01 A, R = 100 Ohm, therefore V = 0.01 x 100 = 1 V

The D layer is formed soon after sunrise and disappears at sunset and absorbs low frequencies, roughly below 4 MHz therefor during times of darkness low frequencies go through and are not absorbed by the D-layer and is refracted by the F layer for long distant propagation.

Also use µW, mW, kW and MW for Power, Ohm is the unit of Resistance, Ampere is the unit of current. Volt is the unit of Potential difference or voltage.

During day time D, E, F₁ and F₂ layers are present. At night D and E dissapears and F₁ and F₂ merged as one single F layer.

V = I x R, I = 200mA = 0.2 A, R = 25 Ohm therefore V = 0.2 x 25 = 5 V

This is act as a parallel resonance LRC circuit. at resonance impedance = maximum . Almost infinity. Practically this is use as traps in multi band antenna system.

P = V² / R , therefore V² = P R = 4 x 25 = 100, therefore V = 10

Class-A output is full cycle and Class-B out put is a half cycle. Therefore Class-AB output is more than a half cycle.

For parallel combination of capacitors use C = C₁ + C₂ + C₃, and for series combination use 1/C = 1/C₁ + 1/C₂ + 1/C₃

The figure shown a PNP transistor. 1 = emitter, 2 = base, 3 = collector. (If the arrow head in opposite direction it is NPN.)

In forward bias P-N junction, current flows from P to N and electrons flow from N to P. In the diode symbol arrow head shows the direction of current.

Consider the theory of loaded dipole.

wave length (m) x frequency (MHz) = 300 given frequency = 7100 kHz = 7.1 MHz, therefore wave length = 300 / 7.1 = 42 m, half wave length = 20m (approx)

Consider the formula X = 2 π f L, When "f" or "L" increases "X" is increasing.

P = V I and you must remember these two also. P = V² / R and P = I² R

:- Small transistorized receiver draws a d.c. current of about 100mA. Therefore you can use a d.c. milli-ammeter .For smaller currents we can use a micro-ammeter and for larger currents we can use an ammeter. For a.c. current we have to use an a.c. ammeter or Clip-on meter. Power meter use to measure the power in Watts. Electro static volt meter is use to measure d.c. high voltages.

If there are resistors and inductors in a circuit, we can define the Power Factor = true power / Apperent power . This is always between 0 and 1. For a pure resistance, PF =1, for a pure inductance, PF = 0

Power Factor defined as (true power) / (Apparent power) ; If φ is the phase angle between voltage and current in an inductor then, power factor = cos φ

In sky wave propagation signals can arrive at the receiving antenna in more than one angle of refraction. The phase difference of the multiple signals create fading. Diversity reception using two rexeivers can overcome this to a great extent.

Equivalent capacitance of two, 1C parallel capacitors is 2C. That 2C and the other 2C in the circuit (left upper) are in series. The effectiv capacitance of these two is equal to 1C because they are in series. This 1C and the other 1C in the circuit (lower side 1C) are in parallel and it is equal to 2C.

When increase the selectivty , the bandwidth is decreases. The result is high selectivity has a narrow bandwith and low selectivity has a wide bandwith. Tuning range is not depend on the bandwidth.

In HF the sky wave is refracted off the ionosphere. There is also radiation along the ground, which is attinuated as by the ground, this is more during the day while at night it could travel quite a distance

In super Heterodyne System, Local Oscillator signal is always higher than the incoming signal and the difference between them is equal to IF. For normal broadcast receivers IF is 455 kHz.

:- A single diode is used for half-wave rectifiers. There are two rectifier systems for Full wave Rectifiers. 1. Diode bridge or bridge rectifier made out of 4 diodes. 2, Two diodes and a center tapped transformer is used. Rectifier vacuum diode tubes were used by old timers.

V = I x R, therefore I = V / R, V = 40 , R = 1000/2 = 500 therefore I = 40 / 500 = 0.08 A = 80 mA

Formula for Attenuation or gain for Voltage is 20 log(V₁ / V₂) = 20 log(10/1) = 20 dB; formula for Power :- 10 log ( P₁ / P₂)

P = V x I, V = 300Volt, I = 30mA = 0.03 A, P = 300 x 0.03 = 9 W