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Note for Basic Electrical Engineering - BEE By Suman Kumar Acharya

  • Basic Electrical Engineering - BEE
  • Note
  • Biju Patnaik University of Technology BPUT - BPUT
  • Electrical and Electronics Engineering
  • 1 Topics
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Suman Kumar Acharya
Suman Kumar Acharya
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Figure 9.2 Electrostatic deflection Let, Ea = voltage of pre-accelerating anode; (volt) e = charge of an electron; (Coulomb) m = mass of electron; (kg) θ = deflection angle of the electron beam vox = velocity of electron when entering the field of deflecting plates; (m/s) Ed = potential difference between deflecting plates; (volt) d = distance between deflecting plates; (m) ld = length of deflecting plates; (m) L = distance between screen and the centre of the deflecting plates; (m) y = displacement of the electron beam from the horizontal axis at time t and D = deflection of the electron beam on the screen in Y direction; (m) The loss of potential energy (PE ) when the electron moves from cathode to accelerating anode; The gain in kinetic energy (KE ) by an electron This is the velocity of the electron in the X direction when it enters the deflecting plates. The velocity in the X direction remains same throughout the passage of electrons through the deflecting plates as there is no force acting in the direction. Suppose ay is the acceleration of the electron in the Y direction, therefore, 3

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As there is no initial velocity in the Y direction [Eq. (9.8)], the displacement y at any instant t in the Y direction is As the velocity in the X direction is constant, the displacement in X direction is given by Substituting the above value of t in Eq. (9.8), we have This is the equation of a parabola. Putting x = ld in Eq. (9.12), we get the value of tan θ. After leaving the deflection plates, the electrons travel in a straight line. The straight line of travel of electron is tangent to the parabola at x = ld and this tangent intersects the X axis at point O’. The location of this point is given by The apparent origin is thus the centre of the deflecting plates, the deflection D on the screen is given by From Eq. (9.16) we conclude the following: For a given accelerating voltage Ea, and for particular dimensions of CRT, the deflection of the electron beam is directly proportional to the deflecting voltage. This means that the CRT may be used as a linear indicating device. The discussions above assume that Ed is a fixed dc voltage. The deflection voltage is usually a time varying quantity and the image on the screen thus follows the variation of the deflections voltage in a linear manner. The deflection is independent of the (e/m) ratio. In a cathode ray tube, in addition to the electrons many types of negative ions such as oxygen, carbon, chlorine etc are present. With electrostatic deflection system, because deflection is independent of e/m, the ions 4

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travel with the electrons and are not concentrated at one point. Hence cathode ray tube with electrostatic deflection system does not produce an ion burn. The deflection sensitivity of a CRT is defined as the deflection of the screen per unit deflection voltage. The deflection factor of a CRT is defined as the reciprocal of sensitivity It is clear from Eq. (9.17), that the sensitivity can be increased by decreasing the value of accelerating voltage Ea. but this has a disadvantage as the luminosity of the spot is decreased with decrease in Ea. On the other hand a high value of Ea, produced a highly accelerated beam and thus produces a bright spot. However, a high accelerating voltage (Ea) requires a high deflection potential (Ed) for a given deflection. Also, highly accelerated beam is more difficult to deflect and is sometimes called hard beam. Example 9.1 An electrically deflected CRT has a final anode voltage of 2000 V and parallel deflecting plates 1.5 cm long and 5 mm apart. If the screen is 50 cm from the centre of deflecting plates, find (a) beam speed, (b) the deflection sensitivity of the tube, and (c) the deflection factor of the tube. Solution Velocity of the beam Deflection sensitivity, Example 9.2 Calculate the maximum velocity of the beam of electrons in a CRT having a anode voltage of 800 V. Assume that the electrons to leave the anode with zero velocity. Charge of electron = 1.6 × 10−19 C and mass of electron = 9.1 × 10−31 kg. Solution Velocity of electron is 5

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Example 9.3 A CRT has an anode voltage of 2000 V and 2 cm long and 5 mm apart parallel deflecting plates. The screen is 30 cm from the centre of the plates. Find the input voltage required to deflect the beam through 3 cm. The input voltage is applied to the deflecting plates through amplifiers having an overall gain of 100. Solution TIME BASE GENERATOR 9.4 Generally, oscilloscopes are used to display a waveform that varies as a function of time. For the waveform to be accurately reproduced, the beam must have a constant horizontal velocity. Since the beam velocity is a function of the deflecting voltage, the deflecting voltage must increase linearly with time. A voltage with this characteristic is called a ramp voltage. If the voltage decreases rapidly to zero with the waveform repeatedly reproduced, as shown in Figure 9.3, the pattern is generally called a sawtooth waveform. During the sweep time, Ts, the beam moves from left to right across the CRT screen. The beam is deflected to the right by the increasing amplitude of the ramp voltage and the fact that the positive voltage attracts the negative electrons. During the retrace time or flyback time, Tr, the beam returns quickly to the left side of the screen. This action would cause a retrace line to be printed on the CRT screen. To overcome this problem the control grid is generally ‘gated off’, which blanks out the beam during retrace time and prevents an undesirable retrace pattern from appearing on the screen. Figure 9.3 Typical sawtooth waveform applied to the horizontal deflection plates In low-cost oscilloscopes the time base is said to be free running, although the time base oscillator may, in fact, be synchronised to the vertical amplifier signal. Unless the time base is so synchronous, the waveform marches across the screen and remains unstable. 6

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