EEAE0042: Analog Electronics Module I PDF

Presented by Dr. Jyoti Kumar Barman Assistant Professor ADBUSOT August 28, 2024 EEAE0042 EEE, ADBUSOT JB 1  Note: The information provided in the slides are taken form class notes of S. Le Gof...

Presented by Dr. Jyoti Kumar Barman Assistant Professor ADBUSOT August 28, 2024 EEAE0042 EEE, ADBUSOT JB 1  Note: The information provided in the slides are taken form class notes of S. Le Goff , School of Electrical and Electronic Engineering , Newcastle University, textbook “Basic Electrical and Electronics Engineering” by D P Kothari & I J Nagrath and various other resources from internet, for teaching/academic use only August 28, 2024 EEAE0042 EEE, ADBUSOT JB 2  Module I (4 lectures) P-N junction diode I-V characteristics of a diode Review of half-wave and full-wave rectifiers Zener diodes clamping and clipping circuits. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 3  P-N junction diode A practical PN junction can only be created by inserting different impurities into different parts of a single crystal. At junction, the extra electrons in the N region will combine with the extra holes in the P region. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 4  P-N junction diode This leaves an area where there are no mobile charges, known as depletion region, around the junction. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 5  P-N junction diode: Application of voltage - Reverse Bias Positive voltage is applied to the N- type material, while negative voltage is applied to the P-type. The positive voltage applied to the N- type material attracts free electrons towards the end of the crystal and away from the junction August 28, 2024 EEAE0042 EEE, ADBUSOT JB 6  P-N junction diode: Application of voltage - Reverse Bias The negative voltage applied to the P- type end attracts holes away from the junction. The result is that all available current carriers are attracted further away from the junction, and the depletion region grows correspondingly larger. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 7  P-N junction diode: Application of voltage - Reverse Bias Therefore, there is no current flow through the crystal because no current carriers can cross the junction. This is known as reverse bias applied to the semiconductor crystal. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 8  P-N junction diode: Application of voltage - Forward bias The negative voltage applied to the N-type end pushes electrons towards the junction The positive voltage at the P-type end pushes holes towards the junction. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 9  P-N junction diode: Application of voltage - Forward bias This has the effect of shrinking the depletion region. At some point, the applied voltage V has become large enough to make the depletion region completely disappear (V becomes equal to the threshold voltage Vd ) Vd  0.7 volt for silicon and Vd  0.3 volt for germanium August 28, 2024 EEAE0042 EEE, ADBUSOT JB 10  P-N junction diode: Application of voltage - Forward bias Current carriers of both types are able to cross the junction into the opposite ends of the crystal. Electrons in the P-type end are attracted to the positive applied voltage, while holes in the N-type end are attracted to the negative applied voltage. This August is the 28, 2024 condition of forward bias. EEAE0042 EEE, ADBUSOT JB 11  Takeaway 1 The conclusion is that an electrical current can flow through the junction in the forward direction, but not in the reverse direction. This is the basic property of a semiconductor diode. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 12  Note 1 It is important to realize that holes exist only within the crystal. A hole reaching the negative terminal of the crystal is filled by an electron from the power source and simply disappears. At the positive terminal, the power supply attracts an electron out of the crystal, leaving a hole behind to move through the crystal toward the junction again. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 13  I-V characteristics of a diode The Shockley diode equation, named after transistor co-inventor William Shockley, gives the current–voltage characteristic of a diode in either forward or reverse bias. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 14  I-V characteristics of a diode qV V ( ) ( ) kT VT The equation is given by I  I s {e  1}  I s {e  1} where , Is: Saturation current of the diode (in the range 10-8 to 10-16 A, typically); : Emission coefficient. Also called as Ideality factor. This is an empirical constant that varies from 1 to 2 depending on the fabrication process and semiconductor material and in many cases is assumed to be approximately equal to 1 (and thus omitted). Note:  = 1 for Ge and  = 2 for Si for low current, below the knee of the curve and  = 1 for both Ge and Si for higher level of current beyond the knee q: Electron charge (= 1.602 x 10-19 C); T: Temperature in degrees Kelvin; k: Boltzmann’s 10-23 J/K); EEE, August 28, 2024 constant (= 1.38 X EEAE0042 VT:ADBUSOT Thermal voltage JB ( 26 mV at room temp.15)  I-V characteristics of a diode qV V ( ) ( ) kT VT I  I s {e  1}  I s {e  1} August 28, 2024 EEAE0042 EEE, ADBUSOT JB 16  I-V characteristics of a diode This rather complicated equation is a bit difficult to use for manual circuit analysis. Engineers deal with this problem by simplifying things and using the much simpler model of the diode Vd is called threshold voltage or forward voltage drop of the diode. Vd  0.7 volt for silicon and Vd  0.3 volt for germanium, Vd  0.25 V for a Schottky August 28, 2024 diode. EEAE0042 EEE, ADBUSOT JB 17  Example 1 1. An Si diode has Is = 10 nA operating at 25°C. Calculate ID for a forward bias of 0.6 V. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 18  Equivalent Circuit of Diode: Piecewise Linear Model Ideal diode Piecewise linear model Approximate model August 28, 2024 EEAE0042 EEE, ADBUSOT JB 19  Dynamic Resistance of Diode It is the resistance offered by the diode to the flow of AC through it when we connect it in a circuit which has an AC voltage source as an active circuit element. vac dv Mathematically rac   iac di VT rac  From the V-I characteristics, rac is iD reciprocal of the slope of the curve 1 rac  di dv August 28, 2024 EEAE0042 EEE, ADBUSOT JB 20  Example 2 2. For the diode circuits of Fig. , find the value of I. Use approximate model of the diode. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 21  Example 2 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 22  Rectifier Circuits A dc power supply is required to bias all electronic circuits. A diode rectifier forms the first stage of a dc power supply. Rectification is the process of converting an alternating (ac) voltage into one that is limited to one polarity. Rectification is classified as half-wave or full-wave rectifier. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 23  Half-Wave Rectifier The primary function of a rectifier circuit is to change an AC input voltage into a voltage that is only positive or only negative. In essence, a rectifier eliminates the unwanted polarity of the input waveform. As an illustration, consider the circuit below called half- wave rectifier. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 24  Half-Wave Rectifier Ideal case: The output DC voltage of a half wave rectifier can be calculated with the following two ideal equations Vm Vm Vrms  Vdc  2  August 28, 2024 EEAE0042 EEE, ADBUSOT JB 25  Half-Wave Rectifier  1 V Vdc  [Vm  sin(t )d (t )]  m  0.318Vm 2 0  Vm 1 I dc   0.318 I m  RL August 28, 2024 EEAE0042 EEE, ADBUSOT JB 26  Half-Wave Rectifier DC Power supply performance measures: – DC output voltage – DC output current – Voltage regulation – Ripple factor / Form factor – Power conversion efficiency August 28, 2024 EEAE0042 EEE, ADBUSOT JB 27  Half-Wave Rectifier Voltage regulation – It is a measure of the variation of DC output voltage as a function of DC output current i.e. variation in load. V  VFL % regulation = NL 100% VFL VNL= Voltage across load resistance when minimum current flows through it. VFL = Voltage across load resistance when maximum current flows through it. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 28  Half-Wave Rectifier Ripple factor : Ripple factor is defined as rms value of the ac component of load (output) voltage  dc component of load voltage August 28, 2024 EEAE0042 EEE, ADBUSOT JB 29  Half-Wave Rectifier We find that ripple factor of a half-wave rectifier is quite high, which is unacceptable August 28, 2024 EEAE0042 EEE, ADBUSOT JB 30  Half-Wave Rectifier Form factor= rms value/ average Efficiency (): The ratio of dc output value Vm power to ac input power is known as  2  1.57 rectifier efficiency (). Vm  Peak factor= peak value/ rms value Vm  2 Vm 2 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 31  Half-Wave Rectifier: Some Parameters Relationship between the number of turns of a step-down transformer and the input/output voltages vp N1  vs N2 The peak inverse voltage (PIV) of the diode is the peak value of the voltage that a diode can withstand when it is reversed biased Duty Cycle: The fraction of the wave cycle over which the diode is conducting. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 32  Half-Wave Rectifier: Diode with Threshold voltage Vs< V, diode off, open circuit, no current flows, Vo = 0V Vs> V, diode conducts, current flows, Vo = Vs – V i v0  id R  vs  v vs  v id  R vD V August 28, 2024 EEAE0042 EEE, ADBUSOT JB 33  Half-Wave Rectifier: Diode with Threshold voltage vs  v p sin(t ) Vp V Vs >V Rectified output August 28, 2024 EEAE0042 EEE, ADBUSOT JB 34  Example 3 3. A half-wave rectifier, having a resistive load of 1000 Ω rectifies an alternating voltage of 325 V peak value and the diode has a forward resistance of 100 Ω. Calculate a. peak, average and rms value of current b. dc power output c. ac input power d. efficiency of the rectifier. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 35  Example 3 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 36  Full-wave Rectifier It converts an ac voltage into a pulsating dc voltage using both half cycles of the applied ac voltage.  1 2Vm Vdc  [Vm  sin( t ) d (t )]   0   1 1 V Vrms  [Vm  sin (t ) d (t )] 2  m 2 2  0 2 2Vm 1 I dc   RL August 28, 2024 EEAE0042 EEE, ADBUSOT JB 37  Full-wave Rectifier Vm 1  [ 2  1] 2  0.482 Ripple factor 2Vm  Efficiency () = The maximum efficiency of a full-wave rectifier is 81.2%. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 38  Full-wave Rectifier Form factor = rms value/ average value Vm  2  1.11 2Vm  Peak factor= peak value/ rms value Vm   2 Vm 2 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 39  Example 4 4. A 230 V, 60 Hz voltage is applied to the primary of a 5 : 1 step-down, center-tap transformer used in a full wave rectifier having a load of 900 Ω. If the diode resistance and secondary coil resistance together has a resistance of 100 Ω, determine a) dc voltage across the load b) dc current flowing through the c) dc power delivered to the load d) PIV across each diode e) Ripple voltage and its frequency. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 40  Example 4 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 41  Bridge Rectifier The need for a center tapped transformer in a full-wave rectifier is eliminated in the bridge rectifier The average values of output voltage and load current for bridge rectifier are the same as for a center-tapped full wave rectifier. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 42  Advantages of the bridge rectifier The ripple factor and efficiency of the rectification are the same as for the full-wave rectifier. The PIV across either of the non-conducting diodes is equal to the peak value of the transformer secondary voltage, Vm The bulky center tapped transformer is not required. Transformer utilisation factor is considerably high. The bridge rectifiers are used in applications allowing floating output terminals, i.e. no output terminal is grounded. PIV rating required for the diodes in a bridge rectifier is only half of that for a center tapped full-wave rectifier. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 43  Comparison of Rectifiers August 28, 2024 EEAE0042 EEE, ADBUSOT JB 44  Capacitor Filter An inexpensive filter for light loads is found in the capacitor filter which is connected directly across the load August 28, 2024 EEAE0042 EEE, ADBUSOT JB 45  Capacitor Filter: Analysis Charge acquired: Vr , pp C How to calculate the ripple factor ? Hint: If the time T2 is equal to half the periodic time , the ripple waveform I dc  T2 will be triangular in nature Charge lost: Applying principle of conservation of charge : Vr , pp C  I dc  T2 If the value of the capacitor is fairly large, or the value of the load resistance is very large, then it can be assumed that the time T2 is equal to half the periodic time of the waveform. T2=T/2=1/2f, then we have 1 C  I dc  2 fVr , pp August 28, 2024 EEAE0042 EEE, ADBUSOT JB 46  Zener diodes Avalanche Breakdown Zener Breakdown As the applied reverse bias increases, thermally When the P and N regions are heavily doped, generated carriers while traversing the junction direct rupture of covalent bonds takes place acquire a large amount of kinetic energy. because of the strong electric fields, at the These electrons disrupt covalent bonds by junction of PN diode creates new electron-hole colliding with immobile ions and create new pairs electron-hole pairs. As a result of heavy doping of P and N regions, These new carriers again acquire sufficient the depletion region width becomes very small energy from the field and collide with other and for an applied voltage of 6 V or less, the immobile ions thereby generating further field across the depletion region becomes very electron–hole pairs. high, of the order of 107 V/m. This cumulative process results in flow of large This results in breakdown of the junction and amount of current at the same value of reverse the phenomenon is known as Zener breakdown. bias and results in Avalanche breakdown Avalanche effect is predominant in case of lightly doped diodes. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 47  Zener diodes:V-I Characteristics of Zener Diode When the Zener diode is connected, in forward bias, diode acts as a normal diode. But Zener breakdown voltage occurs when the reverse bias voltage is greater than a predetermined voltage. At the knee voltage, once the diode breaks down, the voltage Vz remains nearly constant. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 48  Zener diodes August 28, 2024 EEAE0042 EEE, ADBUSOT JB 49  Application of Zener Diode Zener diode as voltage regulator Zener diode in over-voltage protection Zener diode in clipping circuits August 28, 2024 EEAE0042 EEE, ADBUSOT JB 50  Example 5 The circuit of Fig. has a zener diode connected across the load. a) For RL = 180 Ω, determine all currents and voltages. b) Repeat part (a) for RL = 450 Ω. c) Find the value of RL for the zener to draw maximum power. d) Find the minimum value of RL for the zener to be just in on-state. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 51  Example 5 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 52  Example 5 August 28, 2024 EEAE0042 EEE, ADBUSOT JB 53  Diode clippers Positive clipper Negative clipper Biased clipper and Combination clipper August 28, 2024 EEAE0042 EEE, ADBUSOT JB 54  Positive clipper Input (a) Series (b) Shunt clippers Output August 28, 2024 EEAE0042 EEE, ADBUSOT JB 55  Negative clipper Input (a) Series (b) Shunt clippers Output August 28, 2024 EEAE0042 EEE, ADBUSOT JB 56  Biased clipper: positive clipper Biased Clipper (a) Series, (b) Shunt, and (c) Input and output signal waveforms August 28, 2024 EEAE0042 EEE, ADBUSOT JB 57 Biased clipper: positive clipper (Series) In the biased series positive clipper as shown in Fig. the diode does not conduct as long as the input voltage is greater than +VR and hence, the output remains at +VR. When the input voltage becomes less than +VR, the diode conduct and acts as a short circuit. Hence, all the input signal having less than +VR as well as negative half cycle of the input wave will appear at the output. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 58  Biased clipper: positive clipper (Shunt) In the biased shunt positive clipper as shown in Fig. the diode conducts as long as the input voltage is greater than +VR and the output remains at +VR until the input voltage becomes less than +VR. When the input voltage is less than +VR, the diode does not conduct and acts as an open switch. Hence all the input signal having less than + VR as well as negative half cycle of the input wave will appear at the output. The clipping level can be shifted up or down by varying the bias voltage VR. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 59  Biased positive clipper with reverse polarity of the battery August 28, 2024 EEAE0042 EEE, ADBUSOT JB 60  Biased negative clipper Biased negative clipper (a) Series, (b) Shunt, and (c) Input and output signal waveforms August 28, 2024 EEAE0042 EEE, ADBUSOT JB 61  Biased negative clipper with reverse polarity of the battery Biased negative clipper with reverse polarity of VR (a) Series, (b) Shunt, and (c) Input and output signal waveforms August 28, 2024 EEAE0042 EEE, ADBUSOT JB 62  Combination clipper August 28, 2024 EEAE0042 EEE, ADBUSOT JB 63  Combination clipper Combination of a biased positive clipper and a biased negative clipper. When the input signal voltage Vi ≥ + VR1, diode D1 conducts and acts as a closed switch, while diode D2 is reverse biased and D2 acts as an open switch. Hence, the output voltage cannot exceed the voltage level of + VR1 during the positive half cycle. When the input signal voltage Vi ≤ - VR2 diode D2 conducts and acts as a closed switch, while diode D1 is reverse biased and D1 acts as an open switch. Hence the output voltage Vo cannot go below the voltage level of –VR2 during the negative half cycle. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 64  Two level slicer ?? August 28, 2024 EEAE0042 EEE, ADBUSOT JB 65  Example 6 Determine Vo for the network of Fig. for the waveform given. Assume the diodes are ideal. Sol: When the input voltage Vi ≥ 5V, the diode D1 conducts when Vi ≤ –3V, diode D2 conducts. Hence, the positive half cycle of the input voltage is clipped at +5V and the negative half cycle is clipped at –3V, as shown in Fig. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 66  CLAMPERS Clamping network shifts (clamps) a signal to a different dc level, i.e. it introduces a dc level to an ac signal. The clamping network is also known as dc restorer and clamping may be in positive as well as in the negative side of the input waveform. These circuits find application in television receivers to restore the dc reference signal to the video signal. The clamping network has the various circuit components like a diode, a capacitor and a resistor. The time constant for the circuit τ = RC must be large so that the voltage across the capacitor does not discharge significantly when the diode is not conducting. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 67  Example 7 Determine Vo for the clamping circuit shown in Fig. for the given sinusoidal input signal. Sol: During the positive half of the input signal, the diode conducts and acts like a short circuit, Vo = 0 V. The capacitor is charged to 12V with polarities shown and it behaves like a battery. During the negative half of the input signal, the diode does not conduct and acts like an open circuit. Hence, the output voltage, Vo = –12 V –12 V = –24 V. This gives negatively clamped voltage and the resultant output waveform is shown in fig. August 28, 2024 EEAE0042 EEE, ADBUSOT JB 68  Thank You Happy Learning August 28, 2024 EEAE0042 EEE, ADBUSOT JB 69