analogue circuits for university students engineering, Lecture notes of Engineering

Download analogue circuits for university students engineering and more Lecture notes Engineering in PDF only on Docsity! Chapter 2: Diode Applications 1 2.1 Introduction The analysis of electronic circuits can follow one of two paths: 1.using the actual characteristic 2.applying an approximate model for the device. The use of appropriate approximate model can avoid an unnecessary level of mathematical complexity. The result obtained using a series of approximation are lightly different from those using actual characteristics. These tolerance contribute to the general belief that a response determined through a set of approximations can often be “as accurate” as one that employs the full characteristics. Ideal Equivalent Circuit ’p + YD Ideal characteristics Ty. zy ee D \ 10mA 9 OS \ O———_e—> e—_—__o \ — \ 1 \ 'p r \ \ (a) -20v | > 07 Vv \ Vp . / \ On-off Switch o—o0 += o——o oN \ leom Actual charac < =OmA Ip Mahe s. s (GO) r24e7 (b) Conduction in one direction + Vp - ee <a Ip ~~ Iaeat diode Simplified Equivalent Circuit (rsgere Ideal characteristics 10 mA # Ip alle ee 0.7V Actual characteys6tics Ideal diode 0 Ve=07V Vy silicon semiconductor diode Piecewise-Linear Equivalent Circuit 7 Because rav is small enough to be ignored, the popular models are simplified equivalent circuit and ideal circuit. Not Perpendicular Fig. 2.3 (a) Circuit; (b) characteristics. 2.2 Load-Line Analysis Example2.1 Employing the diode characteristics of Fig. 2.3b, determine: a.VDQ and IDQ. b.VR. 2.2 Load-Line Analysis Example2.1 Employing the diode characteristics of Fig. 2.3b, determine: a.VDQ and IDQ. b.VR. Fig. 2.4 Solution to Example 2.1.   Actual result: VDQ =0.78V， IDQ=18.5mA 2.2 Load-Line Analysis Example2.2 Repeat Example2.1,using the approximate equivalent model for silicon semiconductor diode to determine: a.VDQ and IDQ. b.VR. Fig. 2.5 Solution to Example 2.1 using the diode approximate model.   Actual result: VDQ =0.78V， IDQ=18.5mA 2.3 Equivalent Model Analysis Equivalent model Analysis steps: 2.Replace the diode using equivalent model For ideal equivalent model 3. Determine the state of circuits • VD = 0V • VR = E • ID = IR =E / R E<0E>0 • VD = E • VR = 0V • ID = IR =0 A Diodes ideally behave as open circuits Diodes ideally behave as short circuits 2.3 Equivalent Model Analysis Equivalent model Analysis steps: 1. Determine the state of each diode based on the equivalent model. For simplified equivalent model: If the direction of applied voltage is a “match” with the arrow in the diode symbol, and applied voltage is greater than the “turn-on” voltage (VK) of each diode, conduction through the diode will occur and the device in the “on” state, otherwise, diode is in the “off” state. E>0.7 or =0.7 match E<0.7 not match 2.3 Equivalent Model Analysis Equivalent model Analysis steps: 2.Replace the diode using equivalent model For simplified equivalent model 3. Determine the state of circuits • VD = 0.7V • VR = E-0.7V • ID = IR =(E-0.7) / R E<0.7V • VD = E • VR = 0V • ID = IR =0 A E>0.7V 2.4 AND/OR Gates Example 2.6 Determine Vo for the network of Fig2.19 Fig. 2.19 Positive logic OR gate. Vo =10-0.7=9.3V 2.4 AND/OR Gates Example 2.7 Determine Vo for the network of Fig2.22 Fig. 2.22 Positive logic AND gate. Exercise 1 Determine Vo for these networks as follows: D2 ON， D1 OFF VO＝ 5V 。 D1ON， D2OFF， VO=0V 。 D1ON, D2OFF VO=1.98V 。 D1 D2 D2 D1 D2 VO VO VO D1E1 E2 E1 2.5 Half-Wave Rectification The diode only conducts when it is in forward bias, therefore only half of the AC cycle passes through the diode. •Using an ideal diode equivalent where Vm = the peak AC voltage.       ttdVV mdc sin Average value of output voltage: m m V. V 3180  The process of removing one-half the input signal to establish a dc level is called half-wave rectification. • Using a silicon diode equivalent If Vm>>VK, the DC output voltage is 0.318(Vm- VK ) 2.5 Half-Wave Rectification • Fig. 2.30 Resulting vo for the circuit of Example 2.8. The DC output voltage is -0.318Vm=- 0.318(20V)=-6.36V Use the ideal diode model Example 2.8 Sketch the output vo and determine the dc level of output for the network 2.6 Full-Wave Rectification Bridge Rectifier • Four diodes are required • VDC = 0.636 Vm 31 Clippers are networks that employ diodes to “clip” away a portion of an input signal without distort the remaining part of the applied waveform. 2.7 Clippers  There are two general categories of clippers: series and parallel. Square waveform Triangular waveform The half-wave rectifier is an example of the simplest form of diode clipper---a resistor and a diode. 32 2.7 Clippers  Fig. 2.49 Series clipper with a dc supply. Vi -V>0, i.e. Vi >V, D is in “on” state Vo=Vi-V Vi <V or =V, D is in “off” state Vo=0 Ideal equivalent model 35 Using simplified equivalent model for silicon ① Vi > 2.7V, D is in “on” state ， ②V i ≤ 2.7V时 , D is in “off” state。 E2V R Vi Vo VK 2.7 Clippers  E2V Si R Vi Vo E2V R Vi Vo VK t Vi/V 0 -5 5 2.7 Vo=E+VK=2.7 V Vo = V i 36 t0 Vo/V 2.7 -5 t Vi/V 0 -5 5 Using simplified equivalent model ① Vi > 2.7V, D is in “on” state ，Vo=E+0.7=2.7 V ②V i ≤ 2.7V时 , D is in “off” state， Vo = V i 。 2.7 E2V D R Vi Vo 2.7 Clippers  2.7 Clippers eager Simple Series Clippers (Ideal Diodes) POSITIVE NEGATIVE 2.9 Zener Diodes Zener diode is usually used as an regulator to ensure that output voltage of a supply remains fairly constant. 2.9 Zener Diodes • When Vi  Vz • The Zener is on • Voltage across the Zener is Vz • Zener current: IZ • The Zener Power: PZ = VZIZ • When VK<Vi < Vz • The Zener is off • The Zener acts as an open circuit 2.9 Zener Diodes Basic Zener regulator Zener diode is usually used as an regulator to ensure that output voltage of a supply remains fairly constant. RL is fixed ， Vi↑ →Iz↑ →keep Vo unchanged VO Vi may have ripple wave. RL may have some fluctuations. Example 2.17. (a)Determine VL,VR,IZ,and PZ. (b) Repeat part(a) with RL=3k Fixed Vi, Fixed RL Solution: (b) VV RR R V i L L 12   VVV ZL 10 VVVV LiR 6 mA.IZ 672 mW.IVP ZZZ 726 mA. k V R V I L L L 3333 10    mA k V R V I RR 61 6    3k Fixed Vi, Variable RL Example 2.17. (a)Determine the range of RL and IL that will result in VRL being maintained at 10V. (b) Determine the maximum wattage rating of diode. Solution: Due to Vz, there is a specific range of resistor values RL(and therefore load current IZ) that will ensure that the Zener is in the “on” state. i L L ZL VRR R VV            250 1050 101 VV Vk VV RV R Zi Z minL Fixed Vi, Variable RL Example 2.17. (a)Determine the range of RL and IL that will result in VRL being maintained at 10V. (b) Determine the maximum wattage rating of diode. Solution: VVVV ZiR 401050   k. mA V I V R minL Z maxL 2518 10 mA k V R V I RR 401 40    mAmAmAIII ZMRminL 83240  (b) mWmAV IVP ZMZmaxZ 3203210    k.RL 251250 mAImA L 408  mA V R V I minL Z maxL 40250 10   