Chapter Monostable Multivibrators Solution uuu Solution u u u uuu - PDF document

Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. Monostable Multivibrators 1. A monostable multivibrator is used as a voltage-to-time converter. Find the time 5.0Time period 5.01ln1001.01010 = 0.11 ms 9.09 kHz.0.1110 Design a collector-coupled monostable multivibrator using an the monostable multivibrator 100.29.8 V5 mA5 mACCCE © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. (sat)2minmin22min5 mA0.125 mA1.51.50.1250.187 mA100.70.18710  T = 0.69RC 1100.6949.701029.1 nF0.6949.7 For the value of (cut-off) Fig.2.1 Circuit to calculate VBB (sat)also0.22BBBBBBBBBBBIfRRR To find R1 = 2 = © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. (sat)15 mA0.5 mA10100.20.20.5 mACEBB A collector-coupled monostable multivibtransistors has the following parameters: BB = 4 = 0.001 (sat) waveforms. Fig.8p.3 Circuit of the monostable table state: assume is ON and in saturation, Fig.3.1. t in the stable state and whether 2min © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. (sat)2minmin22min90.18.9 V4.45 mA2 K2 K90.38.7 V=0.87 mA10 K10 K4.45 mA0.11 mACCCE is in saturation, as assumed. The voltage at the base of 1(sat)121220100.1(9)102010200.06632.934 VBCEBBVVV R  In the stable state VC1 = 9 V, VC2 = 0.1 V VB1 = –2.934 V, VB2 = 0.3 V VA = 8.7 V In the quasi-stable state (on the application of a trigger): 1 is ON and in saturation Q2 is OFF, Fig.3.2. is really ON and in saturation? To verify this, calculate 1min or not. If this is satisfied, © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. (sat)1minmin90.14.45 mA108.98.71.76 mA4.451.766.21 mACCCE I RIRVIII 6.21 mA=0.155 mA Fig.3.3 Circuit to calculate IB1 at t = 0+ © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. 12311min90.32100.725 mA0.390.465 mA0.7250.4650.26 mAIII, in the quasi-stable state, is in saturation. 0.1 V,0.3 V9(0.725)(2)9(1.76)(10)917.6CCCCBCCRVVIRVVIR Alternately (sat)0.390.1CCCEVVIRVVV changes as a function of time. = –8.6 V and decays exponentially. © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. goes into the ON state, Fig.3.4. Fig.3.4 Equivalent circuit at t = T+ (sat)90.10.30.10.22CCCEbbCVVVV  3.95 mA22'9(3.95)(2)(3.95)(0.2)0.3CCCBBBbbVVIRVIrV At the end of the quasi-stable state =1.1 V and jumps to The waveforms are now plotted in Fig.3.5. © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. Fig. 3.5 Waveforms d monostable multivibrator circuit shown in Fig. 8.1, = 20. In the quasi-stable state, in the active region with collector current of 2 mA. Find the time period and the value oB(min) The circuit, when is ON in the quasi-stable state, is given in Fig. 8p.4. ON in the Quasi-stable state. © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. I2= 0.909 mA.110VVVRRRR BBBBThe collector current of 1(min)0.1 mA. From Fig 8p.4, 0.1 mA = 0.909 mA 809.01.0909.0 10210110lnlnln10100.01100.18218.2 s.CCCCCCCCVIRVVIRVVV An emitter-coupled monostable multivibrator in Fig.8p.2 has the following parameters: silicon transistors with =100 are used. A trigger is applied at Assume that is ON at is indeed OFF and is in saturation. Assume that is in the active region and –. +. Fig.8p.2 Emitter-coupled monostable multivibrator © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. (a) At = 0– (i.e. in the stable state) are indicated in Fig. 5.1. Fig. 5.1 Circuit to calculate the currents and voltages in the stable states Writing the KVL equations 2min(min)22(min)222112Henceisinsaturation(0.051)31.0533.15 V2.83.15ENENBCEBEBNENVVIIRVVV is negative with respect to emitter by 0.35 V. As such is in the OFF 22(sat)3.150.23.35 V 3.150.73.85 VCNCCENENCNENCEBNENVVVVVV Voltage across the capacitor terminals = 6 V – 3.85 V = 2.15 V. of the quasi-stable state) © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. = 0, a trigger is applied to drive into the active region and into the OFF state. it in the quasi-stable state = 2.8 – 0.6 = 2.2 V 11112.2 V0.73 mA(1)(1)0.73 mA()0.7 mA0.7 mA50332.15 V0.7 mA3502.15 VCBCCIIII I IIIIIV (1)(2)Solving for © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G. 1110.08 mA,0.65 mA6(0.65)361.954.05 V6(0.08)(50)CNCCCENENBNCCRVVIRVVIR(c) Just prior to the completion of the quasi-stable state ( 62.7 V3.3 V50 K50 K0.066 mACCBNVVT1110.70.0660.634 mA6(0.634)5CNCCCIIIVVIR remain unaltered. Voltage across the capacitor at is in the active region. once again goes into the OFF state and goes into ON state and into saturation. As a . These overshoots are accounted for by taking the base spreading resistance into © Dorling Kindersley India Pvt. Ltd 2010 Pulse and Digital Circuits Venkata Rao K., Rama Sudha K. and Manmadha Rao G.© Dorling Kindersley India Pvt. Ltd 2010 Writing the KVL equations of the two loops (sat)22(6.13)CCABCCCCEBCVVVIIVVII'''22222(sat)'''222'3.94 V6.13and5.8V360.226 mA,0.854 mA()(0.2260.854)3(1.08)33.24 V3.240.23.44 V3.240.7(0.226)(0.1)3.9ENBCECNENCEBNENBbbVIIRVVVVVVIrCNCCThe waveforms can now be plotted