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Electronics II Lecture 3(b): Transistor Bias Circuits A/Lectr. Khalid Shakir Dept. Of Electrical Engineering College of Engineering Maysan University Electronic Devices, 9th edition Thomas L. Floyd Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 1-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014 Linear Operation Bias establishes the operating point (Q-point) of a transistor amplifier; the ac signal I (mA) moves above and below Load line this point. C For this example, the dc base current is 300 A. When the input causes the base current to vary between 200 A and 400 A, the collector current varies between 20 mA and 40 mA. Electronic Devices, 9th edition Thomas L. Floyd Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University Ic 40 ICQ Ib A Q 0 400 µ A 40 30 µ A= I 300 B 20 BQ 200 20 µ A 0 1.2 3.4 VCEQ VCE (V) 5 .6 Vce 2-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014 Waveform Distortion A signal that swings outside the active region will be clipped. For example, the bias has established a low Qpoint. IC Input signal ICQ As a result, the signal will be clipped because it is too close to cutoff. Q Cuto f VCC 0 VCE Cuto f Vce VCEQ Electronic Devices, 9th edition Thomas L. Floyd Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 3-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014 Waveform Distortion High Q- point. The signal will be clipped because it is too close to saturation. Electronic Devices, 9th edition Thomas L. Floyd Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University Input signal too large. The signal will be clipped from both sides. 4-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014 Q. A signal that swings outside the active area will be a. clamped b. clipped c. unstable d. all of the above Electronic Devices, 9th edition Copyright @2013 by Dept. of Electrical Engineering, Thomas L. Floyd College of Engineering, Maysan University 5-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rightsElectronics reserved. II- Lecture 3(b)/1st Semester 013/014 Voltage-Divider Bias A practical way to establish a Q-point is to form a voltagedivider from VCC. R1 and R2 are selected to establish VB. If the divider is stiff, IB is small compared to I2. Then, +VCC RC I2 R2 Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 6-11 RE © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014 Q. A stiff voltage divider is one in which a. there is no load current b. divider current is small compared to load current c. the load is connected directly to the source voltage d. loading effects can be ignored © 2012 Pearson Education. Upper Saddle River, NJ, 07458. Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 7-11 Electronics II- Lecture 3(b)/1st Semester 013/014 Voltage-Divider Bias +VCC +15 V R1 27 k βDC = 200 Determine the base voltage for the circuit. R2 V CC VB R1 R2 12k 15V 4.62 V 27 k 12 k RC 1.2 k R2 12 k RE 680 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 8-11 Electronics II- Lecture 3(b)/1st Semester 013/014 Voltage-Divider Bias +VCC +15 V What is the emitter voltage, VE, and current, IE? R1 27 k VE is one diode drop less than VB: 4.62 V VE = 4.62 V – 0.7 V = 3.92 V RC 1.2 k βDC = 200 3.92 V Applying Ohm’s law: R2 12 k RE 680 VE 3.92V IE 5.76 mA RE 680 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 9-11 Electronics II- Lecture 3(b)/1st Semester 013/014 Q. Assuming a stiff voltage-divider for the circuit shown, the emitter voltage is +VCC +15 V a. 4.3 V R1 20 k b. 5.7 V RC 1.8 k βDC = 200 c. 6.8 V d. 9.3 V R2 10 k RE 1.2 k VB = 10 / (10 + 20) * 15 = 5 V VE = 5 – 0.7 = 4.3 V Electronic Devices, 9th edition Thomas L. Floyd Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 10-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. st Electronics II- Lecture 3(b)/1 Semester 013/014 Q. For the circuit shown, the dc load line will intersect the y-axis at (neglecting IB) +VCC +15 V a. 5.0 mA R1 20 k b. 10.0 mA RC 1.8 k βDC = 200 c. 15.0 mA d. none of the above R2 10 k RE 1.2 k @ VCE=0: IC = (15 – 0) / (1.8k + 1.2k) = 15 / 3k = 5 mA Copyright @2013 by Dept. of Electrical Engineering, College of Engineering, Maysan University 11-11 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved. Electronics II- Lecture 3(b)/1st Semester 013/014