1 / 20

ANALOG ELECTRONIC CIRCUITS 1

ANALOG ELECTRONIC CIRCUITS 1. EKT 204 Basic BJT Amplifiers (Part 2). Basic Common-Emitter Amplifier. The basic common-emitter circuit used in previous analysis causes a serious defect : If BJT with V BE =0.7 V is used, I B =9.5 μA & I C =0.95 mA

jacie
Télécharger la présentation

ANALOG ELECTRONIC CIRCUITS 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ANALOG ELECTRONIC CIRCUITS 1 EKT 204 Basic BJT Amplifiers (Part 2)

  2. Basic Common-Emitter Amplifier • The basic common-emitter circuit used in previous analysis causes a serious defect : • If BJT with VBE=0.7 V is used, IB=9.5 μA & IC=0.95 mA • But, if new BJT with VBE=0.6 V is used, IB=26 μA & BJT goes into saturation; which is not acceptable  Previous circuit is not practical • So, the emitter resistor is included: Q-point is stabilized against variations in β, as will the voltage gain, AV • Assumptions • CC acts as a short circuit • Early voltage = ∞ ==> ro neglected due to open circuit

  3. Common-Emitter Amplifier with Emitter Resistor inside transistor CE amplifier with emitter resistor Small-signal equivalent circuit (with current gain parameter, β)

  4. Common-Emitter Amplifier with Emitter Resistor • ac output voltage • Input voltage loop • Input resistance, Rib • Input resistance to amplifier, Ri • Voltage divider equation of Vin to Vs Remember: Assume VA is infinite,  ro is neglected

  5. Common-Emitter Amplifier with Emitter Resistor Cont.. • So, small-signal voltage gain, AV • If Ri >> Rs and (1 + β)RE >> rπ Remember: Assume VA is infinite,  ro is neglected

  6. VCC RC R1 vO RS CC vs R2 RE CE RS Vo Vs R1|| R2 r gmV ro RC Common-Emitter Amplifier with Emitter Bypass Capacitor Emitter bypass capacitor, CE provides a short circuit to ground for the ac signals Emitter bypass capacitor is used to short out a portion or all of emitter resistance by the ac signal. Hence no RE appear in the hybrid-π equivalent circuit Small-signal hybrid-π equivalent circuit

  7. DC & AC LOAD LINE ANALYSIS • DC load line • Visualized the relationship between Q-point & transistor characteristics • AC load line • Visualized the relationship between small-signal response & transistor characteristics • Occurs when capacitors added in transistor circuit

  8. Common Emitter Amplifier with emitter bypass capacitor Example 1 Common-emitter amplifier with emitter bypass capacitor

  9. DC Load Line Solution... • KVL on C-E loop

  10. AC Load Line Solution... • KVL on C-E loop AC equivalent circuit

  11. DC & AC Load Lines Full solution

  12. AC LOAD LINE ANALYSIS Determine the dc and ac load line. VBE=0.7V, β=150, VA=∞ Example 2

  13. DC Load Line • To determine dc Q-point, KVL around B-E loop

  14. AC Load Line Small signal hybrid-π equivalent circuit

  15. DC & AC Load lines Full solution

  16. Maximum Symmetrical Swing • When symmetrical sinusoidal signal applied to the input of an amplifier, the output generated is also a symmetrical sinusoidal signal • AC load line is used to determine maximum output symmetrical swing • If output is out of limit, portion of the output signal will be clipped & signal distortion will occur

  17. Maximum Symmetrical Swing • Steps to design a BJT amplifier for maximum symmetrical swing: • Write DC load line equation (relates of ICQ & VCEQ) • Write AC load line equation (relates ic, vce ; vce = - icReq, Req = effective ac resistance in C-E circuit) • Generally, ic = ICQ – IC(min), where IC(min) = 0 or some other specified min collector current • Generally, vce = VCEQ – VCE(min), where VCE(min) is some specified min C-E voltage • Combination of the above equations produce optimum ICQ & VCEQ values to obtain maximum symmetrical swing in output signal

  18. Maximum Symmetrical Swing Example 3 Determine the maximum symmetrical swing in the output voltage of the circuit given in Example 2. Solution: • From the dc & ac load line, the maximum negative swing in the Ic is from 0.894 mA to zero (ICQ). So, the maximum possible peak-to-peak ac collector current: • The max. symmetrical peak-to-peak output voltage: • Maximum instantaneous collector current:

  19. Self-Reading • Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007 • Chapter 6: Basic BJT Amplifiers • Page: 397-413, 415-424.

  20. Exercise • Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007 • Exercise 6.5, 6.6, 6.7,6.9 • Exercise 6.10 , 6.11

More Related