Schuler Electronics Instructor CH07 amplifiers part 2.ppt
- 1. Electronics Principles & Applications Sixth Edition Chapter 7 More About Small-Signal Amplifiers ©2003 Glencoe/McGraw-Hill Charles A. Schuler
- 2. INTRODUCTION • Amplifier Coupling • Voltage Gain • FET Amplifier • Negative Feedback • Frequency Response
- 3. VCC These two points are at different dc voltages. Capacitive coupling is convenient in cascade ac amplifiers.
- 4. VCC Direct coupling is required for dc gain.
- 5. VCC The darlington is a popular dc arrangement.
- 6. VCC P S 10:1 10 W ZRATIO = TRATIO 2 = 102 = 100 ZCOLLECTOR = 100 x 10 W = 1000 W Transformer coupling offers the advantage of impedance matching.
- 7. VCC Transformer coupling can be used in bandpass amplifiers to achieve selectivity. fR Gain
- 8. Amplifier Coupling Quiz Capacitive coupling is not useful for _________ amplifiers. dc Dc frequency response requires ________ coupling. direct Transformer coupling offers the advantage of _________ matching. impedance Tuned transformer coupling provides frequency _____________. selectivity A darlington amplifier is an example of _________ coupling. direct
- 9. RB1 E B C RL VCC RB2 RE = 12 V 2.7 kW 22 kW = 2.2 kW More about solving the practical circuit for its ac conditions: = 220 W Zin = ?
- 10. RB1 E B C RL VCC RB2 RE = 12 V 2.7 kW 22 kW = 2.2 kW Zin is a combination of RB1, RB2, and rin of the transistor. = 220 W rin = b (RE + rE) rin = 150 (220 W + 9.03 W) rin = 34.4 kW Note: rin = brE when RE is bypassed. Determine rin first:
- 11. RB1 E B C RL VCC RB2 RE = 12 V 2.7 kW 22 kW = 2.2 kW = 220 W Zin = 1 RB2 1 rin 1 + RB1 1 + + + Zin = 1 2.7 kW 1 34.4 kW 1 22 kW 1 Zin = 2.25 kW RB1, RB2, and rin act in parallel to load the input signal.
- 12. RB1 VCC RB2 RE = 12 V 2.7 kW 22 kW RL= 2.2 kW = 220 W Load = 2.2 kW What happens when an amplifier is loaded? RL and the Load act in parallel. RP = 1.1 kW
- 13. RB1 RB2 RE VCC = 12 V 2.7 kW 22 kW RL= 2.2 kW = 220 W Load = 2.2 kW There are two saturation currents for a loaded amplifier. RP = 1.1 kW ISAT(DC) = VCC RL + RE = 4.96 mA ISAT(AC) = VCC RP + RE = 9.09 mA
- 14. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 VCE in Volts IC in mA 20 mA 0 mA 100 mA 80 mA 60 mA 40 mA There are two load lines for a loaded amplifier. DC TEMPORARY AC The DC load line connects VCC and ISAT(DC). A temporary AC load line connects VCC and ISAT(AC).
- 15. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 VCE in Volts IC in mA 20 mA 0 mA 100 mA 80 mA 60 mA 40 mA 5.3 V DC AC TEMP. AC The quiescent VCE is projected to the DC load line to establish the Q-point. The AC load line is drawn through the Q-point, parallel to the temporary AC load line.
- 16. 0 2 4 6 8 10 12 14 16 18 2 4 6 8 10 12 14 VCE in Volts IC in mA 20 mA 0 mA 100 mA 80 mA 60 mA 40 mA 5.3 V AC The AC load line shows the limits for VCE and if the Q-point is properly located. With loaded amplifiers, the Q-point is often closer to saturation.
- 17. RB1 RB2 RE VCC = 12 V 2.7 kW 22 kW RL= 2.2 kW = 220 W Load = 2.2 kW What about voltage gain for a loaded amplifier? RP = 1.1 kW AV = RP RE + rE AV = 1.1 kW 220 W + 9.03 W = 4.8
- 18. VCC Zin of the 2nd stage loads the 1st stage. When analyzing cascade amplifiers, remember: 2nd 1st
- 19. Amplifier ac Conditions Quiz Emitter bypassing _________ an amplifier’s input impedance. decreases Loading at the output of an amplifier ________ its voltage gain. decreases A loaded amplifier has two load lines: dc and ___________. ac The clipping points of a loaded amplifier are set by its _______ load line. ac In a cascade amplifier, the Zin of a stage _______ the prior stage. loads
- 20. Drain Source Gate VDD = 20 V VGS = 1.5 V RG CC RL = 5 kW Input signal Common-source JFET amplifier. Fixed bias ISAT = 20 V 5 kW = 4 mA Phase-inverted output
- 21. 0 2 4 1 VDS in Volts ID in mA 5 10 15 20 25 3 -2.5 -2.0 -1.5 -1.0 -0.5 0 N-channel JFET characteristic curves V GS in Volts Load line The Q-point is set by the fixed bias. 8 VP-P 1 VP-P AV = 8
- 22. 0 2 4 1 VDS in Volts ID in mA 5 10 15 20 25 3 -2.5 -2.0 -1.5 -1.0 -0.5 0 Determining forward transfer admittance: Yfs = DID DVGS V GS in Volts VDS 1.6 mA = 1.6 mS
- 23. D S G VDD = 20 V VGS = 1.5 V RG CC RL = 5 kW When the forward transfer admittance is known, the voltage gain can be determined using: AV = Yfs x RL = 1.6 mS x 5 kW = 8 This agrees with the graphic solution.
- 24. D S G VDD VGS = ID x RS RG CC RL RS Source bias eliminates the need for a separate VGS supply. IS = ID This resistor also provides ac negative feedback which decreases the voltage gain.
- 25. Vin - BVout A(Vin - BVout) BVout A = open loop gain Summing junction Vin Vout A B Feedback A negative feedback model B = feedback ratio Vout = A(Vin - BVout) Vout = AVin - ABVout AVin Vout 1 = - AB AVin Vout AB +1 = Vin Vout AB +1 A = Vin Vout AB +1 A = AB +1 A Vin Vout A simplified model
- 26. D S G VDD RG CC RL RS = 5 kW = 800 W The feedback ratio (B) for this circuit is easy to determine since the source and drain currents are the same. B = 800 W 5 kW = 0.16
- 27. AB +1 A Vin Vout Use the simplified model: A(WITH NEG. FEEDBACK) = 8 (8)(0.16) + 1 = 3.51
- 28. CS D G VDD RG CC RL RS The source bypass capacitor will eliminate the ac negative feedback and restore the voltage gain.
- 29. JFET Amplifier Quiz In a common-source amplifier, the input signal goes to the _______. gate In a common-source amplifier, the input to output phase relationship is ____. 180o The voltage gain of a C-S amplifier is equal to Yfs x _________. load resistance Source bias is produced by current flow through the _______ resistor. source An unbypassed source resistor _______ the voltage gain of a C-S amp. decreases
- 30. Amplifier Negative Feedback • DC reduces sensitivity to device parameters • DC stabilizes operating point • DC reduces sensitivity to temperature change • AC reduces gain • AC increases bandwidth • AC reduces signal distortion and noise • AC may change input and output impedances
- 31. 0.707 Amax A f The frequency response curve of an ac amplifier Bandwidth The gain is maximum in the midband. Amax Midband The bandwidth spans the -3 dB points which are called the break frequencies. -3dB
- 32. 50 W 10 mF 10 mF 1 kW 100 W 1k W 6.8 kW The emitter bypass capacitor in this amplifier has a significant effect on both gain and bandwidth.
- 33. Gain in dB 0 50 Frequency 10 Hz 100 MHz BW1 BW2 Gain and bandwidth with and without the emitter bypass
- 34. Amplifier Frequency Response • The lower break frequency is partly determined by coupling capacitors. • It is also influenced by emitter bypass capacitors. • The upper break frequency is partly determined by transistor internal capacitance. • Both break frequencies can be influenced by negative feedback.
- 35. REVIEW • Amplifier Coupling • Voltage Gain • FET Amplifier • Negative Feedback • Frequency Response