Lecture 9Common Base (CB) AmplifierSmall-Signal Analysis of CB CoreTradeoff between Gain and HeadroomSimple CB Stage ExampleInput Impedance of a CB StageCB Stage with Source ResistancePractical Example of a CB StageOutput Impedance of a CB StageOutput Impedance: CE vs. CB StagesAv of CB Stage with Base Resistance (VA = ∞)Voltage Gain: CE vs. CB StagesRin of CB Stage with Base Resistance (VA = ∞)Input Impedance Seen at Emitter vs. BaseInput Impedance ExampleEE105 Fall 2007 Lecture 9, Slide 1 Prof. Liu, UC BerkeleyLecture 9OUTLINE•BJT Amplifiers (cont’d)–Common-base topology–CB core–CB stage with source resistance–Impact of base resistanceReading: Chapter 5.3.2ANNOUNCEMENTS•Friday discussion section (103) has been moved back to 5 Evans.•HW#5 is now posted.•List of frequently misunderstood/forgotten points has been updated.EE105 Fall 2007 Lecture 9, Slide 2 Prof. Liu, UC BerkeleyCommon Base (CB) Amplifier•The base terminal is biased at a fixed voltage; the input signal is applied to the emitter, and the output signal sensed at the collector.EE105 Fall 2007 Lecture 9, Slide 3 Prof. Liu, UC BerkeleySmall-Signal Analysis of CB Core•The voltage gain of a CB stage is gmRC, which is identical to that of a CE stage in magnitude and opposite in phase. CmvRgA EE105 Fall 2007 Lecture 9, Slide 4 Prof. Liu, UC BerkeleyTradeoff between Gain and Headroom •To ensure that the BJT operates in active mode, the voltage drop across RC cannot exceed VCC-VBE.TBECCCTCvVVVRVIAEE105 Fall 2007 Lecture 9, Slide 5 Prof. Liu, UC BerkeleySimple CB Stage Example2.1722301301CmvRgAVCC = 1.8VIC = 0.2mAIS = 5x10-17 A = 100k7.67 ,k3.22 2010 Choose if 354.1212111212RRRRVAIIIIVRRRVVCCBBCCbEE105 Fall 2007 Lecture 9, Slide 6 Prof. Liu, UC BerkeleyInput Impedance of a CB Stage•The input impedance of a CB stage is much smaller than that of a CE stage.AminVgR if 1EE105 Fall 2007 Lecture 9, Slide 7 Prof. Liu, UC BerkeleyCB Stage with Source Resistance•With the inclusion of a source resistance, the input signal is attenuated before it reaches the emitter of the amplifier; therefore, the voltage gain is lowered. –This effect is similar to CE stage emitter degeneration.SmCvRgRA1EE105 Fall 2007 Lecture 9, Slide 8 Prof. Liu, UC BerkeleyPractical Example of a CB Stage•An antenna usually has low output impedance; therefore, a correspondingly low input impedance is required for the following stage.EE105 Fall 2007 Lecture 9, Slide 9 Prof. Liu, UC BerkeleyOutput Impedance of a CB Stage•The output impedance of a CB stage is equal to RC in parallel with the impedance looking into the collector. 121||||)||(1outCoutEOEmoutRRRrRrrRgREE105 Fall 2007 Lecture 9, Slide 10 Prof. Liu, UC BerkeleyOutput Impedance: CE vs. CB Stages•The output impedances of emitter-degenerated CE and CB stages are the same. This is because the circuits for small-signal analysis are the same when the input port is grounded.EE105 Fall 2007 Lecture 9, Slide 11 Prof. Liu, UC BerkeleyAv of CB Stage with Base Resistance (VA = ∞)•With base resistance, the voltage gain degrades. BCoutPBCmoutBPCmoutCmoutRrRvvRrRgrvRrrvvRgvvRvgv EinBCoutCmoutmEinPmRvRrRvRgvgrRvvvgrv1 :P nodeat KCL 111BEmCBECinoutRRgRRRrRvvEE105 Fall 2007 Lecture 9, Slide 12 Prof. Liu, UC BerkeleyVoltage Gain: CE vs. CB Stages•The magnitude of the voltage gain of a CB stage with source and base resistances is the same as that of a CE stage with base resistance and emitter degeneration.EE105 Fall 2007 Lecture 9, Slide 13 Prof. Liu, UC BerkeleyRin of CB Stage with Base Resistance (VA = ∞)•The input impedance of a CB stage with base resistance is equal to 1/gm plus RB divided by (+1). This is in contrast to a degenerated CE stage, in which the resistance in series with the emitter is multiplied by (+1) when seen from the base.1111BmBxxinxxBmxmRgRrivRivRrrgrivgrvKCLxBvRrrvEE105 Fall 2007 Lecture 9, Slide 14 Prof. Liu, UC BerkeleyInput Impedance Seen at Emitter vs. BaseCommon Base Stage Common Emitter StageEE105 Fall 2007 Lecture 9, Slide 15 Prof. Liu, UC BerkeleyInput Impedance Example•To find RX, we have to first find Req, treat it as the base resistance of Q2 and divide it by (+1).
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