EE105 - Spring 2007Microelectronic Devices and CircuitsLecture 11Bipolar Amplifiers (Part 2)2Common Base (CB) Amplifier In common base topology, where the base terminal is biased with a fixed voltage, emitter is fed with a signal, and collector is the output.3CB Core The voltage gain of CB stage is gmRC, which is identical to that of CE stage in magnitude and opposite in phase. CmvRgA=4Tradeoff between Gain and Headroom To maintain the transistor out of saturation, the maximum voltage drop across RCcannot exceed VCC-VBE..CvCTCC BETIARVVVV=−=5Simple CB Example1217.222.367.7vmCAgRRKRK===Ω=Ω6Input Impedance of CB The input impedance of CB stage is much smaller than that of the CE stage.1inmRg=7Practical Application of CB Stage To avoid “reflections”, need impedance matching. CB stage’s low input impedance can be used to create a match with 50 Ω.8Output Impedance of CB Stage The output impedance of CB stage is similar to that of CE stage.||out O LRrR=9CB Stage with Source Resistance With an inclusion of a source resistor, the input signal is attenuated before it reaches the emitter of the amplifier; therefore, we see a lower voltage gain. This is similar to CE stage emitter degeneration; only the phaseis reversed.1CvSmRARg=+10Practical Example of CB Stage An antenna usually has low output impedance; therefore, a correspondingly low input impedance is required for the following stage.11Realistic Output Impedance of CB Stage The output impedance of CB stage is equal to RCin parallel with the impedance looking down into the collector.[]()111(||) ||||out m E O Eout C outRgR r r R rRRRππ=+ +=12Output Impedance of CE and CB Stages The output impedances of CE, CB stages are the same if both circuits are under the same condition. This is because when calculating output impedance, the input port is grounded, which renders the same circuit for both CE and CB stages.13Fallacy of the “Old Wisdom” The statement “CB output impedance is higher than CE output impedance” is flawed.14CB with Base Resistance With an addition of base resistance, the voltage gain degrades.11out CBinEmvRRvRgβ≈+++15Comparison of CE and CB Stages with Base Resistance The voltage gain of CB amplifier with base resistance is exactlythe same as that of CE stage with base resistance and emitter degeneration, except for a negative sign.16Input Impedance of CB Stage with Base Resistance The input impedance of CB with base resistance is equal to 1/gmplus RBdivided by (β+1). This is in contrast to degenerated CE stage, in which the resistance in series with the emitter is multiplied by (β+1) when seen from the base.111BXBXmrRvRigπββ+=≈+++17Input Impedance Seen at Emitter and Base18Input Impedance Example To find the RX, we have to first find Req, treat it as the base resistance of Q2and divide it by (β+1).2111111BXmmRRggββ⎛⎞=+ +⎜⎟++⎝⎠19Bad Bias Technique for CB Stage Unfortunately, no emitter current can flow.20Still No Good The input signal is shorted to ground. The circuit still does not amplify.21Proper Biasing for CB Stage()1||111in EmoutmCSinmEERRgvgRRvgRR==++22Reduction of Input Impedance Due to RE The reduction of input impedance due to REis bad because it shunts part of the input current to ground instead of to Q1(and RC) . 23Creation of Vb Resistive divider lowers the gain. To remedy this problem, a capacitor is inserted from base to ground to short out the resistor divider at the frequency of interest.24Example of CB Stage with Bias For the circuit shown above, RE>> 1/gm. R1and R2are chosen so that Vbis at the appropriate value and the current that flows thru the divider is much larger than the base current. Capacitors are chosen to be small compared to 1/gmat the required frequency.25Emitter Follower (Common Collector Amplifier) 26Emitter Follower Core When the input is increased by ΔV, output is also increased by an amount that is less than ΔV due to the increase in collector current and hence the increase in potential drop across RE. However the absolute values of input and output differ by a VBE.27Small-Signal Model of Emitter Follower As shown above, the voltage gain is less than unity and positive. 11111AoutEinEmEVvRrvRgRπβ=∞=≈++⋅+28Unity-Gain Emitter Follower The voltage gain is unity because a constant collector current (= I1) results in a constant VBE, and hence Vout follows Vin exactly.1=vA∞=AV29Analysis of Emitter Follower as a Voltage Divider∞=AV30Emitter Follower with Source Resistance11AoutESinEmVvRRvRgβ=∞=+++31Input Impedance of Emitter Follower The input impedance of emitter follower is exactly the same as that of CE stage with emitter degeneration. This is not surprising because the input impedance of CE with emitter degeneration does not depend on the collector resistance. (1 )AXEXVvrRiπβ=∞=++32Emitter Follower as Buffer Since the emitter follower increases the load resistance to a much higher value, it is suited as a buffer between a CE stage and a heavy load resistance to alleviate the problem of gain degradation.33Output Impedance of Emitter Follower Emitter follower lowers the source impedance by a factor of β+1 Æ improved driving capability. EmsoutRgRR ||11⎟⎟⎠⎞⎜⎜⎝⎛++=β34Emitter Follower with Early Effect Since rOis in parallel with RE, its effect can be easily incorporated into voltage gain and input and output impedance equations.()( )OEmsoutOEinmSOEOEvrRgRRrRrRgRrRrRA||||11||111||||⎟⎟⎠⎞⎜⎜⎝⎛++=++=+++=βββπ35Current Gain There is a current gain of (β+1) from base to emitter. Effectively speaking, the load resistance is multiplied by (β+1) as seen from the base.36Emitter Follower with Biasing A biasing technique similar to that of CE stage can be used for the emitter follower. Also, Vbcan be close to Vccbecause the collector is also at Vcc.37Supply-Independent Biasing By putting a constant current source at the emitter, the bias current, VBE, and IBRBare fixed regardless of the supply value.38Summary of Amplifier Topologies The three amplifier topologies studied so far have different properties and are used on different occasions. CE and CB have voltage gain with magnitude greater than one, while follower’s voltage gain is at most one.39Amplifier Example I The keys in solving this problem are recognizing the AC ground between R1and R2, and Thevenin
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