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Common 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

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Common 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 9/6/2018 Indian Institute of Technology Jodhpur, Year 2018 Analog Electronics (Course Code: EE314) Lecture 13‐14: CB and CE Amplifiers Course Instructor: Shree Prakash Tiwari Email: [email protected] Webpage: http: //home.iitj.ac.in/ ~sptiwari/ Course related documents will be uploaded on http://home.iitj.ac.in/~sptiwari/EE314/ Note: The information provided in the slides are taken form text books for microelectronics (including Sedra & Smith, B. Razavi), and various other resources from internet, for teaching/academic use only 1 Common 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. 1 9/6/2018 Small‐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. Av gm RC Tradeoff between Gain and Headroom • To ensure that the BJT operates in active mode, the voltage drop across RC cannot exceed VCC‐VBE. IC VCC VBE Av RC VT VT 2 9/6/2018 Simple CB Stage Example VCC = 1.8V IC = 0.2mA -17 IS = 5x10 A = 100 Simple CB Stage Example VCC = 1.8V IC = 0.2mA -17 IS = 5x10 A = 100 R2 Vb 1.354V VCC if I1 I B R1 R2 V Choose I 10I 20A CC 1 B R R 1 1 2 Av gm RC 2230 17.2 R 22.3k, R 67.7k 130 1 2 3 9/6/2018 Input Impedance of a CB Stage • The input impedance of a CB stage is much smaller than that of a CE stage. 1 Rin if VA gm CB 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. R A C v 1 RS gm 4 9/6/2018 Practical Example of a CB Stage • An antenna usually has low output impedance; therefore, a correspondingly low input impedance is required for the following stage. Output 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. Rout 1 1 g m (RE || r ) rO RE || r Rout 2 RC || Rout1 5 9/6/2018 Output 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. Av of CB Stage with Base Resistance (VA = ∞) • With base resistance, the voltage gain degrades. vout vout gmv RC v gmRC v vout vP r RB r RB r r gmRC v v out r R P R B C v out r R v v v v 1 v R B in P in out C KCL at node P: gmv gm r RE r gmRC RE v R R out C C 1 R vin r 1 RE RB B RE gm 1 6 9/6/2018 Voltage 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. Rin 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. v KCL gmv ix r 1 r gm vx ix r r RB r vx r RB 1 RB R v vx in r RB ix 1 gm 1 7 9/6/2018 Input Impedance Seen at Emitter vs. Base Common Base Stage Common Emitter Stage Input Impedance Example • To find RX, we have to first find Req, treat it as the base resistance of Q2 and divide it by (+1). 1 R 1 1 1 R B R B Req x gm1 1 gm2 1 gm1 1 8 9/6/2018 Biasing of CB Stage • RE is necessary to provide a path for the bias current IE to flow, but it lowers the input impedance. 1 R 1 g E R R || R m E in E 1 gm 1 gmRE RE gm Rin RE vX vin vin Rin RS RE 1 gmRE RS vout vout vX Av vin vX vin vout RE gmRC vin RE 1 gmRE RS Reduction of Input Impedance Due to RE • The reduction of input impedance due to RE is undesirable because it shunts part of the input current to ground instead of to Q1 (and RC). Choose RE >> 1/gm , i.e. ICRE >> VT 9 9/6/2018 Creation of Vb • A resistive voltage divider lowers the gain. • To remedy this problem, a capacitor is inserted between the base and ground to short out the resistive voltage divider at the frequency of interest. Example of CB Stage with Bias VCC = 2.5V Design a CB stage for A = 10 and R = 50. -16 v in IS = 5x10 A • Rin = 50≈ 1/gm if RE >> 1/gm = 100 VA = ∞ Choose RE = 500 • Av = gmRC = 10 RC = 500 • IC = gm∙VT = 0.52mA • VBE=VTln(IC/IS)=0.899V • Vb = IERE + VBE = 1.16V • Choose R1 and R2 to provide Vb and I1 >> IB, e.g. I1 = 52A • CB is chosen so that (1/(+1))(1/CB) is small compared to 1/gm at the frequency of interest. 10 9/6/2018 Emitter Follower (Common Collector Amplifier) Emitter Follower Core • When the input voltage (Vin) is increased by Vin, the collector current (and hence the emitter current) increases, so that the output voltage (Vout) is increased. • Note that Vin and Vout differ by VBE. 11 9/6/2018 Unity‐Gain Emitter Follower • In integrated circuits, the follower is typically realized as shown below. – The voltage gain is 1 because a constttant collec tor current (= I1) results in a constant VBE; hence Vout = Vin . VA Av 1 Small‐Signal Model of Emitter Follower • The voltage gain is less than 1 and positive. V A v vin vout v vout KCL at emitter: gmv r RE vin vout vout gmvin vout r RE v 1 R out E r 1 1 vin 1 RE 1 RE gm 12 9/6/2018 Emitter Follower as a Voltage Divider V A Emitter Follower with Source Resistance VA v R out E 1 R vin S RE gm 1 13 9/6/2018 Input Impedance of Emitter Follower • The input impedance of an emitter follower is the same as that of a CE stage with emitter degeneration (whose input impedance does not depend on the resistance between the collector and VCC). VA vx Rin r (1 )RE ix Effect of BJT Current Gain • There is a current gain of (+1) from base to emitter. • Effectively, the load resistance seen from the base is multiplied by (+1). 14 9/6/2018 Emitter Follower as a Buffer • The emitter follower is suited for use as a buffer between a CE stage and a small load resistance, to alleviate the problem of gain degradation. R r (1 )R Av gm RC Rspeaker in1 2 2 speaker Av gm RC Rin1 Output Impedance of Emitter Follower • An emitter follower effectively lowers the source impedance by a factor of +1, for improved driving capability. • The follower is a good “voltage buffer” because it has high input impedance and low output impedance. 1 R s Rout ||RE gm 1 15 9/6/2018 Emitter Follower with Early Effect • Since rO is in parallel with RE, its effect can be easily incorporated into the equations for the voltage gain and the input and output impedances. RE || rO Av RS 1 RE || rO 1 gm Rin r 1 RE || rO Rs 1 Rout || RE || rO 1 gm Emitter Follower with Biasing • A biasing technique similar to that used for the CE stage can be used for the emitter follower. • NtNote tha t VB can be biase d to be close to VCC because the collector is biased at VCC. 16 9/6/2018 Supply‐Independent Biasing • By putting an independent current source at the emitter, the bias point (IC, VBE) is fixed, regardless of the supply voltage value. Summary of Amplifier Topologies • The three amplifier topologies studied thus far have different properties and are used on different occasions. • CE and CB stages have voltage gain with magnitude greater than one; the emitter follower’s voltage gain is at most one. 17 9/6/2018 Amplifier Example #1 • The keys to solving this problem are recognizing the AC ground between R1 and R2, and using a Thevenin transformation of the input network. CE stage Small-signal Simplified small-signal equivalent circuit equivalent circuit v R || R R out 2 C 1 R || R 1 vin 1 S R1 RS RE 1 g m Amplifier Example #2 • AC grounding/shorting and Thevenin transformation are needed to transform this complex circuit into a simple CE stage with emitter degeneration. v R R out C 1 R || R 1 vin S 1 R1 RS R2 1 g m 18 9/6/2018 Amplifier Example #3 • First, identify Req, which is the impedance seen at the emitter of Q2 in parallel with the infinite output impedance of an ideal current source. • Second, use the equations for a degenerated CE stage with RE replaced by Req.
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