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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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://h ome.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.

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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

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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  20A  CC 1 B R  R 1 1 2 Av  gm RC   2230  17.2  R  22.3k, R  67.7k 130 1 2

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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

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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

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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

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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

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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

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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

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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 = 52A • CB is chosen so that (1/(+1))(1/CB) is small compared to 1/gm at the frequency of interest.

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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.

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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  gmvin vout  r RE v 1 R out   E r 1 1 vin  1  RE   1 RE gm

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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

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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).

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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

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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.

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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.

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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

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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.

Rin  r 1  r 2  R1  R A  C v 1 1 R   1 1 R1 g m1 g m 2  1 Req   g m2  1

Amplifier Example #4

• Note that CB shorts out R2 and provides a ground for R1, at the frequency of interest.  R1 appears in parallel with RC; the circuit simplifies to a simple CB stage with source resistance.

R || R A  C 1 v 1  R S g m

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Amplifier Example #5

• Note that the equivalent base resistance of Q1 is the parallel connection of RE and the impedance seen at the emitter of Q2.

1 1  1 R    S  Rin     ||RE gm1  1 gm2  1 

Review: BJT Amplifier Design • A BJT amplifier circuit should be designed to 1. ensure that the BJT operates in the active mode, 2. allow the desired level of DC current to flow, and 3. couple to a small‐signal input source and to an output “load”.  Proper “DC biasing” is required! (DC analysis using large‐signal BJT model) • Key amplifier parameters: (AC analysis using small‐signal BJT model)

– Voltage gain Av  vout/vin – Input resistance Rin  resistance seen between the input node and ground (with output terminal floating) – Output resistance Rout  resistance seen between the output node and ground (with input terminal grounded)

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Large‐Signal vs. Small‐Signal Models • The large‐signal model is used to determine the DC operating point (VBE, VCE, IB, IC) of the BJT.

• The small‐signal model is used to determine how the output responds to an input signal.

Small‐Signal Models for Independent Sources

• The voltage across an independent voltage source does not vary with time. (Its small‐signal voltage is always zero.) It is regarded as a short circuit for small‐signal analysis.

Large-Signal Small-Signal Model Model

• The current through an independent current source does not vary with time. (Its small‐signal current is always zero.) It is regarded as an open circuit for small‐signal analysis.

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Comparison of Amplifier Topologies

Common Emitter Common Base Emitter Follower

• Large Av < 0 • Large Av > 0 • 0 < Av ≤ 1 ‐ Degraded by RE ‐Degraded by RE and RS ‐ Degraded by RB/(+1) ‐ Degraded by RB/(+1) ‐ Degraded by RB/(+1) • Large Rin • Moderate Rin • Small Rin (due to RE(+1)) ‐ Increased by RB ‐ Increased by RB/(+1) • Small Rout ‐ Increased by RE(+1) ‐ Decreased by RE ‐ Effect of source • Rout  RC • Rout  RC impedance is reddduced by +1 • r degrades A , R • r degrades A , R o v out o v out ‐ Decreased by RE but impedance seen but impedance seen looking into the collector looking into the collector • ro decreases Av, Rin, can be “boosted” by can be “boosted” by and Rout emitter degeneration emitter degeneration

Common Emitter Stage

VA   VA   v  R out  C 1 R vin B  RE  g m  1

Rin  RB  r  ( 1)RE

Rout  RC Rout  RC ||rO1 gm(RE ||r )

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Common Base Stage

VA   v R R out  C  E 1 R vin B RS  RE  RS RE  gm  1

 1 R   B  V   Rin     RE A  gm  1

Rout  RC Rout  RC ||rO1 gm(RE ||r )

Emitter Follower

V   VA   A

v R v R || r out  E out  E O 1 R 1 R vin S vin S RE   RE || rO   gm  1 gm  1

Rin  r  (1  )RE Rin  r   1 RE || rO  1 R   R 1  R   s ||R R   s   || R || r out  g  1 E out   E O  m    1 gm 

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What next

• Cascode and Current Mirrors with BJT • MOSFET fundamentals and Amplifiers or • MOSFET fundamentals and Amplifiers • Cascode and Current Mirrors with BJT and MOSFET