Single Transistor Amp

Single Transistor and Multiple Transistor Amplifiers

SJSU EE223 by Koorosh Aflatooni

1

Overview ¾ Introduction Modeling

¾ Single Transistor amplifiers

Common emitter/source Common base/gate Common collector/drain Common emitter/source with degeneration

¾ Multiple transistor amplifiers Common collector-common emitter Common collector-common collector Cascode ( Simple cascode ( Active cascode

¾ Differential pairs

DC transfer of common emitter/source pairs DC transfer of common emitter/source pairs with degeneration Small signal characteristics Device mismatch


2

Modeling i1 v1

¾ Two port modeling Express the relation between input and output Superposition of the each source contribution

Two port network

i1 = y11v1 + y12 v2 i2 = y21v1 + y22 v2

y11


i2 v2

y12v2 y21v1

y22

3

Modeling (cont.) i1

¾ Feedback Bilateral Unilateral => y12=0

v1

i2

Zi

Gmv1

¾ Other terms Short-circuit transonductance => Gm=y21 Input impedance => Zi=1/y11 Output impedance => Zo=1/y22

¾ Knowing any two parameters leads to the third parameter v av = 2 v1

i2 =0 = −Gm Z o

Norton to Thevenin

i1 v1

v2

Zo

i2 +

Zi

_

avv1

v2

Zo

¾ The key is to understand the effect of loading on performance SJSU EE223 by Koorosh Aflatooni

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Common Emitter ¾ Large signal Collector current is related to base current Output voltage is defined by considering load line

¾ Small Signal Input resistance

IB =

V IC I S exp i = IB βF  VT

  

V Vo = VCC − RC I S exp i  VT Ri = rπ =

  

βo gm

Transconductance

Gm = g m

Output resistance

Ro = RC || ro

Open circuit voltage gain

av = − g m (ro || RC )


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Common Source ¾ Large signal Output voltage related to input Transition from cutoff => active => triode

¾ Small signal

Vo = VDD −

µCox W 2

L

RD (Vi − Vt )

Input resistance

Ri → ∞

Transconductance

Gm = g m

Output resistance

Ro = RD || ro

Open circuit voltage gain

av = − g m (ro || RD )

2

¾ The maximum voltage gain for CS is proportional to 1/√ID, in contrast to BJT that is independent of current SJSU EE223 by Koorosh Aflatooni

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Common Base ¾ Small signal Modifying π-model to T model to decouple the dependent current source between input-output ports Input resistance Transconductance Output resistance Open circuit voltage gain

Ri = re Gm =

gm r 1+ b rπ

Ro = RC av = g m RC

Compare to common emitter, the input resistance is reduced by (1+b) as well the the current gain


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Common Gate ¾ Large signal Not much interesting

¾ Small signal 1 g m + g mb

Input resistance

Ri =

Transconductance

Gm = g m + g mb

Output resistance Open circuit voltage gain


Ro = RD av = ( g m + g mb )RD

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Common Gate (cont.) ¾ Considering a case with ro => bilateral because of feedback provided by output => input resistance depends on output load ¾ Small signal Input resistance Transconductance

Ri =

ro + RD || RL 1 + ( g m + g mb )ro

( Ro has no effect since this is measured with output shorted

Gm = g m + g mb

Output resistance


Ro = R || (( g m + g mb )ro RS )

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Common Collector ¾ Emitter follower Ideally base-emitter voltage remains constant, independent of collector voltage In reality it is not quite constant It is not unilateral

¾ Small signal (π model) Input resistance Voltage gain Output resistance


Ri = rπ + (β o + 1)(RL || ro ) av =

1 RS + rπ 1+ (β o + 1)(RL || ro ) Ro =

rπ + RS || ro βo +1

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Example ¾ Calculate the input resistance, output resistance, and voltage gain of the emitter follower. Assume RS=RL=1kΩ, β=100, rb=0, ro =>∞, Io=100µA.


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Common Drain ¾ Source follower Ideally, source follows gate voltage In reality, it deviates due to body effect and channel modulation effect

¾ Small signal Input resistance

Ri = ∞

Voltage gain ( Depends on body effect av =

g m ro 1 + ( g m + g mb )ro +

Output resistance Ro =


ro RL

1 g m + g mb +

1 1 + ro RL

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Common Emitter with Emitter Degeneration ¾ Adding the resistance to emitter: Reduces the transconductance, and increases output/input resistances

¾ Small signal Input resistance Transconductance

Output resistance

R    ro + C  βo +1  Ri = rπ + (β o + 1)RE   ro + RC + RE     

  R   1− E   β o ro Gm = g m     1 1   1 + g m RE 1 +  β + g r    o m o   

Ro = (rπ || RE ) + ro [1 + g m (rπ || RE )] SJSU EE223 by Koorosh Aflatooni

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Common Source with Source Degeneration ¾ Small signal Input resistance

Ri = ∞ gm 1 + ( g m + g mb )RS

Transconductance

Gm =

Output resistance

Ro = RS + ro [1 + ( g m + g mb )RS ]


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Single Transistor Summary Configuration

Voltage gain

Current gain

Input resistance

Output resistance

Common Source

Av > 1

-

∞

Moderate to high

Source-Follower

Av ~ 1

-

∞

Low

Common-Gate

Av > 1

Ai ~ 1

Low

Moderate to high

Common-Emitter

Av > 1

Ai > 1

Moderate

Moderate to high

Emitter-Follower

Av > 1

Ai > 1

High

Low

Common-Base

Av > 1

Ai ~ 1

Low

Moderate to high


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Multiple Transistor Amplifiers ¾ In many applications performance of a single stage amplifier is not sufficient to meet various requirements Need to combine multiple transistors to achieve voltage, current, input/output impedance adjustments In general, the overall voltage/current gain is not simply the product of all the stages, but it is a function of loading (input/output resistance) of each stage => need specific analysis Stage 1 Av1 Ri1

Stage n Avn

Stage 2 Av2 Ro1

Ri2

Ro2

Rin

Ron

¾ Some popular combinations Common collector- common emitter & Common collector- common collector Cascode Super source follower Differential pair SJSU EE223 by Koorosh Aflatooni

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Common Collector- Common Emitter ¾ Goal: To achieve higher input resistance and gain

¾ Operation principle: Ibias provides the DC biasing Q2 appears as load on emitter of Q1 => input resistance increases; also gives two stages of current gain Consider a combined transistor

¾ Small circuit analysis Input resistance Transconductance

Current gain Output resistance SJSU EE223 by Koorosh Aflatooni

Ri = rπ 1 + (β o + 1)(rπ 2 || ro )       1 Gm = g m 2    rπ 1  1 +     ( β + 1)r   o π 2   

β c = β o (β o + 1) Ro = ro 2

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Darlington Configuration ¾ Similar to: cc-cc: as discussed cc-ce: but in Darlington collector of Q1 gives feedback path=> reduction of output resistance & increase of input capacitance

¾ BiCMOS version finds many applications High input resistance Large transconductance SJSU EE223 by Koorosh Aflatooni

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Cascode Configuration ¾ Bipolar version Common emitter- common base

¾ Small circuit analysis Input resistance Transconductance Voltage gain Output resistance


Ri = rπ 1 Gm = g m1

Av = −

βo η

    g m 2 ro1   Ro = ro 2 1 +  g m 2 ro1  + 1  β o  

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Cascode Configuration (cont.) ¾ MOSFET version: Common source- common gate Output resistance can be tuned => limited by power supply voltage and signal swing

¾ Small circuit analysis Input resistance Transconductance

Ri → ∞ Gm ≈ g m1

Output resistance Ro = ( g m 2 + g mb 2 )ro1ro 2 SJSU EE223 by Koorosh Aflatooni

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Cascode Configuration (cont.) ¾ Active cascode Using an amplifier to to provide negative feedback and increases the output resistance Only works at frequencies amplifier has gain


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Super Source Follower ¾ Goal: Reduce output resistance of source follower => useful if you need to drive a resistive load

¾ Small signal Output resistance


Ro =

1 1 (g m1 + g mb1 ) g m 2 ro1

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Differential Pair ¾ Goal: to eliminate the common sources (e.g., noise sources) and amplify differential input signal ¾ Analysis Large signal (Bipolar: Linear region 26mV around zero (MOSFET:

Small signal


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BJT Differential Pair Large Signal Analysis ¾ Assuming

2

Rtrail very large and ro can be ignored

¾ Steps: Write KCL for input signals to emitter of transistors Relate Ic1 and Ic2 to Itrail Relate output voltages to input voltages

1

¾ Highlights Useful range ~ Linear range ~
av =

α f I Train RC VT

¾ How to increase the useful range? Emitter degeneration SJSU EE223 by Koorosh Aflatooni

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MOSFET Differential Pair Large Signal Analysis ¾ Assuming

2

Rtrail very large and ro can be ignored

¾ Steps:

Write KCL for input signals to emitter of transistors Relate Id1 and Id2 to Itrail Relate output voltages to input voltages

¾ Highlights

1 2 I Trail W  k '  L

Useful range ~ Linear range

≤

Voltage gain

≈ k ' I Trail RD

¾ How to increase the useful range?

W/L Over-drive


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Example ¾ Compare the forward transconductance of a MOSFET differential gain against a bipolar differential gain? (assume ITrail=500µA and k’=100µA/V2, W/L=1,αF=1) I c1 =

4 I Trail I k' W 2 Vid I d 1 = Trail + − Vid 2 4 L k ' (W / L)

MOSFET

g m (max) =

∂I d 1 k ' I Trail W |Vid =0 = ∂Vid 4 L

g m (max) = 35µA / V


BJT

α F I Trail

 v  1 + exp − i1   VT  ∂I α I g m (max) = c1 |Vi1=0 = F Trail 2VT ∂Vi1

g m (max) = 9766 µA / V

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Differential Pair Small Signal Analysis ¾ Breaking analysis to: Differential mode vod = Adm vid + Acm− dm vic voc = Adm −cm vid + Acm vic Common mode Ideally we like Adm-cm=0 and Acm-dm=0, in reality they are not A Common mode rejection CMRR = dm Acm ratios (CMRR) (Other important ratios


Adm Acm− dm

Vid/2 -Vid/2 -Vic

Adm Adm −cm

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Differential Pair Small Signal Analysis (cont.) ¾ In a balanced differential pair, increase if current in path 1, means current in path 2 decreases by same amount

1

2

( Voltage across Rtrail stay constant ( Dropping Rtrail makes no difference

¾ Voltage gain

Adm = − g m (R ro )

The gmb has no effect since source to ground stays at a constant potential SJSU EE223 by Koorosh Aflatooni

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Differential Pair Small Signal Analysis (cont.) 1

¾ Due to symmetry, we could assume no current flows between two sections

2

Breaking the circuit into two sections

¾ Each of these sections present a degenerate source follower configuration Common mode gain

Degenerate Source follower

Gm =

In case ro>0

gm 1 + ( g m + g mb )RS

Acm = −Gm RD =

Acm = −Gm RD =

− g m RD 1 + ( g m + g mb )2 RTrail

− gm RD {2 RTrail + ro [1 + ( g m + g mb )2 RTrail ]} 1 + ( g m + g mb )2 RTrail

¾ Note: increase of Rtrail leads to increase of CMRR CMRR ≈ 1 + 2( g m + g mb )RTrail SJSU EE223 by Koorosh Aflatooni

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Example ¾ Find the differential-mode gain, common-mode gain, and differential-mode input resistance for a bipolar differential pair? (assume ITrail=20µA, RTrail=10MΩ, RC=100kΩ, VEE=VCC=5V, βο=150, and neglect rb, ro, and rµ.  20 µA  100 KΩ = 78 Adm = − g m RC = −  VT 

Acm =

− gm RC = −0.005  1  1 + g m RTrail 1 +   βo 


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Summary ¾ Review of various single and two state amplifiers, including differential pairs ¾ End of chapter problems: 3-2, 3-4, 3-7, 3-9, 3-14, 3-16, 3-24


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Single Transistor Amp

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