Click here to load reader

Lecture 20 Bipolar Junction Transistors (BJT): Part 4 Small Signal

  • View
    220

  • Download
    1

Embed Size (px)

Text of Lecture 20 Bipolar Junction Transistors (BJT): Part 4 Small Signal

  • Lecture 20Lecture 20

    Bipolar Junction Transistors (BJT): Part 4

    Small Signal BJT Model

    Reading:

    Jaeger 13.5-13.6, Notes

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Further Model Simplifications(useful for circuit analysis)( y )

    Ebers-MollForward Active

    Mode

    Neglect Small Terms

    TEB

    TEB

    TCB

    TEB

    VV

    SCRV

    V

    FFCV

    V

    RV

    V

    FFC eIIIeIIeIeII

    0000 111

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Modeling the Early Effect (non-zero slopes in IV curves)

    iB1< iB2< iB3IC Base width changes due to changes in the base-

    iB3 (theory)

    collector depletion width with changes in VCB.

    This changes T, which

    iB1 (theory)iB2 (theory)

    VA VCE

    g T,changes IC, DC and BF

    A

    Major BJT Circuit Relationships

    TEB

    TEB

    TEB

    VV

    FO

    S

    F

    CB

    A

    CEFOF

    A

    CEVV

    SCV

    V

    SC eIi

    iVv

    Vv

    eIieIi

    11

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Small Signal Model of a BJT

    Just as we did with a p-n diode, we can break the BJT up into a large signal analysis and a small signal analysis and linearize the non-linear behavior of the Ebers-Moll model.

    S ll i l M d l l f l f F d i dSmall signal Models are only useful for Forward active mode and thus, are derived under this condition. (Saturation and cutoff are used for switches which involve very large voltage/current swings from the on to off states.)

    Small signal models are used to determine amplifier characteristics (Example: Gain = Increase in the magnitude of a signal at the output of a circuit relative to its magnitude atof a signal at the output of a circuit relative to it s magnitude at the input of the circuit).

    Warning: Just like when a diode voltage exceeds a certainWarning: Just like when a diode voltage exceeds a certain value, the non-linear behavior of the diode leads to distortion of the current/voltage curves (see previous lecture), if the i t / t t d t i li it th f ll Eb M ll d l

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    inputs/outputs exceed certain limits, the full Ebers-Moll model must be used.

  • Consider the BJT as a two-port Network

    i1 iTwo Port Network

    +V1

    -

    i1 i2+V2

    -

    i1=y11v1 + y12v2 ib=y11vbe + y12vce

    General y-parameter Network BJT y-parameter Network

    i2=y21v1 + y22v2 ic=y21vbe + y22vce

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Consider the BJT as a two-port Network

    iib=y11vbe + y12vceic=y21vbe + y22vce

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Consider the BJT as a two-port Network

    o is most

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    often taken as a constant, F

  • Alternative Representations

    CT

    Cm

    V

    IVI

    yg 21

    1

    40

    Transconductance

    CEA

    m

    o

    C

    To

    VVgI

    Vy

    r

    11

    1

    1 Input Resistance

    C

    CEAo I

    VVy

    r

    22

    1Output Resistance

    Y-parameter Model Hybrid-pi Model

    v1

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Alternative Representations

    bobmbem iirgvg

    Voltage Controlled Current source version of Hybrid pi

    Current Controlled Current source version of Hybrid pisource version of Hybrid-pi

    Modelsource version of Hybrid-pi

    Model

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Single Transistor Amplifier Analysis: Summary of Procedure Steps to Analyze a Transistor Amplifier

    Important!

    1.) Determine DC operating point and calculate small signal parameters (see next page)

    Step

    Step 1

    2.) Convert to the AC only model.DC Voltage sources are shorts to groundDC Current sources are open circuitsLarge capacitors are short circuits

    Step 2

    Large capacitors are short circuitsLarge inductors are open circuits

    3.) Use a Thevenin circuit (sometimes a Norton) where necessary. Ideally the

    Step 3

    Norton) where necessary. Ideally the base should be a single resistor + a single source. Do not confuse this with the DC Thevenin you did in step 1.

    Step 4

    4.) Replace transistor with small signal model5.) Simplify the circuit as much as necessary

    Step 5

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    necessary. 56.) Calculate the small signal parameters (r, gm, ro etc) and then gains etc

  • Single Transistor Amplifier AnalysisDC Bias Point

    Important!Step 1 detail

    Thevenin

    Vbe Ie

    Ib

    RTH

    3V=IERE+Vbe+IBRTH3V=IC((o+1)/o) Re+0.7V+IBRTH3V IC((o 1)/o) Re 0.7V IBRTH

    3V=IB o((o+1)/o) Re+0.7V+IBRTH3V= IB(100+1)1300+0.7+ IB7500

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    B B

    IB=16.6 uA, IC= IB o=1.66 mA, IE=(o+1) c/ o=1.67 mA

  • Single Transistor Amplifier AnalysisCalculate small signal parameters

    I CStep 6 detail Important!

    gIV

    yr

    SIVI

    yg

    o

    C

    To

    CT

    Cm

    15061

    0664.040

    11

    21

    Transconductance

    Input Resistance

    KIV

    IVV

    yr

    gIy

    C

    A

    C

    CEAo

    mC

    2.451

    22

    11

    Output Resistance

    RTHVTH=0.88 VS

    rRL= RC| R3| ro

    Vbe gmVbe

    Vout880

    vvandR

    rvvandRvgv SThThbeLbemout 88.0

    rR

    rRg

    vv

    vv

    vv

    vv

    GainVoltageA

    rRg

    ThLm

    S

    th

    th

    be

    be

    out

    S

    outv

    SThTh

    ThbeLbemout

    88.0

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    VVA

    A

    v

    v

    ThSthbeS

    /139

    88.01506880

    1506000,100||4300||200,450664.0

    For Extra Examples see: Jaeger section 13.6, and

    pages 627-630 (top of 630)

  • Supplement Added to describe more details of theImportant!

    Supplement Added to describe more details of the Solution of this Problem

    Bipolar Junction Transistors (BJT): Part 5

    D t il f A lifi A l iDetails of Amplifier Analysis

    Reading:

    Jaeger 13.5-13.6, Notes

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Detailed Example: Single Transistor Amplifier Analysis

    Important!

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    Notes on slides 14-25 were prepared by a previous student.

  • Step 1: Determine DC Operating PointRemove the Capacitors

    Important!

    Remove the Capacitors

    Because the impedance of a capacitor is Z = 1/(jC), capacitors have infinitehave infinite impedance or are open circuits in DC ( = 0).

    Inductors (not present in this circuit) have an impedance Z = jLimpedance Z = jL, and are shorts in DC.

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Step 1: Determine DC Operating PointDetermine the DC Thevenin Equivalent

    Important!Determine the DC Thevenin Equivalent

    Replace all connections to the transistor with their Thevenin equivalents.

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Step 1: Determine DC Operating PointCalculate Small Signal Parameters

    Important!

    Calculate Small Signal Parameters

    Identify the type of transistor (npn in this example) and draw the base, collector, and emitter currents in their

    di i d h i di l+

    ICproper direction and their corresponding voltage polarities.Applying KVL to the controlling loop (loop 1):

    -

    IB

    ( p )VTHB IBRTHB VBE IERE = 0

    Applying KCL to the transistor:

    + -+

    VBEB

    IE

    pp y gIE = IB + IC

    Because IC = IB,

    +

    -BE

    1IE = IB + IC = IB + IB = IB(1+)

    Substituting for IE in the loop equation:

    -

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

    VTHB IBRTHB VBE IB(1+)RE = 0

  • Step 1: Determine DC Operating PointPlug in the Numbers

    Important!

    Plug in the Numbers

    VTHB IBRTHB VBE IB(1+)RE = 0VTHB VBE IB(RTHB + (1+ )RE) = 0

    +

    ICVTHB VBE IB(RTHB + (1+ )RE) 0VTHB = 12R1/(R1+R2) = 3 VRTHB = R1 || R2 = 7.5 kAssume VBE = 0.7 V

    -

    IB

    BEAssume for this particular transistor is given to be 100. + -

    +VBE3 0 7 I (7500 + (1+100)*1300) 0 B

    IE

    +

    -BE3 0.7 IB(7500 + (1+100)*1300) = 0IB = 16.6 AIC = IB = 1.66 mAI = I + I = 1 676 mA

    1 -IE = IB + IC = 1.676 mA

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Step 1: Determine DC Operating PointCheck Assumptions: Forward Active?

    Important!

    Check Assumptions: Forward Active?

    VC = 12 ICRC = 12 - (1.66 mA)(4300) = 4.86 VV = I R = (1 67 mA)(1300) = 2 18 V

    +

    ICVE = IERE = (1.67 mA)(1300) = 2.18 VVB = VTHB IBRTHB = 3 (16.6A)(7500) = 2.88 V

    -

    Check: V

    IB

    + -+

    VBE

    Check:For an npn transistor in forward active:

    VC > VB

    VC

    VE

    VB

    B

    IE

    +

    -BE4.86 V > 2.88 V

    VB VE = VBE = 0.7 V

    E

    1 -2.88 V 2.18 V = 0.7 V

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Step 2: Convert to AC-Only ModelShort the Capacitors and DC Current Sources

    Important!

    Short the Capacitors and DC Current Sources

    DC voltage sources are h t ( ltshorts (no voltage

    drop/gain through a short circuit).

    DC current sources areDC current sources are open (no current flow through an open circuit).

    Large capacitors are shorts (if C is large, 1/jC is small).

    Large inductors are Large inductors are open (if L is large, jL is large).

    ECE 3040 - Dr. Alan DoolittleGeorgia Tech

  • Step 2: Convert to AC-Only Model