Modern optical telecommunications systems often employ optical fibers for carrying signals. Low-cost, high-performance, highly integrated fiber optic interface circuits are becoming increasingly necessary to meet the demands of high-speed digital data communication. One standard circuit function for fiber optic communication systems is a transimpedance amplifier that is used within an optical receiver.
FIG. 1 depicts an architectural diagram of a typical optical receiver. The light from an optical fiber 101 impinges on photodiode detector 102, producing a current IPD. Transimpedance amplifier 103 converts the very small currents indicative of optical signals applied to photodetector 102 into a large signal voltage, VTIA, which is further amplified by post amplifier 104 to voltage VPA for output to digital circuitry.
FIG. 2 depicts a schematic diagram of the transimpedance-input stage of a transimpedance amplifier in the prior art. An input current is injected into node 201. This current falls across input resistance RS and reverse-biased diode D1, both connected to power source VSS. Node 201 is electrically connected to the gate terminal of NMOS transistor MB and the source of NMOS transistor M1. The gate terminal of transistor M1 is electrically connected to the drain terminal of transistor M2 and one terminal of resistor RB. The output of the circuit is terminal 202, which is electrically connected to the drain of transistor M1, and one terminal of resistor R1. The second terminal of resistor R1 is electrically connected to the second terminal of resistor RB and to power supply VDD.
The feedback of the voltage at point 2 between transistor MB and resistor RB to the gate of transistor M1 serves to keep the biasing of the output voltage at point 3, VOUT, stable. When the current of M1 increases, the voltage at point 3 increases and the drain current of MB increases. This causes the voltage at point 2 to decrease and this, in turn, causes the drain current of M1 to decrease to its previous value.
FIG. 3 depicts the small signal analysis of the circuit of FIG. 2 in the prior art. This analysis is for the circuit at mid-band, where parasitic capacitances are ignored. From FIG. 3a, the sum of the two transistor gate-to-source voltages is equal to the voltage drop across resistor RB, orVGS1+VGSB=−gmBVGSBRB  (Eq. 1)
FIG. 3a is rearranged in FIGS. 3b and 3c. FIG. 3b shows the dependent current source between nodes VIN and VOUT in FIG. 3a can be split into two dependent current sources to ground. From FIG. 3c-1 and equation (1), since VGS1=−VGSB (1+gmBRB), then the value of dependent current source is −gm1VGSB(1+gmBRB). Since the voltage across the current source is proportional to the current through the source by a factor of VGSB, the current source, with polarity reversed, can be replaced by an equivalent resistor of value 1/(gm1(1+gmBRB)). Also from FIG. 3c-1, VGSB=VIN, soVGSB=RS∥1/(gm1(1+gmBRB)IIN  (Eq. 2)ZIN=RS∥1/(gm1(1+gmBRB)≈1/(gm1(1+gmBRB)  (Eq. 3)Thus, the input impedance is low, an ideal characteristic for an input current source represented by the photodetector.
From FIG. 3c-2:
                                                                        V                OUT                            =                            ⁢                                                -                                      g                    m1                                                  ⁢                                  V                  GS1                                ⁢                R1                                                                                        =                            ⁢                                                -                  R1                                ⁢                                                                  ⁢                                                      g                    m1                                    ⁡                                      (                                          -                                                                        V                          GSB                                                ⁡                                                  (                                                      1                            +                                                                                          g                                sB                                                            ⁢                              RB                                                                                )                                                                                      )                                                                                                                          =                            ⁢                                                R1I                  IN                                ⁢                                                      g                    m1                                    [                                      RS                    ⁢                                                                                        1                        /                                                  (                                                                                    g                              m1                                                        ⁡                                                          (                                                              1                                +                                                                                                      g                                    mB                                                                    ⁢                                  RB                                                                                            )                                                                                )                                                                    ]                                        ⁢                                          (                                              1                        +                                                                              g                            mB                                                    ⁢                          RB                                                                    )                                                                                                                                              ≈                            ⁢                                                I                  IN                                ⁢                RB                                                                        (                  Eq          .                                          ⁢          4                )            so that the transimpedance is:VOUT/IIN≈RB  (Eq. 5)As the circuit operates at higher and higher frequencies, the parasitic capacitances of the transistors must be taken into consideration. They will have the effect of inducing poles into the transimpedance equation 4, thus reducing transimpedance gain.Once the input current has been converted into a voltage by the transimpedance amplifier, it is often desirable to further amplify the voltage output. FIG. 4 depicts a typical common-source voltage amplifier circuit of the prior art. FIG. 4a is the circuit with input resistance RS connected to the gate of an NMOS transistor. The drain of the transistor is connected to a resistor RD at the output VOUT. RD is also connected to power supply VDD. FIG. 4b is the high frequency model of the common-source amplifier taking parasitic capacitance into effect across the terminals of the transistor. We obtain the transfer function of the voltage amplifier using nodal analysis:
                                                                        V                X                            -                              V                IN                                                    R              S                                +                                    V              X                        ⁢                          sC              gs                                +                                    (                                                V                  X                                -                                  V                  OUT                                            )                        ⁢                          sC              gd                                      =        0                            (                  Eq          .                                          ⁢          5                )                                                                    (                                                V                  OUT                                -                                  V                  X                                            )                        ⁢                          sC              gd                                +                                    g              m                        ⁢                          V              X                                +                                    V              OUT                        ⁡                          (                                                1                                      R                    D                                                  +                                  sC                  db                                            )                                      =        0                            (                  Eq          .                                          ⁢          6                )            From (6), VX can be expressed as
                              V          x                =                                            V              OUT                        ⁡                          (                                                sC                  gd                                +                                  1                                      R                    D                                                  +                                  sC                  db                                            )                                                          g              m                        -                          sC              gd                                                          (                  Eq          .                                          ⁢          7                )            By substituting &7) into equation (5), the transfer function of the amplifier is:
                                                                                                              V                    OUT                                                        V                    IN                                                  ⁢                                  (                  s                  )                                            =                                                                                                                      ⁢                                                                    (                                                                  sC                        gd                                            -                                              g                        m                                                              )                                    ⁢                                      R                    D                                                                                                              R                      S                                        ⁢                                          R                      D                                        ⁢                    ξ                    ⁢                                                                                  ⁢                                          s                      2                                                        +                                                            [                                                                                                                                  R                              S                                                        ⁡                                                          (                                                              1                                +                                                                                                      g                                    m                                                                    ⁢                                                                      R                                    D                                                                                                                              )                                                                                ⁢                                                      C                            gd                                                                          +                                                                              R                            s                                                    ⁢                                                      C                            gs                                                                          +                                                                              R                            D                                                    ⁡                                                      (                                                                                          C                                gd                                                            +                                                              C                                db                                                                                      )                                                                                              ]                                        ⁢                    s                                    +                  1                                                                                                                        And                ⁢                                                                  ⁢                ξ                            =                                                                    C                    gs                                    ⁢                                      C                    gd                                                  +                                                      C                    gd                                    ⁢                                      C                    db                                                  +                                                      C                    gs                                    ⁢                                      C                    db                                                                                                          (                  Eq          .                                          ⁢          8                )            By manipulating equation (8), the denominator can be expressed as:
                    D        =                                            (                                                s                                      ω                    p1                                                  +                1                            )                        ⁢                          (                                                s                                      ω                    p2                                                  +                1                            )                                =                                                    s                2                                                              ω                  p1                                ⁢                                  ω                  p2                                                      +                                          (                                                      1                                          ω                      p1                                                        +                                      1                                          ω                      p2                                                                      )                            ⁢              s                        +            1                                              (                  Eq          .                                          ⁢          9                )            The coefficient of s is approximately equal to 1/ωp1 if ωp2 is much higher in frequency. It follows from equation (8) and (9) that the poles are located at:
                              ω          p1                =                  1                                                                      R                  S                                ⁡                                  (                                      1                    +                                                                  g                        m                                            ⁢                                              R                        D                                                                              )                                            ⁢                              C                gd                                      +                                          R                S                            ⁢                              C                gs                                      +                                          R                D                            ⁡                              (                                                      C                    gd                                    +                                      C                    db                                                  )                                                                        (                  Eq          .                                          ⁢          10                )                        and                                                                ω          p2                =                                                                              R                  S                                ⁡                                  (                                      1                    +                                                                  g                        m                                            ⁢                                              R                        D                                                                              )                                            ⁢                              C                gd                                      +                                          R                S                            ⁢                              C                gs                                      +                                          R                D                            ⁡                              (                                                      C                    gd                                    +                                      C                    db                                                  )                                                                        R              S                        ⁢                                          R                D                            ⁡                              (                                                                            C                      gs                                        ⁢                                          C                      gd                                                        +                                                            C                      gs                                        ⁢                                          C                      db                                                        +                                                            C                      gd                                        ⁢                                          C                      db                                                                      )                                                                        (                  Eq          .                                          ⁢          11                )            This analysis shows that the amplifier gain falls off at about 40 dB per decade of frequency at high frequencies, i.e. frequencies that are likely to be in the range of the data of the fiber optic receiver of FIG. 1.
To achieve higher bandwidth for the amplifier of FIG. 4 in the prior art, an inductance is introduced in series with the load resistor RD. FIG. 5 shows what is called a shunt peaked amplifier of the prior art. FIG. 5a shows shows capacitance Co which includes all the capacitance at the output node. When the amplifier is used for a wideband application, inductance LS in series with resistance RD extends the bandwidth of the amplifier that is normally limited by the time constant RDCo, by introducing a zero as shown by the small signal analysis represented by the circuit of FIG. 5b as follows:
                                          V            OUT                                V            IN                          =                                                            g                m                            ⁡                              (                                                      R                    D                                    +                                      j                    ⁢                                                                                  ⁢                    ω                    ⁢                                                                                  ⁢                                          L                      S                                                                      )                                                    1              +                              j                ⁢                                                                  ⁢                ω                ⁢                                                                  ⁢                                  R                  D                                ⁢                                  C                  o                                            -                                                ω                  2                                ⁢                                  L                  S                                ⁢                                  C                  o                                                              ⁢                                          ⁢                                          =                                                    g                m                            ⁢                                                R                  D                                ⁡                                  (                                      1                    +                                          j                      ⁢                                                                                          ⁢                      ωτ                      ⁢                                                                                          ⁢                      m                                                        )                                                                    1              +                              j                ⁢                                                                  ⁢                ωτ                            ⁢                                                          -                                                ω                  2                                ⁢                                  τ                  2                                ⁢                m                                                                        (                  Eq          .                                          ⁢          12                )            Where τ=RDCo and m=LS/(RD2Co)
In order to bring the resonance frequency close to the bandwidth of the fiber optic receiver of FIG. 1, the value of LS would be unrealistically high and therefore too large to fit on a typical integrated circuit.