Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼...

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Transcript of Transistor Circuits I - Cleveland Institute of Electronics continued •h FB = α = 𝐼...

  • Transistor Circuits I

    Common-Base, DC operation

  • The humble transistor


    Emitter (E)

    Collector (C)

    Base (B)

  • Transistor basics

    Emitter to base junction is forward biased (normally)

    Collector to base junction is reverse biased (normally)

    Transistors are current operated devices, so KCL should be applied first:

    IE = IC + IB

  • Basics continued

    Leakage current: ICBO (Emitter open)

    Usually is considered negligible, but can affect things when IC is small

  • Basics continued

    hFB = =

    ( 1)

    hFE = =

    ( has no definitive limit)

    The symbols hFB and hFE are BJT parameters; the h means Hybrid parameter. The subscript FB means Forward current transfer ratio of the common, or grounded, Base circuit, whereas FE means Forward current transfer ratio of the common, or grounded, Emitter circuit.

  • Basic configuration of Common-Base

  • First circuit

    If VEE = 20V and VEB is negligible, find IE when RE equals (a) 80k, (b) 40k, (c) 20k, (d) 10k, (e) 5k, and (f) 1k.

  • Work for first circuit

    Formula is =



    (a) =20V

    80k= 0.25mA = 250A;

    (b) =20V

    40k= 0.5mA = 500A;

    (c) =20V

    20k= 1mA;

    (d) =20V

    10k= 2mA;

    (e) =20V

    5k= 4mA;

    (f) =20V

    1k= 20mA

  • Use approximation for VEB If VEB = 0.7V, rework previous values to determine IE in each instance.

    Formula is now =





    (a) =19.3V

    80k= 241.25A;

    (b) =19.3V

    40k= 482.5A;

    (c) =19.3V

    20k= 965A;

    (d) =19.3V

    10k= 1.93mA;

    (e) =19.3V

    5k= 3.86mA;

    (f) =19.3V

    1k= 19.3mA

  • Proving the point (Measuring (a))

  • Measuring (b)

  • Measuring (c)

  • Measuring (d)

  • Measuring (e)

  • Measuring (f)

  • PNP versions

  • Second circuit

    If RE = 10k and VEB = 0V, what are the values of IE, IC and VCB when RC equals (a) 5k, (b) 8k, (c) 10k, (d) 12k. Assume hFB = = 1.


    18V -22V

  • Points to ponder

    Since IE will remain constant, use the value found from calculating through the derivation

    of Ohms Law: =



    10k= 1.8mA.

    Additionally, we are treating as 1, which

    means using the formula

    = 1 translates into

    IE and IC being equal. Therefore, the value of IC for all changes of RC remains a constant 1.8mA (so long as saturation is avoided)

  • Additional points

    Following KVL, VCC = VRC + VCB. Since VRC = ICRC (Ohms Law), substituting this into the original formula gives us VCC = ICRC + VCB. Reworking the equation to solve for VCB yields VCB = VCC ICRC (You can use the absolute value for VCC when calculating, but remember to include polarity when writing the final answer).

  • Work through for circuit

    (a) = = 22 1.8mA 5k =22 9 = 13V (Remember the polarity of VCC is negative)

    (b) = = 22 1.8mA 8k =22 14.4 = 7.6V

    (c) = = 22 1.8mA 10k =22 18 = 4V

    (d) = = 22 1.8mA 12k =22 21.6 = 0.4V

  • Third circuit

    If VCC = 24V, find (a) the saturation current, IC(sat). If VEB = 0.7V, what are the values of IE, IC, and VCB when RE equals (b) 20k, (c) 10k, and (d) 5k. Assume DC = 1.

  • IC(sat) and some constants in our calculations

    (a) () =



    5k= 3mA

    VRE = VEE VEB = 20 0.7 = 19.3V (technically this should be written as -19.3V since VEE is negative in polarity)

  • Calculating the rest

    (b) =



    20k= 965A = 965A;

    VCB = VCC ICRC = 24 (965A)(8k) = 24 7.72 = 16.28V

  • And the beat goes on

    (c) =



    10k= 1.93mA =

    1.93mA; VCB = VCC ICRC = 24 (1.93mA)(8k) = 24 15.44 = 8.56V

  • And on

    (d) =



    5k= 3.86mA =

    3mA (this cannot be exceeded); VCB = VCC ICRC = 24 (3mA)(8k) = 24 24 = 0V

  • Measure to make sure (a)

  • (b)

  • (c)

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