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Calculate the common operating point parameters for the transistor circuit shown in Fig. 4.118.

Computer Science

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Okay, I have the diagram now (Fig. 4.118). Here are the calculations for the common operating point parameters:

Given:
VCC=16 VV_{CC} = 16 \, \text{V}
RB=510 kΩ=510×103 ΩR_B = 510 \, \text{k}\Omega = 510 \times 10^3 \, \Omega
RC=1.8 kΩ=1.8×103 ΩR_C = 1.8 \, \text{k}\Omega = 1.8 \times 10^3 \, \Omega
β=120\beta = 120
The emitter is grounded, so VE=0 VV_E = 0 \, \text{V}.
Assume VBE=0.7 VV_{BE} = 0.7 \, \text{V} for a silicon transistor.

Part a· 1 step

VBV_B

  1. Since the emitter is grounded (VE=0 VV_E = 0 \, \text{V}), the base voltage VBV_B is equal to VBEV_{BE}.

    VB=VBE+VEV_B = V_{BE} + V_E VB=0.7 V+0 VV_B = 0.7 \, \text{V} + 0 \, \text{V} VB=0.7 VV_B = 0.7 \, \text{V}

    The base voltage is 0.7 V\boxed{\text{0.7 V}}.

Part b· 4 steps

IBQI_{BQ}

  1. Apply Kirchhoff's Voltage Law (KVL) to the base-emitter loop.

    VCC−IBQRB−VBE=0V_{CC} - I_{BQ} R_B - V_{BE} = 0

  2. Rearrange to solve for IBQI_{BQ}.

    IBQ=VCC−VBERBI_{BQ} = \frac{V_{CC} - V_{BE}}{R_B}

  3. Substitute the given values.

    IBQ=16 V−0.7 V510×103 ΩI_{BQ} = \frac{16 \, \text{V} - 0.7 \, \text{V}}{510 \times 10^3 \, \Omega} IBQ=15.3 V510×103 ΩI_{BQ} = \frac{15.3 \, \text{V}}{510 \times 10^3 \, \Omega}

  4. Calculate IBQI_{BQ}.

    IBQ=30×10−6 A=30 μAI_{BQ} = 30 \times 10^{-6} \, \text{A} = 30 \, \mu\text{A}

    The base current is 30 \muA\boxed{\text{30 \mu A}}.

Part c· 3 steps

ICQI_{CQ}

  1. Use the relationship between collector current and base current.

    ICQ=βIBQI_{CQ} = \beta I_{BQ}

  2. Substitute the calculated IBQI_{BQ} and given β\beta.

    ICQ=120×(30×10−6 A)I_{CQ} = 120 \times (30 \times 10^{-6} \, \text{A})

  3. Calculate ICQI_{CQ}.

    ICQ=3600×10−6 A=3.6 mAI_{CQ} = 3600 \times 10^{-6} \, \text{A} = 3.6 \, \text{mA}

    The collector current is 3.6 mA\boxed{\text{3.6 mA}}.

Part d· 4 steps

VCV_C

  1. Apply KVL to the collector-emitter loop.

    VCC−ICQRC−VC=0V_{CC} - I_{CQ} R_C - V_C = 0

  2. Rearrange to solve for VCV_C.

    VC=VCC−ICQRCV_C = V_{CC} - I_{CQ} R_C

  3. Substitute the values.

    VC=16 V−(3.6×10−3 A)×(1.8×103 Ω)V_C = 16 \, \text{V} - (3.6 \times 10^{-3} \, \text{A}) \times (1.8 \times 10^3 \, \Omega) VC=16 V−6.48 VV_C = 16 \, \text{V} - 6.48 \, \text{V}

  4. Calculate VCV_C.

    VC=9.52 VV_C = 9.52 \, \text{V}

    The collector voltage is 9.52 V\boxed{\text{9.52 V}}.

Part e· 2 steps

VCEQV_{CEQ}

  1. Use the definition of VCEQV_{CEQ}.

    VCEQ=VC−VEV_{CEQ} = V_C - V_E

  2. Substitute the calculated VCV_C and VE=0 VV_E = 0 \, \text{V}.

    VCEQ=9.52 V−0 VV_{CEQ} = 9.52 \, \text{V} - 0 \, \text{V} VCEQ=9.52 VV_{CEQ} = 9.52 \, \text{V}

    The collector-emitter voltage is 9.52 V\boxed{\text{9.52 V}}.

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