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For a BJT circuit with Vcc=12V, Rb=240kΩ, Rc=2.2kΩ, β=100, and Vbe=0.7V, calculate Ib, Ic, Vce, and confirm the operating region.

Computer Science

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2 parts
a

47.083 μA47.083\,\mu A

4.7083 mA4.7083\,mA

1.64174 V1.64174\,V

c
101

Step-by-step solution

5 steps

Here are the calculations for the BJT circuit:

Part a· 3 steps

Find IBI_B, ICI_C, VCEV_{CE}

  1. Calculate the base current IBI_B.

    The formula for the base current in a fixed-bias configuration is:

    IB=VCC−VBERBI_B = \frac{V_{CC} - V_{BE}}{R_B}

    Substitute the given values:

    IB=12 V−0.7 V240×103 ΩI_B = \frac{12\,V - 0.7\,V}{240 \times 10^3\,\Omega} IB=11.3 V240×103 ΩI_B = \frac{11.3\,V}{240 \times 10^3\,\Omega} IB=4.7083×10−5 AI_B = 4.7083 \times 10^{-5}\,A IB=47.083 μAI_B = 47.083\,\mu A

  2. Calculate the collector current ICI_C.

    The formula for the collector current is:

    IC=β⋅IBI_C = \beta \cdot I_B

    Substitute the values:

    IC=100⋅(47.083×10−6 A)I_C = 100 \cdot (47.083 \times 10^{-6}\,A) IC=4.7083×10−3 AI_C = 4.7083 \times 10^{-3}\,A IC=4.7083 mAI_C = 4.7083\,mA

  3. Calculate the collector-emitter voltage VCEV_{CE}.

    The formula for the collector-emitter voltage is:

    VCE=VCC−IC⋅RCV_{CE} = V_{CC} - I_C \cdot R_C

    Substitute the values:

    VCE=12 V−(4.7083×10−3 A)⋅(2.2×103 Ω)V_{CE} = 12\,V - (4.7083 \times 10^{-3}\,A) \cdot (2.2 \times 10^3\,\Omega) VCE=12 V−10.35826 VV_{CE} = 12\,V - 10.35826\,V VCE=1.64174 VV_{CE} = 1.64174\,V

    The calculated values are:

    IB=47.083 μAI_B = \boxed{47.083\,\mu A} IC=4.7083 mAI_C = \boxed{4.7083\,mA} VCE=1.64174 VV_{CE} = \boxed{1.64174\,V}

Part b

Confirm the operating region

To confirm the operating region, we compare VCEV_{CE} with VCE,satV_{CE,sat} (typically 0.2 V0.2\,V) and check if ICI_C is less than the saturation current IC,satI_{C,sat}.

First, calculate the maximum possible collector current (saturation current) if the transistor were in saturation:

IC,sat=VCC−VCE,satRCI_{C,sat} = \frac{V_{CC} - V_{CE,sat}}{R_C}

Assuming VCE,sat=0.2 VV_{CE,sat} = 0.2\,V:

IC,sat=12 V−0.2 V2.2×103 ΩI_{C,sat} = \frac{12\,V - 0.2\,V}{2.2 \times 10^3\,\Omega} IC,sat=11.8 V2.2×103 ΩI_{C,sat} = \frac{11.8\,V}{2.2 \times 10^3\,\Omega} IC,sat=5.3636×10−3 AI_{C,sat} = 5.3636 \times 10^{-3}\,A IC,sat=5.3636 mAI_{C,sat} = 5.3636\,mA

Now, compare the calculated ICI_C and VCEV_{CE}:

  • Our calculated IC=4.7083 mAI_C = 4.7083\,mA is less than IC,sat=5.3636 mAI_{C,sat} = 5.3636\,mA.
  • Our calculated VCE=1.64174 VV_{CE} = 1.64174\,V is greater than VCE,sat=0.2 VV_{CE,sat} = 0.2\,V.

Since VBE=0.7 VV_{BE} = 0.7\,V (forward-biased base-emitter junction) and VCE>VCE,satV_{CE} > V_{CE,sat}, the transistor is operating in the active region.

Part c101

Calculate the stability factor SS and comment on its significance.

  1. Calculate the stability factor SS.

    For a fixed-bias configuration, the stability factor SS (with respect to ICOI_{CO}) is given by:

    S=1+βS = 1 + \beta

    Substitute the given β\beta:

    S=1+100S = 1 + 100 S=101S = \boxed{101}

  2. Comment on its significance.

    The stability factor SS indicates how much the collector current ICI_C changes due to variations in the reverse saturation current ICOI_{CO} (which is highly temperature-dependent). A high value of SS (like 101) means that the collector current ICI_C is very sensitive to changes in ICOI_{CO}. This implies that the operating point of the fixed-bias circuit is unstable with respect to temperature variations, making it prone to thermal runaway.

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