Solving a Classic Example of a Transistor Bias Circuit: Exam Walkthrough

For a 12V supply with a 22kΩ/4.7kΩ base divider, a 2.2kΩ collector resistor, and a 1kΩ emitter resistor, the exact DC operating point (Q-point) is I_C = 1.35 mA and V_CE = 7.68 V. This places the transistor firmly in the forward-active region, ideal for linear amplification. Below is the complete, step-by-step derivation of these values, the common exam traps to avoid, and the physical component selection framework.

The Problem Statement: Voltage-Divider Bias Example of a Transistor

Method Selection and The "Stiff Divider" Trap

To solve this example of a transistor biasing network, you must use Thevenin's Theorem combined with Kirchhoff's Voltage Law (KVL). Thevenin's theorem allows us to reduce the R₁/R₂ voltage divider and the V_CC supply into a single equivalent voltage source (V_TH) and series resistance (R_TH) driving the base.

Step-by-Step Algebraic Solution

Follow these numbered steps to derive the exact Q-point. Do not skip the algebraic substitutions, as partial credit is often tied to showing the KVL loop setup.

1. Calculate Thevenin Resistance (R_TH):
   R_TH = R₁ || R₂ = (R₁ × R₂) / (R₁ + R₂)
   R_TH = (22,000 × 4,700) / (22,000 + 4,700) = 103,400,000 / 26,700 = 3,872 Ω (3.87 kΩ)
2. Calculate Thevenin Voltage (V_TH):
   V_TH = V_CC × [R₂ / (R₁ + R₂)]
   V_TH = 12 × [4,700 / 26,700] = 2.112 V
3. Write the Base-Emitter KVL Equation:
   Starting from V_TH and moving through the base-emitter junction to ground:
   V_TH - I_BR_TH - V_BE - I_ER_E = 0
4. Substitute I_E with I_B:
   We know I_E = (β + 1)I_B. Substitute this into the KVL equation to solve for a single variable (I_B):
   2.112 - I_B(3,872) - 0.7 - I_B(101)(1,000) = 0
   1.412 = I_B(3,872 + 101,000)
   1.412 = I_B(104,872)
5. Solve for Base Current (I_B):
   I_B = 1.412 / 104,872 = 13.46 μA
6. Calculate Collector and Emitter Currents:
   I_C = β × I_B = 100 × 13.46 μA = 1.346 mA
   I_E = (β + 1) × I_B = 101 × 13.46 μA = 1.360 mA
7. Write the Collector-Emitter KVL Equation:
   V_CC - I_CR_C - V_CE - I_ER_E = 0
   12 - (1.346 mA × 2,200) - V_CE - (1.360 mA × 1,000) = 0
   12 - 2.961 - V_CE - 1.360 = 0
   12 - 4.321 = V_CE
8. Solve for V_CE:
   V_CE = 7.679 V

Sanity Check and Independent Verification

Before moving to the next exam question, run a rapid sanity check on your order of magnitude and units:

• Voltage Bounds: V_CE must be strictly between 0 V (saturation) and V_CC (cutoff). Our answer, 7.68 V, is perfectly centered around V_CC/2 (6 V), confirming the circuit is biased for maximum symmetrical AC swing.
• Current Bounds: The absolute maximum I_C (if V_CE = 0 and R_E = 0) would be 12V / 2.2kΩ = 5.45 mA. Our 1.35 mA is well below this physical limit.
• Unit Check: μA multiplied by unitless β yields mA. mA multiplied by kΩ yields Volts. The dimensional analysis holds.

Decision Path: Picking the Physical Transistor Part

Knowing the theoretical Q-point is only half the battle. For a practical build, you must select a physical component that can handle the calculated stress. Based on I_C = 1.35 mA and V_CE = 7.68 V, the power dissipated by the transistor is P_D = V_CE × I_C = 10.37 mW. Use the decision matrix below to pick your part.

Final Recommendation: Select the ON Semiconductor 2N3904. Our calculated P_D of 10.37 mW is less than 2% of its 625 mW maximum rating, and the 1.35 mA collector current is well under its 200 mA limit. The 2N2222 is unnecessary overkill for this specific signal-level bias point, and the TIP31C would be a waste of board space and budget. For authoritative datasheet limits, always cross-reference the ON Semiconductor 2N3904 datasheet and foundational texts like Electronics Tutorials on BJT Biasing.

Frequently Asked Questions

What happens to V_CE if the transistor heats up?

As the silicon die heats up, the base-emitter voltage drop (V_BE) decreases by roughly 2 mV/°C, and β increases. Both effects cause I_C to rise. Because I_C rises, the voltage drop across R_C increases, which forces V_CE to drop. The emitter resistor (R_E) provides negative feedback to stabilize this: as I_E rises, V_E rises, which reduces the forward bias V_BE and throttles the current back down. This is why R_E is critical for thermal stability.

Can I use the approximate method if β is 300?

If β = 300, the term 0.1(β × R_E) becomes 30 kΩ. Since our R_TH is 3.87 kΩ, the condition R_TH ≤ 0.1(βR_E) is strongly satisfied. In that specific scenario, the approximate method (ignoring I_B loading) will yield an answer within 1-2% of the exact Thevenin method. However, in an exam setting, unless the prompt explicitly says "use approximate analysis," default to the exact method to guarantee full credit.