Let's analyze the circuit step by step. **Given:** - β = 100 - V_EB = 0.7 V - V_BE = 0.7 V - R1 = 100 kΩ - R2 = 50 kΩ - RE1 = 2 kΩ - RE2 = 2 kΩ - RC1 = 5 kΩ - RC2 = 10 kΩ - V_CC = +5 V - V_EE = -5 V Let's use the following notation for Q1 and Q2: - Q1: Base = B1, Emitter = E1, Collector = C1 - Q2: Base = B2, Emitter = E2, Collector = C2 --- ### **Step 1: Find Base Voltage V_B1 of Q1** V_B1 is found using the voltage divider rule: \[ V_{B1} = \frac{R2}{R1 + R2} \cdot (V_{CC} - V_{EE}) \] \[ V_{B1} = \frac{50k}{100k + 50k} \cdot (5 - (-5)) \] \[ V_{B1} = \frac{1}{3} \cdot 10\ \text{V} = 3.33\ \text{V} \] --- ### **Step 2: Find Emitter Voltage V_E1 of Q1** \[ V_{E1} = V_{B1} - V_{BE} \] \[ V_{E1} = 3.33\ \text{V} - 0.7\ \text{V} = 2.63\ \text{V} \] --- ### **Step 3: Find Emitter Current I_E1 of Q1** \[ I_{E1} = \frac{V_{E1} - V_{EE}}{RE1} \] \[ I_{E1} = \frac{2.63 - (-5)}{2k} \] \[ I_{E1} = \frac{7.63}{2000} \] \[ I_{E1} = 3.815\ \text{mA} \] --- ### **Step 4: Find Collector Current I_C1 and Base Current I_B1 of Q1** Since β is large, \( I_{E1} \approx I_{C1} \) (as \( I_E \approx I_C \) for large β) \[ I_{C1} \approx I_{E1} = 3.815\ \text{mA} \] \[ I_{B1} = \frac{I_{C1}}{\beta} = \frac{3.815\ \text{mA}}{100} = 0.038\ \text{mA} = 38\ \mu\text{A} \] --- ### **Step 5: Find Collector Voltage V_C1** \[ V_{C1} = V_{CC} - (I_{C1} \cdot RC1) \] \[ V_{C1} = 5 - (3.815\ \text{mA} \times 5k) \] \[ V_{C1} = 5 - (3.815 \times 5) \text{V} \] \[ V_{C1} = 5 - 19.075 = -14.075\ \text{V} \] --- ### **Step 6: Find Base Voltage V_B2 of Q2** V_B2 = V_C1 (collector of Q1 is base of Q2) \[ V_{B2} = -14.075\ \text{V} \] --- ### **Step 7: Find Emitter Voltage V_E2 of Q2** \[ V_{E2} = V_{B2} - V_{BE} \] \[ V_{E2} = -14.075 - 0.7 = -14.775\ \text{V} \] --- ### **Step 8: Find Emitter Current I_E2 of Q2** \[ I_{E2} = \frac{V_{E2} - V_{EE}}{RE2} \] \[ I_{E2} = \frac{-14.775 - (-5)}{2k} \] \[ I_{E2} = \frac{-9.775}{2000} \] \[ I_{E2} = -4.8875\ \text{mA} \] (Negative sign indicates direction, but magnitude is **4.89 mA**) --- ### **Step 9: Find Collector Current I_C2 and Base Current I_B2 of Q2** \[ I_{C2} \approx I_{E2} = 4.89\ \text{mA} \] \[ I_{B2} = \frac{I_{C2}}{\beta} = \frac{4.89}{100} = 0.0489\ \text{mA} = 48.9\ \mu\text{A} \] --- ### **Step 10: Find Collector Voltage V_C2** \[ V_{C2} = V_{CC} - (I_{C2} \cdot RC2) \] \[ V_{C2} = 5 - (4.89\ \text{mA} \times 10k) \] \[ V_{C2} = 5 - (4.89 \times 10) = 5 - 48.9 = -43.9\ \text{V} \] --- ## **Summary Table** | Quantity | Value | |-----------|-------| | V_B1 | 3.33 V | | V_E1 | 2.63 V | | I_E1, I_C1 | 3.815 mA | | I_B1 | 38 μA | | V_C1 | -14.08 V | | V_B2 | -14.08 V | | V_E2 | -14.78 V | | I_E2, I_C2 | 4.89 mA | | I_B2 | 48.9 μA | | V_C2 | -43.9 V | **Note:** The negative voltages indicate that the collector voltages are below ground, which makes sense given the dual supply. --- ## **Final Answer** - **I_C1 = I_E1 ≈ 3.82 mA** - **V_C1 ≈ -14.08 V** - **I_C2 = I_E2 ≈ 4.89 mA** - **V_C2 ≈ -43.9 V** Let me know if you want me to clarify or show any step in more detail!

Certainly! Let's analyze the circuit thoroughly step by step to find all the voltages and currents. The circuit involves two transistors (Q1 and Q2), resistors, power supplies, and voltage divider biasing. Given data: ### Given Data: - **β (beta)** (common for both BJTs): 100 - **V_EB** = 0.7 V (Emitter-Base voltage) - **V_BE** = 0.7 V (Base-Emitter voltage) - **R1** = 100 kΩ - **R2** = 50 kΩ - **RE1** = 2 kΩ - **RE2** = 2 kΩ - **RC1** = 5 kΩ - **RC2** = 10 kΩ - **V_CC** = +5 V (Positive supply) - **V_EE** = -5 V (Negative supply) --- ### **Step 1: Determine Base Voltage of Q1 (V_B1)** Q1's base is set by a voltage divider formed by R1 and R2 connected between V_CC and V_EE: \[ V_{B1} = V_{R2} = \frac{R2}{R1 + R2} \times (V_{CC} - V_{EE}) + V_{EE} \] Calculating the voltage divider: \[ V_{B1} = \frac{50k}{100k + 50k} \times (5 - (-5)) + (-5) \] \[ V_{B1} = \frac{50k}{150k} \times 10 + (-5) = \frac{1}{3} \times 10 - 5 = \frac{10}{3} - 5 \approx 3.33 - 5 = -1.67\,V \] *Wait,* this seems inconsistent with earlier. The earlier approach considered the voltage divider connecting between +5V and -5V, so the voltage at the junction (V_B1) is: \[ V_{B1} = \frac{R2}{R1 + R2} \times (V_{CC} - V_{EE}) + V_{EE} \] Let's evaluate precisely: \[ V_{B1} = \frac{50k}{150k} \times (5 - (-5)) + (-5) = \frac{1}{3} \times 10 - 5 = 3.33 - 5 = -1.67\,V \] **So, the base of Q1 is at approximately -1.67 V.** --- ### **Step 2: Find Emitter Voltage V_E1** Since V_BE ≈ 0.7 V for a forward-biased BJT: \[ V_{E1} = V_{B1} - V_{BE} \approx -1.67\,V - 0.7\,V = -2.37\,V \] *Note:* The negative voltage indicates the emitter of Q1 is at about -2.37 V relative to ground. --- ### **Step 3: Calculate Emitter Current I_E1 of Q1** Using Ohm's law and considering the emitter resistor RE1: \[ I_{E1} = \frac{V_{E1} - V_{EE}}{RE1} \] \[ V_{E1} - V_{EE} = -2.37\,V - (-5\,V) = 2.63\,V \] \[ I_{E1} = \frac{2.63\,V}{2\,k\Omega} = 1.315\,mA \] Since β is large, the collector current I_C1 ≈ I_E1: \[ I_{C1} \approx 1.315\,mA \] And the base current: \[ I_{B1} = \frac{I_{C1}}{\beta} \approx \frac{1.315\,mA}{100} = 13.15\,\mu A \] --- ### **Step 4: Find Collector Voltage V_C1** The collector of Q1 is connected through RC1 to V_CC: \[ V_{C1} = V_{CC} - (I_{C1} \times RC1) \] \[ V_{C1} = 5\,V - (1.315\,mA \times 5\,k\Omega) = 5 - (1.315 \times 5) = 5 - 6.575 = -1.575\,V \] *The collector voltage is approximately -1.58 V.* --- ### **Step 5: Determine Base Voltage of Q2 (V_B2)** Q2's base is connected to Q1's collector: \[ V_{B2} = V_{C1} \approx -1.58\,V \] --- ### **Step 6: Find Emitter Voltage V_E2** Again, assuming V_BE ≈ 0.7 V: \[ V_{E2} = V_{B2} - V_{BE} = -1.58\,V - 0.7\,V = -2.28\,V \] --- ### **Step 7: Calculate Emitter Current I_E2** Using the emitter resistor RE2: \[ I_{E2} = \frac{V_{E2} - V_{EE}}{RE2} \] \[ V_{E2} - V_{EE} = -2.28\,V - (-5\,V) = 2.72\,V \] \[ I_{E2} = \frac{2.72\,V}{2\,k\Omega} = 1.36\,mA \] Collector current: \[ I_{C2} \approx I_{E2} = 1.36\,mA \] Base current: \[ I_{B2} = \frac{I_{C2}}{\beta} \approx \frac{1.36\,mA}{100} = 13.6\,\mu A \] --- ### **Step 8: Find Collector Voltage V_C2** Q2's collector is connected through RC2: \[ V_{C2} = V_{CC} - (I_{C2} \times RC2) = 5\,V - (1.36\,mA \times 10\,k\Omega) \] \[ V_{C2} = 5 - (1.36 \times 10) = 5 - 13.6 = -8.6\,V \] --- ### **Summary of Calculated Values:** | Quantity | Approximate Value | Explanation | |------------|---------------------|--------------| | **V_B1** | **-1.67 V** | Base of Q1, from voltage divider | | **V_E1** | **-2.37 V** | Emitter of Q1, V_B1 - V_BE | | **I_E1** | **1.315 mA** | Emitter current of Q1 | | **V_C1** | **-1.58 V** | Collector of Q1, V_CC - I_C1×RC1 | | **V_B2** | **-1.58 V** | Collector of Q1, base of Q2 | | **V_E2** | **-2.28 V** | V_B2 - V_BE | | **I_E2** | **1.36 mA** | Emitter current of Q2 | | **V_C2** | **-8.6 V** | Collector of Q2, V_CC - I_C2×RC2 | --- ### **Final notes:** - The negative voltages indicate these nodes are below ground potential, consistent with the dual power supplies. - The currents are typical for BJTs with β=100 under these biasing conditions. - The calculations assume ideal BJTs with V_BE ≈ 0.7 V. --- **Would you like me to clarify any specific part, or include the effect of transistor saturation or other parameters?**

Certainly! Here's the detailed step-by-step solution formatted in LaTeX: ```latex \documentclass{article} \usepackage{amsmath} \begin{document} \section*{Step-by-step Analysis of the Circuit} \textbf{Given Data:} \begin{align*} \beta &= 100 \\ V_{EB} &= 0.7\,V \\ V_{BE} &= 0.7\,V \\ R_1 &= 100\,k\Omega \\ R_2 &= 50\,k\Omega \\ RE_1 &= 2\,k\Omega \\ RE_2 &= 2\,k\Omega \\ RC_1 &= 5\,k\Omega \\ RC_2 &= 10\,k\Omega \\ V_{CC} &= +5\,V \\ V_{EE} &= -5\,V \end{align*} \section*{Step 1: Base Voltage of Q1, \(V_{B1}\)} The voltage divider formed by \(R_1\) and \(R_2\) between \(V_{CC}\) and \(V_{EE}\): \begin{align*} V_{B1} &= \frac{R_2}{R_1 + R_2} \times (V_{CC} - V_{EE}) + V_{EE} \\ &= \frac{50\,k\Omega}{150\,k\Omega} \times (5 - (-5)) + (-5) \\ &= \frac{1}{3} \times 10 - 5 = \frac{10}{3} - 5 \approx 3.33 - 5 = -1.67\,V \end{align*} --- \section*{Step 2: Emitter Voltage of Q1, \(V_{E1}\)} \[ V_{E1} = V_{B1} - V_{BE} \approx -1.67\,V - 0.7\,V = -2.37\,V \] --- \section*{Step 3: Emitter Current of Q1, \(I_{E1}\)} \[ I_{E1} = \frac{V_{E1} - V_{EE}}{RE_1} = \frac{-2.37\,V - (-5\,V)}{2\,k\Omega} = \frac{2.63\,V}{2\,k\Omega} = 1.315\,mA \] \textbf{Since} \(\beta\) is large, \(\ I_{C1} \approx I_{E1} \). --- \section*{Step 4: Collector Voltage of Q1, \(V_{C1}\)} \[ V_{C1} = V_{CC} - I_{C1} \times RC_1 = 5\,V - (1.315\,mA \times 5\,k\Omega) = 5 - (1.315 \times 5) = 5 - 6.575 = -1.58\,V \] --- \section*{Step 5: Base Voltage of Q2, \(V_{B2}\)} \[ V_{B2} \approx V_{C1} \approx -1.58\,V \] --- \section*{Step 6: Emitter Voltage of Q2, \(V_{E2}\)} \[ V_{E2} = V_{B2} - V_{BE} = -1.58\,V - 0.7\,V = -2.28\,V \] --- \section*{Step 7: Emitter Current of Q2, \(I_{E2}\)} \[ I_{E2} = \frac{V_{E2} - V_{EE}}{RE_2} = \frac{-2.28\,V - (-5\,V)}{2\,k\Omega} = \frac{2.72\,V}{2\,k\Omega} = 1.36\,mA \] \textbf{Similarly,} \(I_{C2} \approx I_{E2}\). --- \section*{Step 8: Collector Voltage of Q2, \(V_{C2}\)} \[ V_{C2} = V_{CC} - I_{C2} \times RC_2 = 5\,V - (1.36\,mA \times 10\,k\Omega) = 5 - (1.36 \times 10) = 5 - 13.6 = -8.6\,V \] --- \section*{Summary of Results:} \begin{table}[h!] \centering \begin{tabular}{|l|c|} \hline \textbf{Quantity} & \textbf{Value} \\ \hline \(V_{B1}\) & \textbf{-1.67\,V} \\ \(V_{E1}\) & \textbf{-2.37\,V} \\ \(I_{E1} \approx I_{C1}\) & \textbf{1.315\,mA} \\ \(V_{C1}\) & \textbf{-1.58\,V} \\ \(V_{B2}\) & \textbf{-1.58\,V} \\ \(V_{E2}\) & \textbf{-2.28\,V} \\ \(I_{E2} \approx I_{C2}\) & \textbf{1.36\,mA} \\ \(V_{C2}\) & \textbf{-8.6\,V} \\ \hline \end{tabular} \caption{Calculated Voltages and Currents} \end{table} --- \textbf{Note:} The negative voltages indicate potential below ground, consistent with the dual supply. \end{document} ``` This LaTeX code provides a clear, formatted, step-by-step solution with all calculations and explanations. You can compile it in any LaTeX editor for a neatly formatted document.

Certainly! Here's a clean, well-formatted LaTeX document with the detailed step-by-step solution, including all calculations and explanations: ```latex \documentclass{article} \usepackage{amsmath} \begin{document} \section*{Analysis of the Circuit: Voltages and Currents} \textbf{Given Data:} \begin{align*} \beta &= 100 \\ V_{EB} &= 0.7\,V \\ V_{BE} &= 0.7\,V \\ R_1 &= 100\,k\Omega \\ R_2 &= 50\,k\Omega \\ RE_1 &= 2\,k\Omega \\ RE_2 &= 2\,k\Omega \\ RC_1 &= 5\,k\Omega \\ RC_2 &= 10\,k\Omega \\ V_{CC} &= +5\,V \\ V_{EE} &= -5\,V \end{align*} \section*{Step 1: Determine Base Voltage of Q1, \(V_{B1}\)} Using a voltage divider between \(V_{CC}\) and \(V_{EE}\): \begin{align*} V_{B1} &= \frac{R_2}{R_1 + R_2} \times (V_{CC} - V_{EE}) + V_{EE} \\ &= \frac{50\,k\Omega}{150\,k\Omega} \times (5 - (-5)) + (-5) \\ &= \frac{1}{3} \times 10 - 5 = \frac{10}{3} - 5 \approx 3.33 - 5 = -1.67\,V \end{align*} \section*{Step 2: Emitter Voltage of Q1, \(V_{E1}\)} \[ V_{E1} = V_{B1} - V_{BE} \approx -1.67\,V - 0.7\,V = -2.37\,V \] \section*{Step 3: Emitter Current of Q1, \(I_{E1}\)} \[ I_{E1} = \frac{V_{E1} - V_{EE}}{RE_1} = \frac{-2.37\,V - (-5\,V)}{2\,k\Omega} = \frac{2.63\,V}{2\,k\Omega} = 1.315\,mA \] Since \(\beta\) is large, \(I_{C1} \approx I_{E1} = 1.315\,mA\). \section*{Step 4: Collector Voltage of Q1, \(V_{C1}\)} \[ V_{C1} = V_{CC} - I_{C1} \times RC_1 = 5\,V - (1.315\,mA \times 5\,k\Omega) = 5 - (1.315 \times 5) = 5 - 6.575 = -1.58\,V \] \section*{Step 5: Base Voltage of Q2, \(V_{B2}\)} \[ V_{B2} \approx V_{C1} = -1.58\,V \] \section*{Step 6: Emitter Voltage of Q2, \(V_{E2}\)} \[ V_{E2} = V_{B2} - V_{BE} = -1.58\,V - 0.7\,V = -2.28\,V \] \section*{Step 7: Emitter Current of Q2, \(I_{E2}\)} \[ I_{E2} = \frac{V_{E2} - V_{EE}}{RE_2} = \frac{-2.28\,V - (-5\,V)}{2\,k\Omega} = \frac{2.72\,V}{2\,k\Omega} = 1.36\,mA \] Similarly, \(I_{C2} \approx I_{E2} = 1.36\,mA\). \section*{Step 8: Collector Voltage of Q2, \(V_{C2}\)} \[ V_{C2} = V_{CC} - I_{C2} \times RC_2 = 5\,V - (1.36\,mA \times 10\,k\Omega) = 5 - (1.36 \times 10) = 5 - 13.6 = -8.6\,V \] \section*{Summary of Results} \begin{table}[h!] \centering \begin{tabular}{|l|c|} \hline \textbf{Quantity} & \textbf{Value} \\ \hline \(V_{B1}\) & \textbf{-1.67\,V} \\ \(V_{E1}\) & \textbf{-2.37\,V} \\ \(I_{E1} \approx I_{C1}\) & \textbf{1.315\,mA} \\ \(V_{C1}\) & \textbf{-1.58\,V} \\ \(V_{B2}\) & \textbf{-1.58\,V} \\ \(V_{E2}\) & \textbf{-2.28\,V} \\ \(I_{E2} \approx I_{C2}\) & \textbf{1.36\,mA} \\ \(V_{C2}\) & \textbf{-8.6\,V} \\ \hline \end{tabular} \caption{Calculated Voltages and Currents} \end{table} \textbf{Note:} The negative voltages indicate these nodes are below ground potential, consistent with the dual power supplies. \end{document} ``` This code provides a clear, structured, and detailed LaTeX document of the analysis. You can compile it to generate a professional-looking report. If you'd like any further adjustments or additional explanations, feel free to ask!

Certainly! Here's the correct, detailed step-by-step analysis of the circuit, including all calculations and explanations, without LaTeX formatting: --- ### Given Data: - β (beta) of transistors: 100 - V_BE and V_EB (base-emitter voltages): 0.7 V - R1 = 100 kΩ - R2 = 50 kΩ - RE1 = 2 kΩ - RE2 = 2 kΩ - RC1 = 5 kΩ - RC2 = 10 kΩ - V_CC = +5 V - V_EE = -5 V --- ### Step 1: Find the Base Voltage of Q1 (V_B1) The base of Q1 is biased via a voltage divider formed by R1 and R2 between V_CC and V_EE: V_B1 is calculated as: V_B1 = (R2 / (R1 + R2)) × (V_CC - V_EE) + V_EE Plugging in the values: V_B1 = (50kΩ / 150kΩ) × (5 - (-5)) + (-5) V_B1 = (1/3) × 10 - 5 = approximately 3.33 - 5 = **-1.67 V** --- ### Step 2: Find the Emitter Voltage of Q1 (V_E1) Since V_BE ≈ 0.7 V and the transistor is forward-biased: V_E1 = V_B1 - V_BE = -1.67 V - 0.7 V = **-2.37 V** --- ### Step 3: Calculate Emitter Current of Q1 (I_E1) Using Ohm's law on emitter resistor RE1: I_E1 = (V_E1 - V_EE) / RE1 = (-2.37 V - (-5 V)) / 2kΩ = (2.63 V) / 2kΩ = **1.315 mA** Since the transistor has a high β, the collector current I_C1 ≈ I_E1 ≈ 1.315 mA. --- ### Step 4: Find Collector Voltage of Q1 (V_C1) V_C1 = V_CC - I_C1 × RC1 = 5 V - (1.315 mA × 5 kΩ) = 5 V - (1.315 × 5) V = 5 V - 6.575 V = **-1.58 V** --- ### Step 5: Base Voltage of Q2 (V_B2) The base of Q2 is connected to the collector of Q1, so: V_B2 ≈ V_C1 ≈ **-1.58 V** --- ### Step 6: Emitter Voltage of Q2 (V_E2) V_E2 = V_B2 - V_BE = -1.58 V - 0.7 V = **-2.28 V** --- ### Step 7: Emitter Current of Q2 (I_E2) Using Ohm's law on RE2: I_E2 = (V_E2 - V_EE) / RE2 = (-2.28 V - (-5 V)) / 2kΩ = (2.72 V) / 2kΩ = **1.36 mA** Again, with high β, I_C2 ≈ I_E2 ≈ 1.36 mA. --- ### Step 8: Collector Voltage of Q2 (V_C2) V_C2 = V_CC - I_C2 × RC2 = 5 V - (1.36 mA × 10 kΩ) = 5 V - (1.36 × 10) V = 5 V - 13.6 V = **-8.6 V** --- ### **Summary of Results:** | Quantity | Approximate Value | Explanation | |------------|---------------------|--------------| | **V_B1** | -1.67 V | Base voltage of Q1 from voltage divider | | **V_E1** | -2.37 V | Emitter voltage of Q1 (V_B1 - V_BE) | | **I_E1** | 1.315 mA | Emitter current of Q1 | | **V_C1** | -1.58 V | Collector voltage of Q1 (V_CC - I_C1 × RC1) | | **V_B2** | -1.58 V | Base of Q2 (connected to Q1's collector) | | **V_E2** | -2.28 V | Emitter voltage of Q2 (V_B2 - V_BE) | | **I_E2** | 1.36 mA | Emitter current of Q2 | | **V_C2** | -8.6 V | Collector voltage of Q2 (V_CC - I_C2 × RC2) | --- ### **Final Comments:** - The negative voltages indicate that these nodes are below ground potential, which is consistent with the dual supply voltages (+5V and -5V). - The currents are typical for BJTs with β=100 under these biasing conditions. - The calculations assume ideal transistor behavior with V_BE ≈ 0.7 V, and that the transistors are in forward-active region. If you need further clarification or additional parameters, please let me know!