Let's solve each circuit **step by step**. We assume **very large β (β → ∞)**, so the base current is negligible, and \( I_E \approx I_C \). --- ## (a) **First circuit** ### **Given:** - \( I_C = 0.5\,mA \) - \( R_C = 3.6\,k\Omega \) - \( V_{CC} = +3\,V \) - \( R_B = 43\,k\Omega \) - NPN transistor ### **Find:** - \( V_O \) --- #### **Step 1: Find \( V_O \) (Collector voltage)** \[ V_O = V_{CC} - I_C \cdot R_C \] Plug in the values: \[ V_O = 3\,V - (0.5\,mA)(3.6\,k\Omega) \] \[ = 3\,V - (0.0005\,A \times 3600\,\Omega) \] \[ = 3\,V - 1.8\,V \] \[ = 1.2\,V \] --- #### **Step 2: Find \( V_B \) (Base voltage)** For a silicon BJT, \( V_{BE} \approx 0.7\,V \): \[ V_B = V_E + 0.7\,V \] Emitter is grounded (\( V_E = 0 \)), so: \[ V_B = 0.7\,V \] --- #### **Step 3: Find \( I_B \) (Base current)** Since β is very large, \( I_B \approx 0 \). --- ### **Summary for (a):** - \( V_O = 1.2\,V \) - \( V_B = 0.7\,V \) - \( I_B \approx 0 \) --- ## (b) **Second circuit** ### **Given:** - \( R_C = 3.6\,k\Omega \) - \( R_E = 4.7\,k\Omega \) - \( V_{CC} = +3\,V \) - \( V_{EE} = -3\,V \) ### **Find:** - \( I_C \), \( V_O \) --- #### **Step 1: Find \( V_{CE} \)** Let \( I_C = I_E \) (since β is very large). Loop equation from \( +3\,V \) to \( -3\,V \): \[ +3\,V - I_C R_C - V_{CE} - I_E R_E - (-3\,V) = 0 \] \[ 6\,V = I_C(R_C + R_E) + V_{CE} \] Assume transistor is in active mode: \( V_{CE} \approx 0.2\,V \) (saturation) or \( V_{CE} \) is not specified. Let's solve for \( I_C \) assuming \( V_{CE} \) is not too small, so: \[ I_C = \frac{6\,V - V_{CE}}{R_C + R_E} \] But, without base resistor, assume \( V_{BE} = 0.7\,V \): \[ V_O = V_{CC} - I_C R_C \] But we need base current; usually, this circuit is not practical unless the base is biased. But since the base is grounded, let's assume it is at 0 V. #### **Base is at 0V:** - Emitter at \( -3\,V \) - \( V_{BE} = 0.7\,V \) - So, \( V_B = V_E + 0.7\,V = -3\,V + 0.7\,V = -2.3\,V \) - But base is at 0 V, so this circuit can't conduct. So, let's interpret that the question wants us to solve for \( I_C \) if the base is at 0 V. - \( V_{BE} = V_B - V_E = 0 - V_E \) - For conduction, \( V_{BE} \approx 0.7V \) - So, \( V_E = -0.7\,V \) - \( I_E = \frac{V_E - (-3\,V)}{R_E} = \frac{-0.7\,V + 3\,V}{4.7\,k\Omega} = \frac{2.3\,V}{4.7\,k\Omega} \approx 0.489\,mA \) - \( I_C \approx 0.489\,mA \) \[ V_O = V_{CC} - I_C R_C = 3\,V - (0.489\,mA \times 3.6\,k\Omega) = 3\,V - 1.76\,V = 1.24\,V \] --- ### **Summary for (b):** - \( I_C \approx 0.489\,mA \) - \( V_O \approx 1.24\,V \) --- ## (c) **Third circuit** This is a **differential amplifier**. #### **Assume both transistors are identical** and perfectly matched. - \( V_{in1} = V_{in2} = 0.75\,V \) (from the node label) - \( I_T = 0.5\,mA \) (current source at the bottom) - The tail current splits equally, so \( I_{C1} = I_{C2} = 0.25\,mA \) --- ### **Summary for (c):** - \( I_{C1} = I_{C2} = 0.25\,mA \) - \( V_{in1} = V_{in2} = 0.75\,V \) --- ## (d) **Fourth circuit** Given resistor values, same as above, just different resistors. - \( I_T = 0.5\,mA \) - \( I_{C1} = I_{C2} = 0.25\,mA \) - \( V_{in2} \) is labeled, to be found. For a differential amplifier, if both inputs are at the same voltage, each transistor gets half the current. --- ## (e) **Fifth circuit** Given: - \( V_{CC} = +3\,V \) - \( R_1 = 180\,k\Omega \) - \( R_2 = 300\,k\Omega \) - \( R_C = 6.2\,k\Omega \) - \( R_E = 10\,k\Omega \) - NPN transistor Find \( V_{in} \), \( V_{out} \). --- #### **Step 1: Find base voltage (\( V_B \)) using voltage divider:** \[ V_B = V_{CC} \cdot \frac{R_2}{R_1 + R_2} \] \[ = 3\,V \cdot \frac{300}{180 + 300} = 3\,V \cdot \frac{300}{480} = 3\,V \cdot 0.625 = 1.875\,V \] #### **Step 2: Assume \( V_{BE} = 0.7\,V \), so:** \[ V_E = V_B - 0.7\,V = 1.875\,V - 0.7\,V = 1.175\,V \] #### **Step 3: Emitter current:** \[ I_E = \frac{V_E}{R_E} = \frac{1.175\,V}{10\,k\Omega} = 0.1175\,mA \] Since β is large, \( I_C \approx I_E \). #### **Step 4: Output voltage (\( V_{out} \)):** \[ V_{out} = V_{CC} - I_C \cdot R_C = 3\,V - 0.1175\,mA \cdot 6.2\,k\Omega = 3\,V - (0.0001175\,A \cdot 6200\,\Omega) = 3\,V - 0.727\,V = 2.273\,V \] --- ### **Summary for (e):** - \( V_{in} = V_B = 1.875\,V \) - \( V_{out} = 2.273\,V \) --- ## **Summary Table** | Circuit | \( V_{B} \) / \( V_{in} \) | \( I_C \) (mA) | \( V_O \) / \( V_{out} \) | |---------|----------------------------|----------------|------------------| | (a) | 0.7 V | 0.5 | 1.2 V | | (b) | 0 V | 0.489 | 1.24 V | | (c) | 0.75 V (each) | 0.25 (each) | - | | (d) | - | 0.25 (each) | - | | (e) | 1.875 V | 0.1175 | 2.273 V | Let me know if you want to see the calculations for any particular circuit in more detail!