Let's break down the problem step by step: **Given:** - \( I_{CQ} = 1\, \text{mA} \) - \( I_{BQ} = 0.01\, \text{mA} \) - \( V_{CEQ} = 8\, \text{V} \) - \( V_A = 115\, \text{V} \) (Early voltage) - \( r_1 = 100\, \Omega \) - Room temperature You are asked **what is \( \frac{1}{h_{oe}} \) of the transistor?** --- ### Step 1: Understanding \( h_{oe} \) In the small-signal hybrid-pi model, - \( h_{oe} \) is the output admittance (the reciprocal of output resistance) looking into the collector at a constant base current. The output resistance \( r_o \) due to Early effect is given by: \[ r_o = \frac{V_A + V_{CEQ}}{I_{CQ}} \] At small signals, \( r_o \approx \frac{V_A}{I_{CQ}} \) (since \( V_{CEQ} \ll V_A \)), but for more accuracy, use the formula above. Also, typically, \[ \frac{1}{h_{oe}} = r_o \] --- ### Step 2: Plug in the values Given: - \( V_A = 115\, \text{V} \) - \( V_{CEQ} = 8\, \text{V} \) - \( I_{CQ} = 1\, \text{mA} = 0.001\, \text{A} \) \[ r_o = \frac{V_A + V_{CEQ}}{I_{CQ}} = \frac{115 + 8}{0.001} = \frac{123}{0.001} = 123,000\, \Omega = 123\, \text{k}\Omega \] --- ### Step 3: Compare with options - **115 kΩ** - **62 kΩ** - **123 kΩ** - **None of the mentioned** **The correct answer is:** \[ \boxed{123\,\text{k}\Omega} \] --- ### **Summary Table** | Quantity | Value | |-------------|--------------| | \( V_A \) | 115 V | | \( V_{CEQ} \) | 8 V | | \( I_{CQ} \) | 1 mA | | \( r_o \) (or \( 1/h_{oe} \)) | 123 kΩ | --- ## **Final Answer** \[ \boxed{123\,\text{k}\Omega} \] This is the value of \( \frac{1}{h_{oe}} \) of the transistor.

Let's analyze why the other options are incorrect and clarify the reasoning process: --- ### Why are the other options incorrect? #### 1. **Option: 115 kΩ** - Calculated \( r_o \) is **123 kΩ**, which is close to 123 kΩ, **not** 115 kΩ. - The slight difference arises from adding \( V_{CEQ} \) to \( V_A \). Since \( V_A = 115\,V \) and \( V_{CEQ} = 8\,V \), the total is \( 123\,V \), leading to a larger \( r_o \). - Therefore, **115 kΩ** underestimates the value, making it incorrect. --- #### 2. **Option: 62 kΩ** - This is about half of the calculated \( r_o \). - Since \( r_o = (V_A + V_{CEQ}) / I_{CQ} = 123\,kΩ \), **62 kΩ** would imply either a much higher collector current or a significantly lower \( V_A + V_{CEQ} \) sum. - Given the data, this value does not match the calculation, so it is **incorrect**. --- #### 3. **Option: None of the mentioned** - Based on the calculation, the correct \( r_o \) is **123 kΩ**, which *is* among the options. - Therefore, "**None of the mentioned**" is **incorrect**. --- ### **Summary:** The precise calculation of the output resistance \( r_o \) considering the Early voltage and collector-emitter voltage gives **123 kΩ**, matching exactly with the third option. **Hence, the correct answer is:** \[ \boxed{123\,\text{k}\Omega} \] --- **In conclusion:** - The other options do not match the calculated \( r_o \). - The calculation directly follows from the Early effect formula. - The provided data points to **123 kΩ** as the correct \( 1/h_{oe} \). **This explains why the correct choice is the third option, and why the others are incorrect.**