Chapter 13: Problem 55
A certain \(n\)p\(n\) silicon transistor at room temperature has \(\beta=100 .\) Find the corresponding values of \(r_{x}\) if \(I_{C Q}=1 \mathrm{mA}, 0.1 \mathrm{mA}\) and \(1 \mu\) A. Assume operation in the active region and that \(V_{T}=26 \mathrm{mV}\).
Short Answer
Expert verified
\( r_x \) values are 2600 Ohm, 26000 Ohm, and 2600000 Ohm for respective \( I_{CQ} \) values.
Step by step solution
01
Understanding the Parameters
We are given a silicon transistor with a current gain \( \beta = 100 \). The transistor operates in the active region. The thermal voltage is given as \( V_T = 26 \text{ mV} \). We need to find \( r_x \) for various \( I_{CQ} \). The parameter \( r_x \) is related to the base spreading resistance and can be expressed as \( r_x = \frac{V_T}{I_B} \) where \( I_B \) is the base current.
02
Calculating Base Current for Different Collector Currents
Using \( I_C = \beta \cdot I_B \), we can express \( I_B = \frac{I_C}{\beta} \). We need to calculate \( I_B \) for each given \( I_C \). For \( I_CQ = 1 \text{ mA} \), \( I_B = \frac{1 \times 10^{-3} \text{ A}}{100} = 10 \times 10^{-6} \text{ A} \). For \( I_CQ = 0.1 \text{ mA} \), \( I_B = \frac{0.1 \times 10^{-3}\text{ A}}{100} = 1 \times 10^{-6} \text{ A} \). For \( I_CQ = 1 \text{ \mu A} \), \( I_B = \frac{1 \times 10^{-6}\text{ A}}{100} = 10 \times 10^{-9} \text{ A} \).
03
Calculating \( r_x \) Using the Formula
Now we substitute the calculated \( I_B \) values into \( r_x = \frac{V_T}{I_B} \) for each scenario. For \( I_CQ = 1 \text{ mA} \), \( r_x = \frac{26 \times 10^{-3}\text{ V}}{10 \times 10^{-6}\text{ A}} = 2600 \text{ Ohm} \). For \( I_CQ = 0.1 \text{ mA} \), \( r_x = \frac{26 \times 10^{-3}\text{ V}}{1 \times 10^{-6}\text{ A}} = 26000 \text{ Ohm} \). For \( I_CQ = 1 \text{ \mu A} \), \( r_x = \frac{26 \times 10^{-3}\text{ V}}{10 \times 10^{-9}\text{ A}} = 2600000 \text{ Ohm} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Current Gain (Beta)
In a Bipolar Junction Transistor (BJT), the current gain, denoted as \( \beta \), plays a crucial role in understanding the transistor's efficiency in amplifying current. This parameter is defined as the ratio of the collector current \( I_C \) to the base current \( I_B \). Essentially, \( \beta \) tells you how much larger the output collector current is compared to the input base current.
- Mathematically, \( \beta = \frac{I_C}{I_B} \).
- This ratio is typically a constant value given for each transistor, and it helps in calculating other parameters like \( I_B \) when \( I_C \) is known.
Active Region Operation
The operation of a BJT in different regions determines its functionality. The active region is where the transistor operates as an amplifier. To ensure active region operation, certain conditions must be met:
- The base-emitter junction should be forward-biased.
- The base-collector junction must be reverse-biased.
Base Spreading Resistance
Base spreading resistance, denoted as \( r_x \), is the resistance present in the base region of a BJT. It impacts the performance of the transistor by affecting the voltage that the base-emitter junction sees due to the current flowing through it. The importance of this resistance becomes pronounced when dealing with high-frequency or high-power applications.For a more practical view:
- Base spreading resistance can cause a reduction in the gain at high frequencies, known as the high-frequency roll-off.
- In the exercise, \( r_x \) is calculated using \( r_x = \frac{V_T}{I_B} \).
- The smaller the base current, the larger the \( r_x \); highlighting its inverse relationship with the base current.
Thermal Voltage (V_T)
The thermal voltage \( V_T \) is a vital parameter in semiconductor physics, particularly for BJTs and diodes. It is a temperature-dependent voltage that impacts how these devices conduct current. At room temperature (approximately 300K), the thermal voltage is typically around 26 mV.The formula for thermal voltage is:\[ V_T = \frac{kT}{q} \]where:
- \( k \) is Boltzmann's constant \( \approx 1.38 \times 10^{-23} \text{ J/K} \)
- \( T \) is the absolute temperature in Kelvin
- \( q \) is the charge of an electron \( \approx 1.6 \times 10^{-19} \text{ C} \)