A parallel-tuned circuit in an oscillator consists of a 40-uH inductor in parallel with a 330-pF capacitor. A varactor with a capacitance of 50 pF is connected in parallel with the circuit. What is the resonant frequency of the tuned circuit and the oscillator operating frequency? •
2. If the varactor capacitance of the circuit in Prob. 1 is decreased to
25 pF, (a) how does the frequency change and (b) what is the new resonant frequency?

To find the resonant frequency of the tuned circuit, we use the formula for the resonant frequency of an LC circuit:

\[
f = \frac{1}{2 \pi \sqrt{L \cdot C}}
\]

where:
- \( L \) is the inductance in henries (H)
- \( C \) is the capacitance in farads (F)

### Problem 1: Initial Resonant Frequency Calculation

1. **Given Values**:
   - Inductance: \( L = 40 \, \mu H = 40 \times 10^{-6} \, H \)
   - Capacitor: \( C_1 = 330 \, pF = 330 \times 10^{-12} \, F \)
   - Varactor Capacitance: \( C_2 = 50 \, pF = 50 \times 10^{-12} \, F \)

2. **Total Capacitance Calculation**:
   The total capacitance for capacitors in parallel is the sum of individual capacitances:

   \[
   C_{\text{total}} = C_1 + C_2 = (330 + 50) \times 10^{-12} \, F = 380 \times 10^{-12} \, F
   \]

3. **Resonant Frequency Calculation**:
   Now, substitute \( L = 40 \times 10^{-6} \, H \) and \( C_{\text{total}} = 380 \times 10^{-12} \, F \) into the resonant frequency formula:

   \[
   f = \frac{1}{2 \pi \sqrt{(40 \times 10^{-6}) \cdot (380 \times 10^{-12})}}
   \]

   Calculating the expression under the square root:

   \[
   L \cdot C_{\text{total}} = (40 \times 10^{-6}) \cdot (380 \times 10^{-12}) = 15.2 \times 10^{-15}
   \]

   Taking the square root:

   \[
   \sqrt{15.2 \times 10^{-15}} = 3.9 \times 10^{-8}
   \]

   Thus, the resonant frequency becomes:

   \[
   f = \frac{1}{2 \pi \cdot 3.9 \times 10^{-8}} \approx 4.08 \times 10^{6} \, \text{Hz} = 4.08 \, \text{MHz}
   \]

### Problem 2: Change in Varactor Capacitance to 25 pF

1. **New Total Capacitance**:
   - New Varactor Capacitance: \( C_2' = 25 \, pF = 25 \times 10^{-12} \, F \)
   - New Total Capacitance: 

   \[
   C_{\text{total}}' = C_1 + C_2' = (330 + 25) \times 10^{-12} \, F = 355 \times 10^{-12} \, F
   \]

2. **New Resonant Frequency Calculation**:
   Substitute \( C_{\text{total}}' = 355 \times 10^{-12} \, F \):

   \[
   f' = \frac{1}{2 \pi \sqrt{(40 \times 10^{-6}) \cdot (355 \times 10^{-12})}}
   \]

   Calculate the expression under the square root:

   \[
   L \cdot C_{\text{total}}' = (40 \times 10^{-6}) \cdot (355 \times 10^{-12}) = 14.2 \times 10^{-15}
   \]

   Taking the square root:

   \[
   \sqrt{14.2 \times 10^{-15}} = 3.77 \times 10^{-8}
   \]

   Thus, the new resonant frequency becomes:

   \[
   f' = \frac{1}{2 \pi \cdot 3.77 \times 10^{-8}} \approx 4.22 \times 10^{6} \, \text{Hz} = 4.22 \, \text{MHz}
   \]

### Changes in Frequency:

(a) The frequency increases from 4.08 MHz to 4.22 MHz.

(b) The new resonant frequency is 4.22 MHz.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the effect of inductance on the resonant frequency?
2. How does adding more capacitors in parallel affect the resonant frequency?
3. What are the applications of resonant circuits in oscillators?
4. How does quality factor (Q) relate to the resonant frequency of an LC circuit?
5. What happens if the varactor capacitance is made very small?

**Tip:** The resonant frequency is inversely proportional to the square root of the product of inductance and capacitance. Thus, reducing the capacitance increases the frequency.

A parallel-tuned circuit in an oscillator consists of a 40-uH inductor in parallel with a 330-pF capacitor. A varactor with a capacitance of 50 pF is connected in parallel with the circuit. What is the resonant frequency of the tuned circuit and the oscillator operating frequency? If the varactor capacitance of the circuit is decreased to 25 pF, how does the frequency change and what is the new resonant frequency?

Calculate the resonant frequency of a parallel-tuned circuit with a 40-uH inductor and a 330-pF capacitor, with a varactor connected in parallel. Learn how changes in varactor capacitance affect the oscillator frequency, suitable for college-level electrical engineering students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation