If crystal oscillator parameters are as follows: R = 5K Ohm, L = 0.5H, Cs = 0.06pF, Cp = 1.25pF, then what is the difference between two frequencies of oscillations?a.27.1KHzb.21.7KHzc.36.0KHzd.25.8KHz

The difference between two frequencies of oscillations can be calculated using the formula for the resonant frequency of a crystal oscillator. The resonant frequency is given by the formula:

f = 1 / (2π √(LC))

where:

• L is the inductance in henries (H)
• C is the capacitance in farads (F)

In this case, we have two capacitances: Cs (series capacitance) and Cp (parallel capacitance). Therefore, we will have two resonant frequencies, one for each capacitance.

First, calculate the resonant frequency for Cs:

f1 = 1 / (2π √(0.5H * 0.06pF)) = 1 / (2π √(0.5 * 0.06 * 10^-12)) = 1 / (2π √(0.03 * 10^-12)) = 1 / (2π √(3 * 10^-14)) = 1 / (2π * 5.48 * 10^-8) = 2.9 * 10^6 Hz = 2.9 MHz

Next, calculate the resonant frequency for Cp:

f2 = 1 / (2π √(0.5H * 1.25pF)) = 1 / (2π √(0.5 * 1.25 * 10^-12)) = 1 / (2π √(0.625 * 10^-12)) = 1 / (2π √(6.25 * 10^-14)) = 1 / (2π * 7.9 * 10^-8) = 2.01 * 10^6 Hz = 2.01 MHz

Finally, calculate the difference between the two frequencies:

Δf = f1 - f2 = 2.9 MHz - 2.01 MHz = 0.89 MHz = 890 KHz

None of the options provided (a. 27.1 KHz, b. 21.7 KHz, c. 36.0 KHz, d. 25.8 KHz) match the calculated difference. Please check the values or the options again.

The frequencies of oscillation for a crystal oscillator can be calculated using the formula for the resonant frequency of an LC circuit, which is f = 1 / (2π√(LC)).

In this case, we have two capacitors in series, so the total capacitance (Ct) is given by the formula 1/Ct = 1/Cs + 1/Cp.

First, calculate the total capacitance:

1/Ct = 1/0.06pF + 1/1.25pF = 16.67PF^-1 + 0.8PF^-1 = 17.47PF^-1

So, Ct = 1 / 17.47PF^-1 = 0.0572pF

Now, calculate the resonant frequency using the total capacitance:

f = 1 / (2π√(0.5H * 0.0572pF)) = 1 / (2π√(0.0286µF)) = 1.88MHz

Then, calculate the resonant frequency using only Cs:

f = 1 / (2π√(0.5H * 0.06pF)) = 1 / (2π√(0.03µF)) = 1.83MHz

The difference between the two frequencies is 1.88MHz - 1.83MHz = 0.05MHz, or 50KHz.

None of the options provided (a. 27.1KHz, b. 21.7KHz, c. 36.0KHz, d. 25.8KHz) match this result. Please check the values or the options again.