Question

A CMOS Schmitt-trigger inverter has thresholds of 1.6 V (low-to-high) and 2.4 V (high-to-low). The capacitor is 47 nF and the resistor is 10 k\(\Omega\). The frequency of the oscillator is \(\underline{\hspace{2cm}}\) Hz. (Round off to 2 decimal places.) 

Hint:

For a Schmitt-trigger RC oscillator, frequency is determined solely by the threshold voltages and the RC time constant.

Answer 1

Correct Answer: 3150

The Schmitt-trigger relaxation oscillator charges and discharges between two threshold voltages:
\[ V_L = 1.6\ \text{V}, V_H = 2.4\ \text{V} \] The inverter output swings between:
\[ 0\ \text{V and } 5\ \text{V} \] During charging from \(V_L\) to \(V_H\):
\[ t_1 = R C \ln\left( \frac{5 - V_L}{5 - V_H} \right) \] During discharging from \(V_H\) to \(V_L\):
\[ t_2 = R C \ln\left( \frac{V_H}{V_L} \right) \] Given: \[ R = 10\,000\ \Omega, C = 47\,\text{nF} \] Compute charging time:
\[ t_1 = (10\,000)(47\times 10^{-9}) \ln\left( \frac{5 - 1.6}{5 - 2.4} \right) \] \[ = 4.7\times 10^{-4} \ln\left( \frac{3.4}{2.6} \right) \] \[ t_1 = 4.7\times 10^{-4} (0.268) = 1.26\times 10^{-4}\ \text{s} \] Compute discharging time:
\[ t_2 = 4.7\times 10^{-4} \ln\left( \frac{2.4}{1.6} \right) \] \[ t_2 = 4.7\times 10^{-4} (0.405) = 1.90\times 10^{-4}\ \text{s} \] Total period:
\[ T = t_1 + t_2 = 3.16\times 10^{-4}\ \text{s} \] Frequency:
\[ f = \frac{1}{T} = \frac{1}{3.16\times 10^{-4}} = 3164.56\ \text{Hz} \] \[ \boxed{3164.56\ \text{Hz}} \]