Oscillator Flashcards

Study Unit 8: Oscillators - Barkhausen criteria and Wien-bridge oscillator

Lecture Outcomes

• Topics for today’s discussion
  • Oscillator theory
  • Barkhausen criteria
  • Sinusoidal oscillator
  • Wien bridge configuration

Oscillator Circuits Introduction - Feedback Concept(s)

• Amplifier with open loop gain ($A_0$)
• Additional negative feedback path ($\beta$)
• Additional frequency selective positive feedback path ($K(\omega)$)
• Diagram:
  • Positive feedback path: $K(\omega)$
  • Negative feedback path: $\beta$
  • Input voltage: $v_{in}$
  • Output voltage: $v_{out}$
  • Summing block output: $v_x$

Oscillator Circuits Barkhausen Criteria

• Analysis:
  • $v<em>n = \beta \cdot v</em>{out}$
  • $v<em>p = K(\omega) \cdot v</em>{out}$
  • $v<em>x = v</em>p - v_n$
• Substitution:
  • $v<em>x = K(\omega) \cdot v</em>{out} - \beta \cdot v_{out}$
  • $v<em>x = [K(\omega) - \beta] \cdot v</em>{out}$
• Also:
  • $v<em>{out} = A</em>0 \cdot v_x$
• Substitution:
  • $v<em>{out} = A</em>0 \cdot [K(\omega) - \beta] \cdot v_{out}$

Oscillator Circuits Barkhausen Criteria - cont.

• Analysis conclusion:
  • $v<em>{out} = A</em>0 \cdot [K(\omega) - \beta] \cdot v<em>{out}$ is only possible when $A</em>0 \cdot [K(\omega) - \beta] = 1$
  • Since $A_0 \rightarrow \infty$, we have $[K(\omega) - \beta] \rightarrow 0$.
  • Therefore, $K(\omega) = \beta$
• Since $\beta$ is real and $K$ is complex, both magnitude and phase play a role.
• The Barkhausen criteria states that:
  • Magnitude: $|\frac{K(\omega)}{\beta}| = 1$
  • Phase: $\angle \frac{K(\omega)}{\beta} = 0^\circ$

Oscillator Circuits Wien-Bridge Oscillator

• Circuit Components:
  • Resistors: R, R1, R2
  • Capacitors: C
• Impedance Definitions:
  • $Z_1 = R || \frac{1}{sC}$
  • $Z_2 = R + \frac{1}{sC}$
• Analysis:
  • $v<em>p = K(\omega) \cdot v</em>{out} = \frac{Z<em>1}{Z</em>1 + Z<em>2} v</em>{out}$
  • $K(\omega) = \frac{Z<em>1}{Z</em>1 + Z_2} = \frac{j\omega RC}{[1 - (\omega RC)^2] + 3j\omega RC}$
  • $v<em>n = \beta \cdot v</em>{out} = \frac{R<em>1}{R</em>1 + R<em>2} v</em>{out}$
• Barkhausen criteria: $K(\omega) - \beta = 0$, i.e. $\frac{j\omega RC}{[1 - (\omega RC)^2] + 3j\omega RC} = \frac{R<em>1}{R</em>1 + R_2}$
  • Only possible for $[1 - (\omega RC)^2] = 0$
• Design equations:
  • The above requirement leads to: $\omega_0 = \frac{1}{RC}$
  • And $\frac{R<em>1}{R</em>1 + R_2} = \frac{1}{3}$