Analog Electronics - 7 - Electrical Engineering (EE)

At oscillator frequency 
i.e. gain of non-inverting amplifier 
⇒ R_B = 2R_A

The given oscillator is Colpitts’s oscillator

The condition for oscillation is

|Aβ|=1

A = -R₂/R₁

β=C₂/C₁

• The basic amplifier of the circuit is the op-amp A₃.
• The output of amplifier is connected to 3 stage RC filter.
• Voltage followers eliminate loading effect of each stage.

Transfer function of first RC network

RC networks are identical and no loading effect

The amplifier A₃ is connected in inverting mode

The loop gain is

Substitute real part of denominator to zero

At this frequency

Hartley oscillator

 

Amp with +ve F.B

Loop again AB = 1 + j0         

At oscillation

∠AB = 0 (no reactance X_L = X_C)

At frequency of oscillation look lies resistive Network

For ideal op amp

Use voltage division Rule

R + 1 = 21

R = 20 kΩ

R = 2 × 10⁴Ω 

The given circuit is wein bridge oscillator.

 

R₃ = R₄ = 159 KΩ

C₁ = C₂ = 1 nF

 

= 10³ HZ

= 1 KHZ

In a transistor phase shift oscillator the oscillation frequency

 

R, C are the resistor and the capacitor in phase shift circuit

R_C is collector capacitance

⇒ R_C = 2R = 10 kΩ 

The given oscillator is Hartley oscillator, where the oscillation frequency.

 

= 253.3 mH

L_eq = L + 5 mA

⇒ L = 253.3 – 5

= 248. 3 mH

Given figure is of A-stable multi-vibrator,

Time constant is given by

β = feedback factor

τ = time constant of feedback loop:

Option 3 satisfies the condition

The given oscillator is Wien-bridge oscillator with R = 5 kΩ, C = 1 μF 

⇒ frequency of oscillation f = 1/2πRC

= 31.8 Hz