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Oscillator Circuits
49 questions By Tony R. Kuphaldt
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Question 16 of 49
In this Wien bridge circuit (with equal-value components all around), both output voltages will have the same phase angle only at one frequency:
At this same frequency, Vout2 will be exactly one-third the amplitude of Vin. Write an equation in terms of R and C to solve for this frequency.
Reveal answerIt’s your luck day! Here, I show one method of solution:
$$R-j\frac{1}{wC}=2(\frac{1}{\frac{1}{R}+jwC})$$
$$R-j\frac{1}{wC}=\frac{2}{\frac{1}{R}+jwC}$$
$$(R-j\frac{1}{wC})(\frac{1}{R}+jwC)=2$$
$$\frac{R}{R}+jwRC-j\frac{1}{wRC}-j^2\frac{wC}{wC}=2$$
$$1+jwRC-j\frac{1}{wRC}+1=2$$
$$jwRC-j\frac{1}{wRC}=0$$
$$jwRC=j\frac{1}{wRC}$$
$$wRC=\frac{1}{wRC}$$
$$w^2=\frac{1}{R^2C^2}$$
$$w=\frac{1}{RC}$$
$$2\pi f=\frac{1}{RC}$$
$$f=\frac{1}{2\pi RC}$$
Notes:I chose to show the method of solution here because I find many of my students weak in manipulating imaginary algebraic terms (anything with a j in it). The answer is not exactly a give-away, as students still have to figure out how I arrived at the first equation. This involves both an understanding of the voltage divider formula as well as the algebraic expression of series impedances and parallel admittances.
It is also possible to solve for the frequency by only considering phase angles and not amplitudes. Since the only way Vout2 can have a phase angle of zero degrees in relation to the excitation voltage is for the upper and lower arms of that side of the bridge to have equal impedance phase angles, one might approach the problem in this fashion:
$$\theta = tan^{-1} (\frac{X_{series}}{R_{series}})$$
$$\theta = tan^{-1} (\frac{B_{parallel}}{G_{parallel}})$$
$$\frac{X_{series}}{R_{series}} = \frac{B_{parallel}}{G_{parallel}}$$
You might try presenting this solution to your students if imaginary algebra is too much for them at this point.
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Question 17 of 49
The circuit shown here is a Wien-bridge oscillator:
If one side of the Wien bridge is made from a potentiometer instead of two fixed-value resistors, this adjustment will affect both the amplitude and the distortion of the oscillator’s output signal:
Explain why this adjustment has the effect that it does. What, exactly, does moving the potentiometer do to the circuit to alter the output signal? Also, calculate the operating frequency of this oscillator circuit, and explain how you would make that frequency adjustable as well.
Reveal answerThe potentiometer adjusts the Barkhausen criterion of the oscillator. I’ll let you figure out how to make the frequency adjustable.
f = 153.9 Hz
Follow-up question: identify the paths of positive and negative feedback from the Wien bridge to the first amplifier stage.
Notes:One of the advantages of the Wien bridge circuit is its ease of adjustment in this manner. Using high-quality capacitors and resistors in the other side of the bridge, its output frequency will be very stable.
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Question 18 of 49
Identify the type of oscillator circuit shown in this schematic diagram, and explain the purpose of the tank circuit (L1 and C1):
Also, write the equation describing the operating frequency of this type of oscillator circuit.
Reveal answerThis is a Hartley oscillator circuit, and the tank circuit establishes its frequency of operation.
$$f = \frac{1}{2{\pi}\sqrt{L_1C_1}}$$
Follow-up question: calculate the operating frequency of this oscillator circuit if L1 = 330 mH and C1 = 0.15 μF.
Notes:Ask your students to describe the amount of phase shift the tank circuit provides to the feedback signal. Also, ask them to explain how the oscillator circuit’s natural frequency may be altered.
