Hello all,

I have tried to work at this problem for a while but I am having some trouble with it. Here is the given multi-part question and I will discuss what I have done so far.

Design a Wienbridge Oscillator using the circuit configuration shown. Assume that the voltage supply to the op-amp is +/- 10 V and that the diode turn-on voltage, Vd,on is 0.7 V. The oscillation frequency required is 10 kHz and the output amplitude is 5 V. Select R1 = 5 kΩ, R = 10 kΩ and set hte gain of the circuit at the start of oscillations to be 2.2.

a) Determine the value of C and R2.
b) Write an expression for the output voltage given the conditions shown in the figure.
c) Determine the value of R3 that provides the output amplitude of 5 V.
d) Accurately draw two cycles (periods) of the output waveform vs. time. Carefully labelling the drawing.

My solutions:

a)

\(f_{0} = 10 kHz -> w0 = \frac{1}{RC} = 2*pi*f_{0} -> C = 1/(2*pi*f_{0}*R) = 1.59 nF\)

\(Gain = 1 + \frac{R_{2}}{R_{1}} = 2.2 -> R_{2} = 1.2*R_{1} = 6k \) Not sure though...

b) KCL @ V- ?

\(\frac{V_{o} - V_{d,on} - V_{-}}{R_{3}} + \frac{V_{o} - V_{-}}{R_{2}} = \frac{V_{-}}{R1}\)

\(V_{-} = \frac{V_{o}}{3}\) Solve for Vo... Not sure if this is correct as it doesn't take into account the other diode direction...

\(V_{o} = \frac{(30M)V_{d,on}}{(14k)R_{3} + 60M}\)

c) Not sure...

d) 5 Vpp sinewave with a frequency of 10 kHz...

Wondering if anyone can check my solutions and help me out. Thanks!

 

I'd be surprised the circuit could get into oscillation with an initial gain of 2.2. Would one not need a gain of at least 3x ...??

 

Say with an initial gain of 3.3 it would most likely get going OK.

 

haha well this is actually a test question that I got back and did not answer correctly. My prof did not post solutions...

 

I see.

Well you seem to have calculated the 'correct' capacitance value.

You'll find R2 based on a 'sensible' linear region gain - I'm thinking not 2.2X...??

To find the value of R3 a simple approach would be:

The voltage at the amp -ve input will be 5/3 volts. So the current in the 5k to ground will be 1/3 mA.

The voltage drop across R will be (5-5/3) V. The current in R2 will be I_R2=(5-5/3)/R2.

So the current in R3 will 1/3mA - I_R2

The voltage across R3 will be ~(5-5/3-0.7). This will enable you to find R3 since you also know the current.

 

I used the G = 2.2 = 1 + R2/R1 because it said that upon startup so I figured that the diodes would not be on, thus cutting off R3, so it would be a simple gain circuit between R1 and R2. Also, what is the significance of the current through R, if there is a capacitor in series with the other R?

It is exactly the same as this example I found in my Blalock textbook, with the exception that 'their' R2 = 0 in my question.

Thank you again!

 

Does the text not give a worked example from which you can model your answer?

Yes - the diode pair + R3 branch only (rather progressively) comes into play as the output approaches either the positive or negative peak value. So the initial 'linear' gain at low output amplitude (<0.7V) will be (R1+R2)/R1.

 

If your work was marked as incorrect what about your classmates? I'd like to be there when your prof explains how the circuit works with a gain less than 3x.

 

The 2.2 gain is just to start of with and build up the gain to 3, then the diode keeps it from going under or over.

 

The 2.2 gain is just to start of with and build up the gain to 3, then the diode keeps it from going under or over.

You are missing the point.

If the gain is less than 3x in the low output (linear) region then it will decrease as the signal level increases - not increase. If you are notionally adding R3 in parallel with R2 then the stage gain is dynamically varying downwards - not upwards. If it starts out at 2.2 x then it will decrease with increasing output signal excursion. That's the very rationale of including the D1/D2 diode array + R3 circuit branch.

 

Hmm, that seems reasonable, but I'm wondering why our prof would give us this sort of question if it were unsolvable (unrealistic)... I'll email him again.

Thanks

 

It's not unreasonable to allow for a typo to have occurred in the posting of the problem. It's most likely a simple issue of that nature.