Wien bridge oscillator output voltage

[MrAl]

Hi,

Here is a quick formula for the total output period:
Tp=4*(C1*R1*R3*VH)/(R2*VL)

and the frequency is 1/Tp and you can make VL=VH for simplicity, giving us:

Tp=4*C1*R1*R3/R2

and so the frequency is:
F=R2/(4*C1*R1*R3)

Note this last result does not depend on the output voltage levels of the op amp(s) as long as the minus excursion is the same as the positive excursion.

 

[audioguru]

The TL07x opamps do not have equal positive and negative excursions.

 

[MrAl]

The TL07x opamps do not have equal positive and negative excursions.

Hi there audioguru,


That should be ok as long as the two extremes are not too much different.

The theoretically exact frequency formula is:
F=R2/(4*C1*R1*R3)

but of course other slight variations might cause a slight variation in the frequency too, so we still have to regard that formula as an estimate even though it is exact with perfect components. For example we might calculate 20kHz with the formula when we actually measure 19kHz on the real life breadboard. If we want a closer estimation then we need to consider measuring the actual positive and negative excursions, but because there is also the issue of component variation (especially the timing capacitor) then we would also need to measure the capacitance. That makes it seem more practical just to estimate the frequency and then adjust as needed.

Another issue that comes up is the triangle symmetricalness. If we want a symmetrical triangle we need equal positive and negative output voltage excursions. If we cant get that then we need to modify the design a little. Changing the zero voltage reference would solve that, but that involves another adjustment with a pot or something.
We could also use a rail to rail comparator instead of another op amp, but people usually like to use a dual op amp and then use the two internal op amps instead.

 

[Adam2014]

thanks
but , I want the triangular oscillator analysis

 

[MrAl]

Hi,

You'll have to be a bit more descriptive of what you want exactly.

 

[Adam2014]

I want the triangular oscillator analysis to calculate its output
the triangular oscillator in post #35

 

[MrAl]

Hi,

Yes but you must have not read the posts after that one. Check out post #41.

 

[Adam2014]

thanks MrAl
sorry , I want to know how you find the output equation , not just a quick formula ?
if there are steps , or analysis for the circuit ?

 

[MrAl]

Hi,

Yes, the output is like this:

Vout=Vin*t/(C1*R3)+Vc0

where Vc0 is the initial capacitor voltage,
Vin is the output of the first op amp stage,
t is time.

The first op amp stage acts as a comparator, and Vin is that stage's output, and that depends on the power supply voltage and the type of op amp or comparator being used. To approximate you can use Vin equal to either +Vcc or Vin=-Vcc.
As the output of the second stage changes, you have to solve for the point where the first stage comparator changes stage. Can ou do that?
That tells you what the maximum point of the output of the second stage is, that's the tip of the triangle for both plus and minus output.

So the output ramps up according to that equation and then trips the comparator, then ramps down according to that equation and then trips the comparator again, then those two actions repeat over and over again.

 

[Ratchit]

Adam2014,

As requested, here is the analysis for the triangle wave generator. The design equations give a frequency of 24 khz, whereas as my simulator says the freq will be 20 khz. I believe the difference is due to not accounting for the output impedance of U1. Ask if you have any questions.

Ratch

 

[Adam2014]

thanks MrAl and Ratchit
I upload my schematic for the triangular oscillator
in simulation i get -/+ 10 volt on output ,,, but i don't know the output equation ,
with output equation that you gave me i calculate V_out and have a different value !

 

[audioguru]

Simply guess the maximum output voltage of your opamps then use Ohm's Law to calculate the peak output voltages.
The datasheets do not show a typical unloaded peak output voltage of a TL074 or LM301 opamp so I guessed it is +13.5V and -13.5V when the supply is plus and minus 15V.

The opamps have a voltage gain of about 200,000 times so the input voltage when the output switches its polarity is very close to 0V. Then I used Ohm's Law to calculate the output of the triangle wave swings from -11.07V to +11.07V.

You can also calculate how long it takes for the intergrator opamp to charge and discharge its capacitor to determine the frequency.

 

[MrAl]

Hi,

If we take Ratch's result for frequency:
F=(1/4)*R2/(C1*R1*R3)

and insert your schematic values we get:
24587Hz

The period Tp=1/f is then about:
Tp=1/f=4.0672e-5

Now since the voltage ramps up in half the time and down in the other half of the time, the half period Thp is:
Thp=Tp/2=2.0336e-5

Then if the max voltage out of the first op amp is 12 volts max and -12 volts min then the output ramp is:
Vo=Thp*12/(R3*C1)

and doing this with the component values and Thp above we get:
Vo=19.68 volts.

Since Vo is negative for half the time and positive for the other half, the ramp goes from minus half this value to plus half this value. The peak then is half this value:
Vpk=19.68/2=9.84 volts

so the ramp goes form -9.84 volts to +9.84 volts.

This is very close to what you got in sim so you must not be calculating it right that's all.

 

[Adam2014]

thanks MrAl and Audioguru
* MrAl : sorry but the output is +/- 11 volt not 10 like i thought .
* Audioguru : sorry i don't understand what you mean ,,, i can get the relations in post #53 ?

 

[audioguru]

* Audioguru : sorry i don't understand what you mean ,,, i can get the relations in post #53 ?

In post #53, MrAl used the oscillator's frequency to calculate how long and to what voltage the integrator capacitor charges to.

In my post #52, I calculated the peak voltage of the triangle waveform simply using Ohm's Law:
1) I guessed that the maximum output of the opamps is +/- 13.5V because the datasheet does not give the unloaded typical values (only the minimum with a load) and I am too lazy to set up a TL07x opamp and measure it.

2) The left opamp is a comparator whose output switches high, low, high, low over and over. It drives R3 in the right opamp circuit (the integrator) that linearily charges and discharges C1.

3) The comparator opamp switches its output polarity when the non-inverting input passes the 0V of the inverting input which occurs when the current in R1 is the same as the current in R2.

4) We know the output voltage of the comparator (plus or minus 13.5V) and Ohm's Law gives the current in R2 which is the same as the current in R1 at the switching voltage and Ohm's Law again gives the voltage of the peak of the triangle waveform at R1.

 

[MrAl]

thanks MrAl and Audioguru
* MrAl : sorry but the output is +/- 11 volt not 10 like i thought .
* Audioguru : sorry i don't understand what you mean ,,, i can get the relations in post #53 ?

Hi,

No need to be sorry, that just means that the output of the op amp is higher than 12 volts. I used 12 volts because that is a typical value that might be expected, but you can measure it yourself and we can calculate the new more accurate output voltage.
If the output was really 11 volts then that means the input was higher like 13 volts. In real life it could very well be different anyway so it's best to stay away from an exact value. There is no clear value for the output of the op amp on the data sheet, so we are going by their rather vague data set given there.

 

[Adam2014]

MrAl and Audioguru thank you for your helps

for MrAl in your post #49 you said the output is Vout=Vin*t/(C1*R3)+Vc0
i just want to know how you calculate it , is there any analysis ?

 

[MrAl]

MrAl and Audioguru thank you for your helps

for MrAl in your post #49 you said the output is Vout=Vin*t/(C1*R3)+Vc0
i just want to know how you calculate it , is there any analysis ?

That expression comes from knowing beforehand how op amp integrators work, although it can also come from a knowledge of op amp amplifiers in general.

Those integrators very often appear as just a single capacitor in the feedback (output to inverting input) and a single resistor on the input (Vin to inverting terminal) with the non inverting terminal to ground. They appear so much that after a while you dont have to analyze it anymore because you remember that form.

If we look at it as just a general op amp amplifier, then we know that the 'gain' of the op amp is the feedback impedance divided by the input series impedance. So that would be zFB/zIN. Since zFB=1/(s*C) and zIN is just R, we get:
Gain=(1/(s*C))/R=1/(s*C*R)

and it is an inverter so the gain is -1/(s*C*R).

So the output is -Vin/(s*C*R) and for a step input the output is -Vin/(s^2*C*R) in the frequency domain. Using the Inverse Laplace Transform we get to the time domain (there are other ways however) and that gives us:
Vout=-Vin*t/(R*C)

and that is a ramp.

To account for the initial capacitor voltage Vc0 in this case we note that the capacitor voltage is equal to the output voltage so if there was any initial cap voltage it would equal Vout to start with at t=0, so we get finally:
Vout=-Vin*t/(R*C)+Vc0

Another way is to simply model the op ap as a voltage controlled voltage source with known gain A and just do a complete analysis that starts with:
Vout=A*(vp-vn)

where vp is the voltage at the non inverting terminal and vn is the voltage at the inverting terminal and A is the internal gain of the amplifier. The gain A is usually high in an op amp so what we do to simplify the equation a little is after we write the equation we let A go to infinity and that removes A from the equation and leaves us with the same result as above.

Still yet another way is to start in the time domain with a state vector differential equation and then solve it for all time.

So you see there are several approaches.

Also note that the time equation shown above is correct but we must apply it in a non linear way. We must manually limit the time period to 1/(2*f) because at that time the first op amp output polarity changes, so we have to stop and restart with a different Vin and previous Vc0 to get the output for the next half cycle. So for example if we calculated the output to be +11 volts at the end of the first half cycle with Vin=-13, then we have to make Vc0=+11 and Vin=+13 and then do the calculation again to get the next half cycle.