Characterization of an Op-amp Function Generator Circuit

In summary: Next: at what negative V2 will U1 switch its output from +11 to -11V? U1 will switch to -11V when V2 becomes negative. Then the process repeats in the other direction.Yes, that is correct.
  • #1
lpuppy79
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1

Homework Statement


I need to characterize the circuit below in order to compare measured data to theory. Can you determine any the theoretical relationships for the frequency, amplitude, gain of the two voltage outputs? I measured the output voltage of each op-amp while varying both of the variable resistors individually. I can't seem to find any theoretical relationships for this circuit online, and I definitely don't have the knowledge to do it myself.

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Source: http://www.unirioja.es/cu/lzorzano/OPAMPcollection.pdf

Homework Equations


Don't know.

The Attempt at a Solution


Don't know what to do.
 
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  • #2
A classical way to put the relationship into mathematical terms is to represent the electronic components with Laplace transformations in the S-plane. Then the frequency response can be analysed. The Laplace transforms are very powerful and can be applied to very complicated circuits, but it is a subject in itself. If that is not what you are looking for, then there may be a simpler way to analyse it using more basic EE fundamentals.
 
  • #3
Considering I don't understand anything of what you just said, using EE fundamentals would be very appreciated.
 
  • #4
lpuppy79 said:
Considering I don't understand anything of what you just said, using EE fundamentals would be very appreciated.
Ok. Sounds like my answer was way off the mark. Sorry. I hope someone with EE background can help.
 
  • #5
FactChecker said:
Ok. Sounds like my answer was way off the mark. Sorry. I hope someone with EE background can help.
Thanks for trying!
 
  • #6
Yeah, Fourier Transforms will not be needed today... :smile:

@lpuppy79 -- What have you learned about opamps so far? What are the characteristics of an ideal opamp? How do those characteristics help you to write the equations for the transfer function of opamp circuits like this one?
 
  • #7
Also note -- that first opamp is being used as a comparator and not an opamp...
 
  • #8
Ideal Characteristics:
High Input Impedance
Low Output Impedance

Yes, the first is used as a comparator.

Is the second op amp essentially just an integrating circuit?
 
  • #9
lpuppy79 said:
Ideal Characteristics:
High Input Impedance
Low Output Impedance

Yes, the first is used as a comparator.

Is the second op amp essentially just an integrating circuit?

Yes, good!

The first comparator is used to make a square wave output, and the 2nd stage opamp integrates that to make a triangle output. That triangle output is fed back to cause the 1st stage comparator to keep tripping to make the square wave.

How would you go about figuring out the amplitude and frequency of the output triangle wave? Do you know what opamps you will be using in your lab measurements for comparison?
 
  • #10
vtri=-1/(RC)*integral(vsqudt) from 0 to T

ftri=fsqu

I already have taken the measurements. I used a LM358.
 
  • #11
lpuppy79 said:
vtri=-1/(RC)*integral(vsqudt) from 0 to T

ftri=fsqu

I already have taken the measurements. I used a LM358.

What did you use for your power supply rails? +/-12V? What was the square wave amplitude? When an opamp output pegs against the rails, the voltage depends on the opamp output stage. Some have "rail-to-rail outputs", and others like the LM358 are not... :smile:
 
  • #12
+/- 12 V as the power supply rails. The amplitude of both the square and the triangle wave varied based on the resistance of R2 so there was no constant square wave amplitude.
 
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  • #13
lpuppy79 said:
there was no constant square wave amplitude.
Are you sure about this?
 
  • #14
While varying R3 from 100-1000 kiloohms with constant R2 (1 megohm) , the frequency's range was .311 kHz to 2.05 kHz, vsquare ranged 22.9-23.1 V , vtriangle ranged 200-272 mV.
While varying R2 from from 0-1 megaohm with constant R3 (140 kilohms), the frequency's ranged from .02275-1.61 kHz, vsquare range 22.3-23.1 V, vtriangle ranged .253-18.3 V.
 
  • #15
At power-up, assume U1 (left-hand op amp) output V1 latches up to + 11Vdc, causing the integrator U2 output voltage V2 to crank downwards (at what rate? I mean, what is dV2/dt at the output of U2?)
Next: at what negative V2 will U1 switch its output from +11 to -11V? Note that this depends on the relative resistances of (R1+R2) vs. R5.
Then the process repeats in the other direction.
With this info you should be able to compute the pk-pk voltages V1 and V2, and the frequency as determined by dV2/dt.
Your data makes sense. The square wave pk-pk V1 should always be ~22V. The pk-pk of V2 depends on the setting of R2. Once R2 is set the pk-pk voltage V2 should not change as you vary R3. But at a high enough frequency the op amps no longer behave ideally so expect deviations from this statement at high frequencies.
Example: set R2 = 500K, V2 pk-pk should be about 0.36V pk-pk.
 
  • #16
1. In all scenarios, dV2/dt would be -1/[RC]=-1/[(R3+R4)C], correct?
2. V2=R5/(R1+R2+R5)*V1max?
3. If those are correct, I don't understand how to get the frequency or V1pp?
 
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  • #17
lpuppy79 said:
While varying R3 from 100-1000 kiloohms with constant R2 (1 megohm) , the frequency's range was .311 kHz to 2.05 kHz, vsquare ranged 22.9-23.1 V , vtriangle ranged 200-272 mV.
You could say that Vsquare was constant at about 23V when you tie this in with theory. Also, Vtriangle ideally would be constant.

While varying R2 from from 0-1 megaohm with constant R3 (140 kilohms), the frequency's ranged from .02275-1.61 kHz, vsquare range 22.3-23.1 V, vtriangle ranged .253-18.3 V.
Again, ideal op-amps would see Vsquare here not changing, while Vtriangle changes over a wide range.

Did you sketch V1 underneath V2 showing how they are related in time?
 
  • #18
lpuppy79 said:
1. In all scenarios, dV2/dt would be -1/[RC]=-1/[(R3+R4)C], correct?
You need to multiply this by a voltage, that voltage being V1's maximum

2. V2=R5/(R1+R2+R5)*V1max?
Not quite. How did you calculate this?

3. If those are correct, I don't understand how to get the frequency or V1pp?
 
  • #19
lpuppy79 said:
1. In all scenarios, dV2/dt would be -1/[RC]=-1/[(R3+R4)C], correct?
That's correct for {dV2/dt}/V1. You need to include V1 in dV2/dt.
2. V2=R5/(R1+R2+R5)*V1max?
Right. Except what about the sign of V2? remember, V2 is + before the switch.
3. If those are correct, I don't understand how to get the frequency or V1pp?
You need to think more about how the circuit works: V2 ramps down until the + input to U1 crosses zero V. Then U1 output switches from +11 to -11 and V2 changes direction, now going up in voltage.
I have to leave it at that. I practically told you what the pk-pk voltage V1 is already. Figure out how much time passes before V2 hits the voltage necessary to reverse the polarity of U1's + input every half-cycle. That times 2 gives you the period, and you do know how to calculate frequency given the period.
 
  • #20
rude man said:
2. V2=R5/(R1+R2+R5)*V1max?
Right. Except what about the sign of V2? remember, V2 is + before the switch.
Still wrong. (the sign change notwithstanding)
 
  • #21
NascentOxygen said:
Still wrong. (the sign change notwithstanding)
Right.
 
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Related to Characterization of an Op-amp Function Generator Circuit

1. What is an Op-amp Function Generator Circuit?

An Op-amp Function Generator Circuit is a type of electronic circuit that uses an operational amplifier (op-amp) to produce a specific waveform, such as a sine wave or square wave. It is commonly used in electronic testing and measurement equipment.

2. How does an Op-amp Function Generator Circuit work?

An Op-amp Function Generator Circuit works by using the properties of the op-amp to amplify and shape the input signal to produce the desired output waveform. The circuit typically consists of resistors, capacitors, and other electronic components connected in a specific configuration.

3. What are the key components of an Op-amp Function Generator Circuit?

The key components of an Op-amp Function Generator Circuit include an operational amplifier, resistors, capacitors, and a power supply. The exact configuration and values of these components will vary depending on the specific circuit design and desired output waveform.

4. What are the advantages of using an Op-amp Function Generator Circuit?

Some of the advantages of using an Op-amp Function Generator Circuit include its simplicity, versatility, and low cost. It can produce a wide range of waveforms with high precision and stability, making it useful for a variety of applications in electronics and engineering.

5. How is an Op-amp Function Generator Circuit characterized?

An Op-amp Function Generator Circuit is typically characterized by its frequency range, output waveform, amplitude, and distortion levels. These characteristics can be measured and adjusted using various testing and calibration techniques to ensure optimal performance of the circuit.

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