Inverting Op-amp signals and Sawtooth waveform generation

[Callum Plunkett]

I’ve been asked to create a sawtooth wave form with four input voltages. After doing this I then added a second op-amp to both change the range of the wave form from -/+5v to -/+2.5 and DC offset of 2.5v to create a wave form that ranges from 0v – 5v as required by the question. For the most part I think I’m close enough with my result apart from the second op-amp seems to of “flipped” the wave form and I can’t quite figure out why and how to solve the issue.

I've included a picture of my circuit, wave form after the first op-amp and after the second op-amp.

Any help would be appreciatedhttps://www.physicsforums.com/attachments/286448https://www.physicsforums.com/attachments/286449https://www.physicsforums.com/attachments/286450

 

[berkeman]

Hi Callum,

I moved your thread to the schoolwork forums from the EE forum, and something about that move messed up your attachments. Could you please re-attach them or add a new reply with them? Also, they were pretty hard to read (when they still existed) -- Is there some way to make the bigger and darker? Thanks.

 

[.Scott]

Could you redo your pictures. I can't read them.
More resolution is required - and the pictures only occupy the top left 10% of the available canvas.

 

[Callum Plunkett]

Could you redo your pictures. I can't read them.
More resolution is required - and the pictures only occupy the top left 10% of the available canvas.

Yeah sorry about that by the i tried to change it but it would not up date. Ae these better?

 

[berkeman]

Whew, much better, thanks!

So you mention synthesizing a "sawtooth" waveform with the 4 voltage sources. You've done a good initial job of doing that, but do they want it to be more like a straight sawtooth instead of stepped? Can you say what you could change/add to the circuit to make it more like a traditional sawtooth?

 

[Callum Plunkett]

Whew, much better, thanks!

So you mention synthesizing a "sawtooth" waveform with the 4 voltage sources. You've done a good initial job of doing that, but do they want it to be more like a straight sawtooth instead of stepped? Can you say what you could change/add to the circuit to make it more like a traditional sawtooth?

View attachment 286459

This is the full question as i was given.

 

[berkeman]

Ah, in that case it looks like you are doing the correct synthesis.

 

[Callum Plunkett]

Yeah i think I am on the right track with it, so I am quite relieved someone else thinks I am some where near.
Cheers.

 

[Jody]

The last op-amp has positive feedback. Isn't that going to be a problem?

 

[berkeman]

The last op-amp has positive feedback. Isn't that going to be a problem?

Good catch! I saw that the simulation output for X2 looked reasonable, so didn't check the schematic in detail. But yeah, with the opamp inputs inverted like that I don't see how the simulation isn't just hitting the rails...

 

[bob012345]

Good catch! I saw the simulation output for X2 looked reasonable, so didn't check the schematic in detail. But yeah, with the opamp inputs inverted like that I don't see how the simulation isn't just hitting the rails...

Op amps invert the signal so you can just use one more with a unity gain to flip it back.

 

[berkeman]

Op amps invert the signal so you can just use one more with a unity gain to flip it back.

I'm pretty sure you're kidding. An opamp with the inputs reversed so that the feedback is postive is a comparator, not a linear amplifier.

 

[bob012345]

I'm pretty sure you're kidding. An opamp with the inputs reversed so that the feedback is postive is a comparator, not a linear amplifier.

A summer gives the negative of the sum. You add another op amp stage to invert that signal.

 

[berkeman]

A comparator slams to the +/- power supply rails when the difference between its inputs is positive or negative. Are you suggesting that a comparator can be used as a linear amplifier like an opamp that uses negative feedback?

 

[bob012345]

A comparator slams to the +/- power supply rails when the difference between its inputs is positive or negative. Are you suggesting that a comparator can be used as a linear amplifier like an opamp with negative feedback?

No. I just thought the issue was the signal was flipped as in multiplied by a minus sign. I was telling how to make it positive.

 

[berkeman]

No. I just thought the issue was the signal was flipped as in multiplied by a minus sign. I was telling how to make it positive.

Okay, that's fair. The quote of mine that you used was agreeing with the catch by @Joshy about the incorrect connection of the 2nd opamp. The question about the inversion or whatever of the signal was in a different post.

 

[bob012345]

Okay, that's fair. The quote of mine that you used was agreeing with the catch by @Joshy about the incorrect connection of the 2nd opamp. The question about the inversion or whatever of the signal was in a different post.

That's fine. Sorry for the misunderstanding.

 

[bob012345]

This is the full question as i was given.

View attachment 286461

Does the output have to be just like fig 2b (going between 0 and 5V) or just similar in shape? Are other DC biases allowed? I take it the input biases all have to be 5V as given. I ran it in LTSpice and got what you got but it's not clear how literal they want it.

 

[Baluncore]

Did you consider simply adding a DC current to the AC harmonic currents at the summing input?
That way it provides the offset without needing a second op-amp.

 

[DaveE]

A summer gives the negative of the sum. You add another op amp stage to invert that signal.

Op-amps have nearly infinite gain, so they will always have a circuit wrapped around them. It is easy to make a summer with positive gain, negative gain or any mixture you like, as in a differential configuration. They can do lots of different things. For example:

This allows you to easily figure out how to do all sorts of summing, subtracting, level shifting, inverting with 1 amp and nearly zero effort. You just need to keep it balanced for the zero algebra option.

edit: also, IF IT'S BALANCED, the ground for the output is the ground at the lower R4, which can be different from the grounds for the inputs (with some common mode limitations, of course). However, "balanced" is relative; variations in the real resistor values will degrade the CMRR, sometimes significantly. There are other configurations which have much better CMRR, like instrumentation amps.

 

[bob012345]

Here is my version of it. I can't get a decent, readable screenshot of the output but it goes between 0 and 5V.

 

[DaveE]

Good work!

A couple of comments: These aren't corrections, more like refinements, like how the pros would do it slightly differently.

1) It is considered bad design practice* to use the power supply voltages for anything except to power devices. Any connection into the signal path like the offset function you needed in this problem is a bad idea. The power supply voltages (usually) aren't intended to be very precise, they are noisy, and can create some really annoying cross coupling between different parts of the system. For example, if this circuit was used in an audio synthesizer, you would probably have lots of hum, pops & clicks, or and other noise at the output (in addition to your triangle wave) that was injected into your signal path through R6. So, in practice you would add a voltage reference of some sort (Zener diode, IC, etc.) to provide a clean and stable reference voltage for your offset.

* In the real world "bad design practice" means a circuit that mostly works but may have problems like low reliability, poor performance, hard to build, expensive, etc. (i.e there are better ways to do it).

2) It's much harder to buy a 3.5KΩ resistor than a 3.48KΩ resistor. Same for many other common values. Resistors are most often supplied with EIA standard values (usually the 1% tolerance for analog designs, in my experience). This really doesn't matter for homework problems or understanding. Theoretically useless, but how the real world works. Similar to choosing nut and bolt sizes. As an aside, I can usually immediately tell if a schematic was drawn by a very experienced engineer by things like this. You really wouldn't learn about these details until you had to choose resistors from a standard list (your stock room, distributors, etc.). For example, go online to your favorite distributor (I like https://www.digikey.com/en/products/filter/chip-resistor-surface-mount/52) and search for a 3.5KΩ 1% 0805 SMT resistor, see how many choices there are. Then do the same for 3.48KΩ.

3) If you adjusted your offset so that the output didn't go below 0V ever, then there would be no need for a negative power supply, you could replace it with a ground connection. You would have to choose an op-amp that can output 0V though, but these are common.

If you have lots of free time and you want a puzzle to solve, try this design using an LM10 op-amp with it's built-in reference and single supply capability. That might be how I would do it.

edit: These are especially not corrections, since this is clearly an unrealistic exercise. No one actually makes triangle waves by starting with 4 synchronized sinewaves. Even if those sinewaves already existed in my design, I wouldn't use them for this.

 

[Baluncore]

Fourier synthesis of the sawtooth waveform is a difficult exercise because there are so many harmonics, and they fall off too slowly. Saw = f1/1 + h2/2 + h3/3 + h4/4 + h5/5 + , , ,

Notice how, each time another harmonic is added to the synthesis, it inverts and attenuates the error function. Since the number of harmonics is sharply limited, I would attenuate the higher given harmonics so as to eliminate the reversals in slope on the rising ramp. That would slightly increase the fall-time, but it would significantly reduce the error function.

That error function minimisation could also be achieved by including a Miller integrator capacitor across the op-amp feedback resistor, but that might be break the rules of the exercise.

 

[Baluncore]

Here I have replaced the coefficients {10k, 20k, 30k, 40k} with {10k, 22k, 40k, 70k} to take the bumps out of the ramp.

 

[bob012345]

3k7=3700?

 

[Baluncore]

3k7=3700?

Yes. The decimal point is replaced by the multiplier in circuit schematics.

 

[bob012345]

If you have lots of free time and you want a puzzle to solve, try this design using an LM10 op-amp with it's built-in reference and single supply capability. That might be how I would do it.

What would be the advantage of using this?

 

[DaveE]

What would be the advantage of using this?

Maybe none, but it's a good exercise.

Operation from a single power supply is often very desirable. For example, suppose your design operates off of a single battery. If you can design your circuits to only use that single power supply voltage you will save lots of complexity, cost, size, etc. If not, you may also be designing a negative output switching power supply as part of your solution. I would wager that nearly every circuit in your average car operates off of a single +12V power supply.

The value of using a dedicated reference for your offset I previously discussed.

Also, buying fewer components with the features you need integrated within them saves lots of space, cost, and can increase reliability.

 

[Callum Plunkett]

Sorry for the late reply I've been busy with work the past few days and have not had a chance to get back to anyone.

 

[Callum Plunkett]

There is a lot of information here and its a massive help. Electronics isn't something I do that much of at home or at work so it does not come that natural to me.

Thank you guys.

 

[Callum Plunkett]

Did you consider simply adding a DC current to the AC harmonic currents at the summing input?
That way it provides the offset without needing a second op-amp.

I did that in an earlier attempt at the question, I can't remember why now but I decided to ty a different method. Might have to reconsider going back to it.

 

[bob012345]

Maybe none, but it's a good exercise.

Operation from a single power supply is often very desirable. For example, suppose your design operates off of a single battery. If you can design your circuits to only use that single power supply voltage you will save lots of complexity, cost, size, etc. If not, you may also be designing a negative output switching power supply as part of your solution. I would wager that nearly every circuit in your average car operates off of a single +12V power supply.

The value of using a dedicated reference for your offset I previously discussed.

Also, buying fewer components with the features you need integrated within them saves lots of space, cost, and can increase reliability.

I need a spice model for the LM10. It's not in LTspice. An equivalent part in LTspice would suffice. Thanks.

 

[DaveE]

I need a spice model for the LM10. It's not in LTspice. An equivalent part in LTspice would suffice. Thanks.

I think you could use any single supply rail-to-rail op-amp model. Then wire it to be like what's inside the LM10.

edit: Use LT1635 it should be in there.

 

[Averagesupernova]

I just want to add to this thread that I wish I had gotten to know spice a lot sooner than I did. Anyone who tinkers with electronics should try it. Nothing to be afraid of.

 

[bob012345]

I think you could use any single supply rail-to-rail op-amp model. Then wire it to be like what's inside the LM10.

edit: Use LT1635 it should be in there.

Well, here it is with the LT1635. I could not get to 5V because of the limitations of the op amp I believe. I had to add the capacitor also. I tweaked the resistors further but I think it is not in the spirit of the exercise. Looking at the literature, this class of op amps has a lot of sophisticated uses and using it for a simple problem such as this I think is overkill. Still, it's fun to try.

Also, I'm working on a completely different approach so please don't spill the beans just yet.
Thanks.

 

[DaveE]

Well, here it is with the LT1635. I could not get to 5V because of the limitations of the op amp I believe. I had to add the capacitor also. I tweaked the resistors further but I think it is not in the spirit of the exercise. Looking at the literature, this class of op amps has a lot of sophisticated uses and using it for a simple problem such as this I think is overkill. Still, it's fun to try.

Also, I'm working on a completely different approach so please don't spill the beans just yet.
Thanks.

View attachment 286620View attachment 286622

Yes. "Rail-to Rail" is always a lie. Those saturation characteristics graphs are one of the most important parts of the data sheet for a part like this. For a high impedance load you should be able to get within <0.1V or so if you're careful (i.e. don't make it source much current).

I'm not sure the exercise wants it to get to 5V. That would be the peak voltage for lots of harmonics, but with only 4 I think the top will be rounded off below that.

That cap isn't necessary if you choose the right resistor values. You seem to have missed the point about using a part with a built in reference to make the offset. The circuit below does this job with a single power supply, 1 IC and 7 resistors (although in the real world, I'd add a couple of caps too).

 

[DaveE]

Also, I'm working on a completely different approach so please don't spill the beans just yet.

Do you mean the one with a 555 timer and a current source? That would be the standard approach for sawtooth generation. Or just buy a SMPS PWM IC and only use the ramp generator in it.

 

[Baluncore]

I just want to add to this thread that I wish I had gotten to know spice a lot sooner than I did. Anyone who tinkers with electronics should try it. Nothing to be afraid of.

And no one should be afraid to ask us for help during their initial learning curve.

 

[DaveE]

A better version, 1 less resistor:

 

[Callum Plunkett]

Here is my version of it. I can't get a decent, readable screenshot of the output but it goes between 0 and 5V.View attachment 286511

View attachment 286513

How did you come up with the coefficients of 10,20,30,40 and RF of 3.5?

I’ve arbitrarily used 10kohms for my RF with coefficients of 10kΩ, 13.3kΩ, 20kΩ and 40kΩ which should give me gains of 1.25, 2.5, 3.75 and 5.

I was hoping as there are 4 inputs at 5volts each that 5/4=1.25 and multiples of, would give a smoother wave form. However not quite the case.

 

[DaveE]

How did you come up with the coefficients of 10,20,30,40 and RF of 3.5?

I’ve arbitrarily used 10kohms for my RF with coefficients of 10kΩ, 13.3kΩ, 20kΩ and 40kΩ which should give me gains of 1.25, 2.5, 3.75 and 5.

I was hoping as there are 4 inputs at 5volts each that 5/4=1.25 and multiples of, would give a smoother wave form. However not quite the case.
View attachment 286657

Are you asking why the output solution for your circuit is the equation below?
$$ V_o = -R2( \frac{V5}{R4} + \frac{V4}{R3} + \frac{V3}{R1} + \frac{V6}{R5} ) $$

Or are you asking how to choose each of those gain terms (i.e. the resistor values)?

 

[bob012345]

How did you come up with the coefficients of 10,20,30,40 and RF of 3.5?

I’ve arbitrarily used 10kohms for my RF with coefficients of 10kΩ, 13.3kΩ, 20kΩ and 40kΩ which should give me gains of 1.25, 2.5, 3.75 and 5.

I was hoping as there are 4 inputs at 5volts each that 5/4=1.25 and multiples of, would give a smoother wave form. However not quite the case.
View attachment 286657

I just used the given formula with ratio's 1, 1/2, 1/3 and 1/4.

$ν(t) = \frac {2}{π} (sin(wt) + \frac {1}{2} sin(2wt) + \frac {1}{3} sin(3wt) + \frac {1}{4} sin(4wt) )$

Rf should have been exactly $10k * 2/pi *1/2$ which should have been 3.183k in keeping with the exercise but I did not realize that then. Here, I made it simpler using 1k for the lowest resistor. I think this is in keeping with the original exercise.

 

[Callum Plunkett]

Are you asking why the output solution for your circuit is the equation below?
$$ V_o = -R2( \frac{V5}{R4} + \frac{V4}{R3} + \frac{V3}{R1} + \frac{V6}{R5} ) $$

Or are you asking how to choose each of those gain terms (i.e. the resistor values)?

I was wondering how you came up with the gain values for each channel. But it seems I over looked the equation as mentioned by bob012345.

Cheers guys.

 

[DaveE]

I was wondering how you came up with the gain values for each channel. But it seems I over looked the equation as mentioned by bob012345.

Cheers guys.

That's an equation you will want to be able to derive on your own. It's not that difficult with the ideal op-amp assumptions.

 

[bob012345]

I was hoping as there are 4 inputs at 5volts each that 5/4=1.25 and multiples of, would give a smoother wave form. However not quite the case.

This animation shows how it works. As the number of terms grows the approximation to a sawtooth wave gets better.

https://kconrad.math.uconn.edu/math1132s10/sawtooth.htmlOf course, if one is forced to use only four terms, one can get a better fit by optimizing the values as was done in some of the previous posts.