Inverting Op-amp signals and Sawtooth waveform generation

  • #36
DaveE
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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 286620


View 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).

Tri_210726-1_sch480.jpg
 
  • #37
DaveE
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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.
 
  • #38
Baluncore
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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.
 
  • #39
DaveE
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A better version, 1 less resistor:
Tri_210726-2_sch480.jpg
 

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  • #40
Callum Plunkett
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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.
wave2.png
 
  • #41
DaveE
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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)?
 
  • #42
bob012345
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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.

Screen Shot 2021-07-27 at 3.19.47 PM.png
 
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  • #43
Callum Plunkett
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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.
 
  • #44
DaveE
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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.
 
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  • #45
bob012345
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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.html


Of 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.
 

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