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I am designing a low frequency oscillator. An image of the schematic is below. enter image description here

The oscillator part works fine. My intention is to scale the triangle wave at the output of U1 to +/-2.5 V by using U4. In the simulation this works correctly; the gain of U4 is of course -(68/100) = -0.68, or 0.68. In the simulation the triangle wave at the output of U4 is approx. +/- 2.5 V.

However, on the breadboard, Vtri at the output of U4 appears as a triangle wave equal in magnitude to its original wave. If I increase or decrease R9 it has no effect on the gain. I am not sure why. Perhaps the way I am trying to couple the output of U1 to U4 is not working the way I want? I also tried AC coupling the output of U1 to a noninverting amp, but this didn't work either.

I thought maybe somehow my 074 was damaged; I popped a 324 into its place and I got the same result. I am not sure why.

Can anyone suggest a way for me to amplify the output of U1 by 0.68, preferably with my 074 op amp?

Thank you.

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  • \$\begingroup\$ Does your VTri output have opposite polarity to U1's output as your would expect? Maybe post a pic of your breadboard in case there's a construction error. \$\endgroup\$ – brhans Jul 10 '16 at 1:43
  • \$\begingroup\$ There's nothing conceptually wrong with what you're doing, you have a wiring or component value error somewhere. \$\endgroup\$ – John D Jul 10 '16 at 1:48
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The circuit should work as diagrammed. I think you have a breadboard wiring error. There's 4 opamps in a TL074, and you appear to only be using 3. Try using U3 instead of U4. This will force you to rewire the breadboard, perhaps correcting the wiring error in the process.

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) What you want. (b) What might have happened?

Just suggesting you check for silly stuff: if the output from your U4 is in phase with the input it would suggest that you have somehow wired U4 as a unity gain buffer as shown in Figure 1b.

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  • \$\begingroup\$ The error you've posted was the first thing I've checked. I've gone through and re-wired my breadboard, using the U4 (pins 12, 13, and 14). The output stays high, which I think is a chip error, because the noninverting terminal is set at ground. The inverting terminal oscillates between +3 and -3. This should cause the output to go from low to high, which it does not. I replaced the 074 with a 324 and the output hangs at low instead. This is even more confusing. The 074 and 324 are both chips I've had around for some time. I am wondering if they are both damaged. \$\endgroup\$ – Thomas Wilk Jul 11 '16 at 22:01
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What you show should work. Most likely you have a wiring error arond U4. Check the datasheet and note how two of the opamps in the quad package have mirrored pinouts of each other.

That answers your question, but a few other things need to said about your circuit. That's one convoluted way to make a triangle wave oscillator!

  1. The slope of the triangle wave segments are dependent on the voltage that U2 clips at, which is based on its output headroom from each power rail. The headroom of a TL0xx is quite large and poorly specified, so the slope, and therefore the frequency, will not be well determined up front.

  2. I can't even guess what you think R6 and R8 are doing for you. All they do is slightly load U2, which makes no sense.

  3. What's with the mess of R7, R15, RRate1, and RRate2? All you need is a single resistor there. The slope of the triangle segments will be inversely proportional to that resistance. You don't need some fancy attenuator.

  4. R7 and RRate1 would be better replaced by two back to back zeners. Those would clip the large signal from U2 to a smaller and more predictable range. RRate2 would then be smaller, like maybe 10 kΩ. That puts more current thru the zeners, which creates a more accurate clipping threshold.

    You would then use the output of this zener clipper as the hysteresis feedback into U2 thru R5. That then also makes the hysteresis thresholds predicatable, which set the peak voltages of the triangle wave. R15 would stay there, and then would be the slope controlling part. You could make it a variable resistor to get varying frequency at a fixed amplitude.

  5. Make sure C1 is of the right type. Many ceramics exhibit decreased capacitance at higher voltages. That would make the triangle segments non-linear. A polypropylene or polystyrene cap might be appropriate here. The right ceramic could work too, especially if it is rated for significantly higher voltage than you are using it at.

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  • \$\begingroup\$ 1) The frequency is meant to be variable, and does not have to be precise, so I think this spec is okay. Based on my simulation I can get this to be as slow as about 4 Hz. 2) For R6 and R8 I wanted to reduce the square wave, which I would likely AC couple from the node between these two resistors. With those two resistors, the square wave becomes about +/- 2.5 V \$\endgroup\$ – Thomas Wilk Jul 11 '16 at 22:06
  • \$\begingroup\$ 3) I want to use a 100k pot in place of Rrate 1 and 2; R7 is to keep the output of U2 from being shorted to ground, which may not be necessary, or maybe it would stop U2 from oscillating. I am not sure. 4) The Zener tip is very helpful. I think that would be a better way to limit the square wave output amplitude, instead of later trying to voltage divide it with R6 and R8. \$\endgroup\$ – Thomas Wilk Jul 11 '16 at 22:13

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