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All my theory tells me this should oscillate and I should have a triangular wave at \$V_a\$ and a square one at \$V_b\$ But those nodes all come out completely zero.

I don't understand what I have incorrect, could someone please point me in the right direction?enter image description here

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  • \$\begingroup\$ You might want to explain why you think those should oscillate so that one can discover the difference between that and what actually happens. \$\endgroup\$
    – PlasmaHH
    Oct 28, 2016 at 10:24
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    \$\begingroup\$ C is tiny. Try 121 nf or make sure you're using a VERY fast opamp. \$\endgroup\$
    – user16324
    Oct 28, 2016 at 10:24
  • \$\begingroup\$ True, I've tried 121.95 nf as well but it still does nothing. The opamp models are ideal so it has infinite slew anyway. And in response to @PlasmaHH because it charges the capacitor based on current at the output of U1 through R which is the opposite polarity to its initial voltage and so when it hits or passes zero at Va the output should swing opposite and it oscillates accordingly. \$\endgroup\$
    – Supernovah
    Oct 28, 2016 at 10:36
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    \$\begingroup\$ I'm wondering if this is due to the perfection of the simulation - normally there is a small offset voltage when you switch on which is translated by U1 to either a + or - V at its output to kick the whole thing off. Just out of curiosity what would happen if you added a 10M resistor from +V supply to the inverting input of the integrator (U2). \$\endgroup\$
    – JIm Dearden
    Oct 28, 2016 at 10:42
  • \$\begingroup\$ @Supernovah: for that to happen you need to simulate with a nonideal opamp \$\endgroup\$
    – PlasmaHH
    Oct 28, 2016 at 10:42

4 Answers 4

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This circuit has a stable operating point for an output of zero volts. This often happens with oscillators. A common technique is to either inject a current pulse in some node using a piecewise linear current source or force an initial condition that is different from zero.

For example add .IC V(Vb) = 1

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  • \$\begingroup\$ WRONG. The U1 has a bipolar output thus a DC offset on U2 (1st Amp) which then integrates until zero crossing is reached on output which is the midpoint reference between square wave sings Vcc to Vee ( if simulation does not roll off frequency response it is always ASTABLE not DC stable at 0V) also RC integration time have open loop gain breakpoint gain >1 to satisfy conditions for loop gain > 1 oscillation \$\endgroup\$
    – Tony Stewart EE75
    Oct 28, 2016 at 15:05
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    \$\begingroup\$ No, it's exactly as I wrote. The circuit has a trivial DC solution with all signal nodes at zero volts. Therefore measures are needed to move the circuit out of this condition. \$\endgroup\$
    – Mario
    Oct 28, 2016 at 15:11
  • \$\begingroup\$ Vb can never be 0V by definition since Vin (diff)<>0 thus not linear thus saturated, thus U2 will always integrate, unless simulator has 0 input offset ( ideal) \$\endgroup\$
    – Tony Stewart EE75
    Oct 28, 2016 at 15:15
  • \$\begingroup\$ even 1 picovolt of input offset is adequate initial condition for oscillation \$\endgroup\$
    – Tony Stewart EE75
    Oct 28, 2016 at 15:21
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    \$\begingroup\$ I wouldn't call it a fault. High-end simulators would show the same result, in fact it's a sign of high accuracy at least for more complex circuits. However, your models or setup could be slightly unbalanced or you have a "clever" solver that discards trivial solutions. \$\endgroup\$
    – Mario
    Oct 28, 2016 at 15:58
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Your simulator is finding a stable but unrealistic point. You need to give it a kick to get things started. You could put and initial condition on the capacitor or set one nodes to something other than 0. You don't say what opamp you are using. Brian is correct that you will need a very fast opamp to make this work. Your time step may be a problem too. Your oscillator will have a period of about 0.2us so your step should be less than this.

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I think, the problem of your relaxation generator neither is a missing "kick" not any other starting aid. It is simply the limited slew rate of the opamp that does not allow operation as desired. The integrator time constant is app. 180nsec only. Just to test the circuit - try to increase the capacitor by a factor of 1000 and see if it works.

Such a relaxation generator does not need any starting aid at all (assuming real opamp models and finite power supplies) because the integrator will start ramping at t=0.

More than that, these oscillators will not work for ideal opamps (VCVS) because the opamp with pos. feedback must be able to "jump" to a finite voltage. Hence, a real model with fixed supply limits is required.

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This self-starts and seems to work pretty well: (note the reasonably fast unity-gain-stable opamp)

enter image description here

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  • \$\begingroup\$ EM Fields, I have repeated the simulation for fixed supply voltages (+-5V and +-15V). The oscillation did start immediately (without ramping the supply voltages). For a +-5V supply the frequency was 1.9 MHz. However, for +-15V the frequency was only 1.7MHz. This frequency reduction is due to the limited slew rate of the LT1208 unit. \$\endgroup\$
    – LvW
    Oct 28, 2016 at 14:27
  • \$\begingroup\$ @LvW: 1. For any power supply, the transition time from OFF to ON can't be infinitesimal, so the ramping of the supplies is incidental in that it's not being applied as a forcing function, it comes about as a consequence of allowing the simulator to turn on the supplies starting at zero volts, more closely simulating the real world.**2.** I don't quite get why you're pushing the point about the frequency being slew-rate dependent, since that's an inherent quality of any multivibrator, but I do detect what, to me, seems to be a condescending attitude coming from your camp. Am I wrong? \$\endgroup\$
    – EM Fields
    Oct 28, 2016 at 16:58
  • \$\begingroup\$ EM Fields - I agree to part 1 of your comment. However, I must admit I cannot understand how you could come to the conclusion written down in the last part of your reply. I think, my comment contains technical results only - nothing else. How could such information be an indication of a condescending attitude? I cannot understand. It was my only intention to supplement your simulation results because you didn`t mention the oscillation frequency. \$\endgroup\$
    – LvW
    Oct 29, 2016 at 8:54
  • \$\begingroup\$ EM Fields - I like to add the following: My first comment was primarily dedicated to the questioner (supernovah). Thus, it would have been better to address this comment to supernovah instead of to you. Perhaps this has caused your misunderstanding? \$\endgroup\$
    – LvW
    Oct 29, 2016 at 9:03
  • \$\begingroup\$ @LvW: perhaps "condescending" was too strong and I should have used "patronizing" instead. If that's true, I apologize. The point I was trying to make was that using a pair of quick opamps in the circuit would bring it to life, which I proved. You then took it upon yourself to demean the raw opamp for its failure to maintain a constant frequency as its supply voltage changed, even though, as you should know, that feat is impossible to overcome. All I see, so far, is self-aggrandizement. \$\endgroup\$
    – EM Fields
    Oct 29, 2016 at 20:00

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