You cant make a 500 Mhz oscillator with an op amp.
You cant make a 500 Mhz oscillator with an op amp.
Ah sorry my mistake, The hazard of reading posts at 12:40 AM.Check the capitalization; the O.P. wants a 500mHz oscillator (which is, as they said, slow).
Okay, so after thinking about it for a bit longer, here's what I came up with for differentiating between an amplifier and a Schmitt Trigger.
If we take a look at the Inverting Amplifier:
**broken link removed**
Say initially Vout is at the negative supply, if we think about it, Vin can't be more negative than the negative supply, (or else the op-amp is useless), so the direction of current flow can be determined, therefore the voltage division can be determined as:
[tex]V_{-} = (V_{in} - V_{out})(\frac{Rf}{Rf + Rin})[/tex]
With this equation, V- can never be less than zero, thus, Vout will remain negative and not change. For this reason, the amplifier will be stable and not oscillate. The same works if Vout was initially held at the positive supply voltage.
In the case of the Schmitt Trigger, the output voltage doesn't not stay stable and it will oscillate.
This more or less makes sense to me, but I'm still not sure where the ideal op-amp equations come into play with this, but it's a start.
The offset is likely cause by the op amp having different output saturation voltages between the plus and minus outputs (the output doesn't go all the way to the supply rail). If you use a rail-to-rail type op amp, such as the MCP6002 by Microchip, then the output should be more symmetrical.........Results:
After wiring this up on my breadboard and checking the scope, I noticed the follow:
f = 460 mHz
Vpeak(+) = 1.68 V
Vpeak(-) = -2.16 V
Appears to be symmetrical but is offset.
My question is, what is the relation between the +/- supplies and my peak voltages, also how do I get rid of this offset?
Hi,
Usually we can assume certain things about the op amp when used as an amplifier, and we assume other things when used with positive feedback. The trouble is, there is another variable and that is time. If we dont assume that it takes some finite time to reach some voltage level then we can find two solutions to the general circuit you have drawn as an 'inverting Schmitt Trigger" with Vin=1v. If we force the non inverting terminal to be positive when we start out we'll find some normal looking solution, but if we dont do that and we take time into consideration we find that the output swings negative and reached the negative supply rail and stays there. So it depends on the initial conditions as well as the inputs in reality. If the initial non inverting terminal voltage is zero (usually a good assumption) and we apply a 1v signal to the inverting terminal, the output RAMPS according to that initial condition which here means it would ramp negative. Since there is no solution with the non inverting terminal going negative because the non inverting terminal is always more positive, the output would continue to seek a more and more negative value and would reach infinity with an ideal voltage controlled voltage source. But if we assume other initial conditions we could end up with a more normal looking solution.
In real life it becomes bistable.
Very good observation there.
Hey, thanks for the detailed post! I guess it's safe to say that time does have its place in these circuits. The amplifier configurations seem to force their inputs to be equal using negative feedback and the Schmitt Trigger cannot be stabilized, so it goes to its extremes.
I am just having a hard time understanding quantitatively how positive and negative feedback work in these configurations. They both have the same feedback loop, is it just the effect that the amplifier configuration seeks to close in on decrementing the input difference between the input terminals (v+ and v-) until they are roughly equal, while the Schmitt Trigger does the opposite in the sense that its feedback loop causes the voltages to become more and more spread apart in the current ramping direction?
Hi again,
Quantitatively you really only have to apply the initial conditions, but in real life the op amp acts more like a fast integrator anyway and since you are trying to understand it better in both configurations perhaps you should assume that the perfect op amp has a small capacitance from output to inverting input, and the inverting input has some resistance between the actual input and the capacitor (ie an integrator circuit). This will show how time enters the solution, and this would be a lot like it really behaves in real life except for the limited slew rate. If you wanted to you could even show the slew rate by using a value of capacitor that provides the correct ramp rate with a given step change in voltage on the 'input'.
With a capacitor on the non inverting terminal (for example) and some initial bias (initial capacitor voltage) you may get a totally different solution. Some circuits have more than one solution for different initial conditions and this shouldnt be too much of a surprise.
Also keep in mind that with positive feedback the circuit could be considered non linear so the solution(s) may not fit any nice neat textbook example.
The key to getting an exact even +/- triangle swing is setting the exact schmidt trigger thresholds at +2v and -2v.
This can be difficult with the basic circuit as drawn, because the output of the schmidt trigger op amp cannot usually go fully and accurately rail to rail. That skews the schmidt trigger voltage thresholds.
One simple solution to this is to use a non inverting CMOS buffer right at the output of the op amp. Even though the op amp struggles to approach rail to rail, the CMOS buffer will provide a fast hard full rail to rail output voltage swing. Take the feedback for the schmidt trigger from the output of the CMOS buffer, not from the output of the op amp, and you will then get a nice symmetrical triangle wave.
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