Timeline for Struggling to understand how a phase locked loop reaches lock
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 23, 2021 at 19:55 | comment | added | Andy aka | Listen, when you have a phase locked-loop, you don't need the feedback resistor so dump it because, even with a massive DC open-loop gain (circa 100,000 for an op-amp), the slug caused by the integration capacitor will usually mean you can avoid instability. At the very least, make it one hundred times higher and see how things pan out @Matt_LDN | |
| Oct 23, 2021 at 18:54 | history | edited | Neil_UK | CC BY-SA 4.0 |
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| Oct 23, 2021 at 18:43 | comment | added | John D | There's a 10K resistor across the integrator cap, so naturally it's not an ideal integrator and the output will return to zero. Also, the DC gain of the circuit is only 10, so it's not even a very good approximation of an integrator. For simulation purposes try making the 10K resistor 1 meg or so. But you still have to make sure your loop is stable with the additional gain. | |
| Oct 23, 2021 at 18:34 | comment | added | Matt_LDN | Great, thank you for the response. That's the simulation I was using multisim.com/content/DJoiPxiirFBDnN8WtgTLcg/op-amp-integrator When you turn off the input, the output just drops to zero. | |
| Oct 23, 2021 at 18:26 | history | answered | Andy aka | CC BY-SA 4.0 |