Can a first-order PLL be implemented in a system with an analog VCO? - Page 1
Forum section: electronics / beginners

author: tverbeure
author_rank: Regular Contributor
timestamp: July 18, 2023, 05:58:27 am
content: Can somebody verify the steps in my reasoning, and where I'm getting off the rails?In a first order PLL, the phase error is used to vary the phase of the generated signal, while keeping the frequency constant. In a second order PLL, the phase error can change both the phase and the frequency.It is possible for a first order PLL to track an input with a slightly different frequency than the generated signal by constantly updating the phase, but even then, the range of the PLL is quite limited. Meanwhile, the range of a second order PLL is pretty much as large a the range of the VCO.In a pure digital system with a numerical controlled oscillator, it's easy to implement both kinds of PLL.For the first order PLL, the phase update for each step is:phase = phase + phase_increment + alpha * phase_error.phase_increment is a constant, hence the fixed frequency of the PLL.In a second order PLL, you get:phase_increment = phase_increment + beta * phase errorphase = phase + phase_increment + alpha * phase_errorBut how does that work in system with an analog PLL that uses a VCO? It seems to me that for those cases, it's impossible to design a first order PLL, unless you apply some kind of pulse function at the input of the VCO that temporarily increase the frequency to accelerate the phase, and then revert back to the stationary position?Edit: I just realized that the output of the phase detector, before low pass filtering, is... a train of pulses. So if you don't install a low pass filter, I guess that means that gets you a first order PLL right there? And the alpha value for the update is simply the amount by which the amplitude of this pulse gets amplified or attenuated.

author: thermistor-guy
author_rank: Frequent Contributor
timestamp: July 18, 2023, 07:12:36 am
content: Quote from: tverbeure on July 18, 2023, 05:58:27 amYou can think of the loop order as equal to the number of integrators. The VCO counts as one integrator: since it converts a DC voltage to frequency, it effectively convertsa DC voltage to phase and then integrates the phase.An integrator in the path of the phase detector's output, say an analog integrating op-amp stage, counts as another.So if you don't integrate the phase error signal, but just feed it directly into the VCO, you get a first order loop. Pass the phase error signal through aintegrator and then into the VCO, you have a second-order loop.Analog integrator stages can be imperfect (lossy), and you'll see that in practical circuits.Designers pick the order of the loop depending on what steady-state errors are allowed. Third-order loops, not so common, are used when you need tofollow a signal with Doppler shift, for example.Loop orders higher than two are certainly possible, but they get tricky to analyze. There's a lot of non-linear behaviour "under the hood".

author: tverbeure
author_rank: Regular Contributor
timestamp: July 18, 2023, 07:56:52 am
content: I had seen the claim in various publications before that the VCO itself counts as an integrator, but I never quite got why that’s the case. Now I get it, and it makes total sense.