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This is a followup question to this question, I asked previously. My synthesizer PLL still isn't locking and it must be the introduction of the mixer and filter into the loop that's causing the issue (since the PLL works just fine without these). I'm using the LPF determined by the Analog Devices ADISimPLL design tool. This tool is oblivious to the extra stages I've added to the loop, so it's no surprise that the loop is unstable really.

Anyway, the obvious problem with having ADISim calculate the LPF for you is that it doesn't supply an accompanying transfer function (that I'm aware of), so you can't adapt the filter to handle different loop architectures like I'm trying to do. So, I've had to go back to fundamentals and try to figure out the TF for myself. I'm starting with the "bare-bones" PLL loop given by ADISim and plotting a Bode chart to see if it matches that given in the tool. After that, I plan to introduce my additional stages (mixer + LPF) into the loop to try calculate the correct LPF component values. Not sure if that will work out, but that's the plan and I'm learning loads as I go.

So, that's the background and the following is what I've achieved so far. My question is, how have I done?

System spec:

enter image description here

Circuit with LPF suggested by ADISim:

enter image description here

Transfer function calculation overview:

enter image description here

If I expand H(s) by including F(s) I get the following: enter image description here

enter image description here

I've substituted in the values given in the ADISim circuit to arrive at a final transfer function. This is essentially a third-order lowpass. I can model this in Python to see the Bode plot and compare:

import matplotlib.pyplot as plt
G_4=matlab.tf([26826.66,5.7078*10**7],
                        [1.873608*10**-7*np.pi
                          ,0 
                          ,4.4415
                          ,9450.0])


print(G_4)

from matplotlib.ticker import FuncFormatter

def units(x, pos):
    if (x < 1000): 
        return '%1.1f' % (x)
    elif (x < 1e6):
        return '%1.1fK' % (x * 1e-3)
    else: return '%1.1fM' % (x * 1e-6)

formatter = FuncFormatter(units)

fList3=np.logspace(np.log10(100),np.log10(10000),1000)
wList3=2*np.pi*fList3
mag, phase, omega=matlab.bode([G_4],omega=wList3,dB=True,Hz=True,deg=True)
ax = plt.gca()
ax.xaxis.set_major_formatter(formatter)

My closed-loop bode:

enter image description here

Versus the one given in ADISim:

enter image description here

Not quite sure why my passband gain is so high compared with the ADISim version and why my phase is positive, but they seem to agree on general shape and frequency.

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  • \$\begingroup\$ N looks somewhat similar to the difference in amplitudes... \$\endgroup\$
    – user16324
    Oct 16, 2020 at 14:50
  • \$\begingroup\$ I wondered if they'd normalized the graph in some way (or more likely, if I'd stuffed-up my derivations). It does look like the resonant part of their peak is higher (about 20db from the earlier passband, as opposed to my 5db or so). Anyway, I'm more concerned that my method of calculating a C.P based PLL loop is correct. \$\endgroup\$
    – Buck8pe
    Oct 16, 2020 at 15:00
  • \$\begingroup\$ If you're modeling your transfer function as \$H(s) = F_{out} / F_{ref}\$, then you'd expect to have a DC gain of $N$ -- ADI is putting their summing junction somewhere else, or otherwise normalizing the DC gain to 1. I can't figure out where your phase response is getting reversed, though. \$\endgroup\$
    – TimWescott
    Oct 16, 2020 at 15:01
  • \$\begingroup\$ In particular, the bit I'm worried about is Kd and Cp. The filter (minus Cp) is standard enough and covered fairly well in the literature. The gain for a charge-pump based PLL was more difficult to find. \$\endgroup\$
    – Buck8pe
    Oct 16, 2020 at 15:04
  • \$\begingroup\$ I'll include the earlier part of the maxima derivation, in case that helps. \$\endgroup\$
    – Buck8pe
    Oct 16, 2020 at 15:07

2 Answers 2

Reset to default
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I think you have a mistake in \$F(s)\$, I get:

$$F(s) = \frac{1+sR_2C_2}{sC_2(1+sR_1C_p)}$$

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  • \$\begingroup\$ well spotted and exactly right, I figured that out after I had submitted this question, but I never got around to putting in an update. When I made this change and substituted my component values I got exactly the same open and closed-loop plots. \$\endgroup\$
    – Buck8pe
    Feb 13, 2021 at 8:17
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That opamp circuit is DC_blocked.

Thus there no control of the VCO frequency.

You may need some variant of P_I_D (using 2 or 3 of the components) to control both the center frequency, and control the dampening.

I encountered such in the past.

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  • \$\begingroup\$ "DC_blocked"? What does this mean? \$\endgroup\$
    – LvW
    Oct 16, 2020 at 16:51
  • 1
    \$\begingroup\$ Look again. That op-amp circuit is a proportional-integrator stage. It's the opposite of DC blocked. \$\endgroup\$
    – TimWescott
    Oct 16, 2020 at 17:14
  • \$\begingroup\$ One thing that does concern me (although it's not really part of this question necessarily) is the biasing on the non-inv op amp input. In my implementation(s), I've tied this to ground. So, Vr = GND. I've seen ADF4002 implementations where this reference voltage is tied to Vcc/2. \$\endgroup\$
    – Buck8pe
    Oct 16, 2020 at 18:44

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