0
\$\begingroup\$

I'm want to use ADIsimPll to calculate the loop filter properties for a PLL I want to build. I read some things in the programs help topics which I find somehow strange. Maybe you guys can help me out with those.

  1. They define the loop bandwith as the frequency of unity gain for the open loop gain. Wouldn't it make more sense to use the closed loop gain instead?
  2. The open loop gain is defined as \$G(s) = \frac{K_d K_v F(s)}{N s}\$, where F(s) denotes the loop filter transfer function. Why is N in there? Normally, the open-loop-gain is defined as VCO-output/phase-error (or \$\frac{\Theta_O}{\Theta_{REF}}\$ in the diagram below) which is just \$\frac{K_d K_v F(s)}{s}\$. If this N is in there or not changes the loop bandwith drastically. Also the closed loop gain is influenced by this.

Here is a block diagram of the (type of) PLL I'm talking about.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Who knows why indeed? Errr.... what is the part and what document are you reading? Proper links please. \$\endgroup\$
    – Andy aka
    Feb 3, 2018 at 13:31
  • \$\begingroup\$ I'm afraid you have to download the program form.analog.com/form_pages/rfcomms/adisimpll.aspx . I didn't find any documentation online. ADIsimPLL is a often used program for PLL design and simulation.. \$\endgroup\$ Feb 3, 2018 at 13:48
  • \$\begingroup\$ What is the meaning of N ? \$\endgroup\$
    – LvW
    Feb 3, 2018 at 16:23
  • \$\begingroup\$ N is the number by which the vco's frequency is divided befor the pfd \$\endgroup\$ Feb 3, 2018 at 16:52

2 Answers 2

1
\$\begingroup\$
  1. As stated by LvW the gainbandwidth is a good approximation for the closed loop bandwidth in the case of unity feedback.

  2. I have a strong hunch that N is the frequency multiplier factor. To multiply the frequency you need to divide it in the feedback loop, hence the division by N.

  3. When plotting in dB the difference between \$|A|\$ and \$|A|^2\$ will be a factor of 2. If you want to check: a single pole transfer function should roll off 20dB per decade for \$|A|\$. 40dB/decade if they're plotting the squared amplitude.

\$\endgroup\$
3
  • \$\begingroup\$ at 1. But why don't we define the loop bandwidth to be the unity gain bandwith of the closed-loop in the first place? Furthermore the approximation only holds if the phase of G is around \$\pm \pi /2\$... 2. Indeed, N is the frequency multiplier factor. But in the open loop gain we shouldn't care what is feed back to the pfd, but rather what the output is, right? I updated my question to stress my point \$\endgroup\$ Feb 3, 2018 at 18:25
  • \$\begingroup\$ at 3. Ok I see my mistake. dB of a transfer function K is 20lg(|K|) instead of 10lg(|K|) \$\endgroup\$ Feb 3, 2018 at 18:40
  • 1
    \$\begingroup\$ The closed-loop will not really tell much to the designer. Actually, for any design with feedback, the open loop transfer characteristic is much more interesting (it tells you how stable it is with phase and gain margin, how much DC gain you have to reject noise and offsets, etc.) Also, in your case, you have a pure integrator, and so the phase is exactly \$\pi/2\$, so the approximation holds (perfectly). \$\endgroup\$
    – Sven B
    Feb 3, 2018 at 20:59
0
\$\begingroup\$

Answer to 1.): The bandwidth of a closed-loop system is identical to the frequency where the loop gain LG(s) is unity (remember the closed-loop denominator wwhich is [1+LG(s)] ).

An answer to the second question cannot be given without knowing the meaning of "N".

Comment: The gain of the complete loop - if the loop is open - is called "loop gain". Instead, the term "open-loop gain" is the gain of an amplifier without feedback (and not applicable for the PLL) .

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.