0
\$\begingroup\$

I want to calculate the contribution of each phase-locked loop (PLL) module to the total phase noise that has been done in ADIsimPLL.

I'm having trouble calculating the contribution of the loop filter to the phase noise. As is mentioned in the ADIsimPLL help document, the noise comes from the components in the loop filter including thermal noise in the resistors and op-amp noise (the first figure).

I know for a resistor the thermal noise(Jonhson noise) is derived from:

$$v_{n}=\sqrt{4kTRB}$$

To calculate the total noise from each resistor and the op-amp:

$$V_{tot}=\sqrt{v_{1}^{2}+v_{2}^{2}+...+v_{n}^{2}}$$

But the unit of this result is \$V/\sqrt{Hz}\$. What I need is phase noise with the unit \$dBc/Hz\$.

And I also get the transfer function of how the loop filter's noise impact the total phase noise:

$$ H_{LF}(s)=\frac{NK_{v}}{NS+K_{v}K_{d}Z(s)} $$

Phase noise calculation block diagram

The topology and the phase noise results are shown below.

So how to exactly calculate the loop filter's phase noise (\$n_{LF}\$) including thermal noise in resistors and op-amp noise? And how to convert voltage noise to phase noise? The two questions are to get the phase noise curve contributed by loop filter shown in the third picture.

loop filter topology

phase noise results in ADIsimPLL

By the way, the voltage noise of the op-amp is \$0.95nV/\sqrt{Hz}\$ and current noise is \$2.5pA/\sqrt{Hz}\$.

\$\endgroup\$
3
  • \$\begingroup\$ If you have volts/rtHz at the filter, and then Hz/V at the VCO, you have Hz/rtHz at the VCO. If you have VCO frequency, and need the phase, then you integrate. There's a bit of scaling to do - phase to SSB, peak to rms - draw a phasor diagram of the output of the VCO together with its noise driven phase fluctuations. \$\endgroup\$
    – Neil_UK
    Commented Dec 9, 2021 at 6:58
  • \$\begingroup\$ @Neil_UK Thank you for your reply! Before the procedures you mentioned, it is a difficulty to calculate the total voltage noise of the loop filter in the 3rd picture. \$\endgroup\$
    – shang xu
    Commented Dec 9, 2021 at 8:45
  • \$\begingroup\$ Yes, it is difficult, there are a lot of components and, due to the filtering action, they contribute differently at different frequencies. Run a simulation with a voltage noise source on the amplifier input, and several times again with a current noise source in parallel with each resistor, and sum the resulting outputs power-wise. \$\endgroup\$
    – Neil_UK
    Commented Dec 9, 2021 at 9:35

1 Answer 1

2
\$\begingroup\$

To calculate the voltage noise density at the output of your low-pass filter, you must first calculate the individual voltage noise contributions of your resistors using \$v_n=\sqrt{4kTR}\$ (without the bandwidth \$B\$, to get a frequency-dependent result in units of \$V/\sqrt{Hz}\$).

However, since the noise of the individual noise sources is not equally included in the total noise, the individual voltage noise densities must be individually weighted depending on the frequency. The weighting is done by the so-called noise gain:

\$V_{tot} = \sqrt{(\mathrm{NG}_1 \cdot v_1)^2+(\mathrm{NG}_2 \cdot v_2)^2 + ... + (\mathrm{NG}_n \cdot v_n)^2}\$

Finding/calculating the individual noise gains can be time-consuming. It is much easier/faster to do a little noise simulation, e.g. in LTspice:

1

From this noise simulation, the total voltage noise density is approximately \$1~\mathrm{uV}/\sqrt{\mathrm{Hz}}\$ at \$1~\mathrm{kHz}\$, \$100~\mathrm{nV}/\sqrt{\mathrm{Hz}}\$ at \$10~\mathrm{kHz}\$, and so on.

Next, you can multiply the voltage noise density \$V_{tot}\$ (units of \$\mathrm{V}/\sqrt{\mathrm{Hz}}\$) with the VCO gain \$K_v\$ (units of \$\mathrm{Hz}/\mathrm{V}\$) (if necessary frequency dependent). This results in a power spectral density of random frequency fluctuations:

\$S_{f}(f) = (V_{tot} \cdot K_v(f))^2\$

in units of \$\mathrm{Hz}^2/\mathrm{Hz}\$ that can be converted into the power spectral density of random phase fluctuations in units or \$\mathrm{rad}^2/\mathrm{Hz}\$:

\$S_{\varphi}(f) = \frac{S_{f}(f)}{f^2}\$

If \$S_{\varphi}(f)\$ is given in its logarithmic form, i.e. \$10~\log_{10}(S_{\varphi}(f))~\mathrm{dB~rad}^2/\mathrm{Hz}\$, you simply need to substract \$3~\mathrm{dB}\$ to get your final results in units of \$\mathrm{dBc}/\mathrm{Hz}\$ (compare my answer here).

Assuming a typical VCO gain of \$K_v = 100~\mathrm{kHz}/\mathrm{V}\$, the following numbers result at an offset frequency of \$1~\mathrm{kHz}\$:

\$V_{tot}(1~\mathrm{kHz})\ = 1~\mathrm{uV}/\sqrt{\mathrm{Hz}}\$

\$S_{f}(1~\mathrm{kHz})\ = 0.01~\mathrm{Hz}^2/\mathrm{Hz}\$

\$S_{\varphi}(1~\mathrm{kHz})\ = 10^{-9}~\mathrm{rad}^2/\mathrm{Hz} = -80~\mathrm{dB~rad}^2/\mathrm{Hz}\$

\$L(1~\mathrm{kHz})\ = -83~\mathrm{dBc}/\mathrm{Hz}\$

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Charly left out his link to the interesting LTSpice noise sources referenced in this post. \$\endgroup\$
    – qrk
    Commented Aug 29, 2022 at 18:36
  • \$\begingroup\$ @shang xu: It would be interesting to know which VCO is the basis of the calculation of your phase noise plot shown above. I obtain a VCO with a gain of about 14 Hz/V by calculation. I would be interested to know if this is correct or if there is a mistake in my calculation. \$\endgroup\$
    – Charly
    Commented Aug 29, 2022 at 20:00

Your Answer

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

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