I am a bit confused about the operating frequency calculation of ring oscillator.
Below is an excerpt from this PLL tutorial.

Each stage provides pi/n phase shift and so it requires two passes through the ring to acquire the necessary 2pi radian phase shift.

Does this mean that DC phase shift (inversion) is not taken into account here? At the oscillating frequency, each stage provides a phase of pi + pi/n. So after n stages the total phase shift will be n*pi + pi that is equivalent to zero phase shift if n is an odd number.
I don't see how the signal passes two times here. Could anyone explain more?

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    \$\begingroup\$ Consider a rising edge at the left side. After 3 propagation delays it arrives back at the starting point, but because it has passed through an odd number of inverters it is now a falling edge. It takes 2 passes through the chain to recover the original condition. \$\endgroup\$ Commented Jan 5, 2017 at 0:20

1 Answer 1


The text is working backwards from assumptions about oscillation occuring and finding the necessary equation, if so. Unfortunately, it's also using \$n\$ in two contexts. (One where it is used mathematically in the form of \$n\cdot 2\pi\$ and means one thing; and, another where it is used to mean the odd number of stages in the oscillator as applied in \$\frac{1}{2~ n~ T_d}\$ and means a different thing.) So you need to parse out the details a bit.

I'll use \$n\$ in the mathematical sense used in \$n\cdot 2\pi\$, so that this is only stating the obvious: \$\left(n\cdot 2\pi\right)~\operatorname{mod}~\left(2\pi\right) = 0\$, for all integer \$n\$. Which is only about the criteria that needs to be met for oscillation and is pretty much obvious. I'll use \$k\$ to mean the odd number of stages used. (It must be odd if there are to be two passes through the system.)

Given an odd number \$k\$ inverters, one pass through the chain ultimately takes the input and presents it at the output, inverted. So it clearly will take two such passes to meet the \$0^\circ\$ criterion. So the signal must pass through \$2\cdot k\$ stages to meet that criterion. However, it takes \$T_d\$ time per stage (the delay through the stage.) So that means the total time must be \$2~k~T_d\$. That means the frequency is \$\frac{1}{2~k~T_d}\$, by definition.

In general, it is noise that will cause it to oscillate. (You can perturb the simulation using the .IC for initial conditions to create it, so that the simulator doesn't find some middling quiescent point of stability that you don't want to see.) Noise at those frequencies which cannot make it around in exactly \$2\pi\$ will die out. Noise at frequencies which can make it around, just right, will be amplified and may survive for another go-around through \$2~k\$ stages, again (by which time the signal will already be clipped and the following gain will become 1, most likely.) Also in general, the more stages you use the less startup time is required.

There will be jitter in the process. The stages won't all have the exact same \$T_d\$ value and it won't always be the exact same in any one of them through all time, either. They aren't even driving all the same loads (as you will take the output from one of them.) So, expect phase jitter to occur and that it will be both temperature and time dependent.

  • \$\begingroup\$ Have never seen the "noise-start" mechanism used in the context of ring oscillators. By the time its Vdd has risen enough to establish recognizable logic states, each inverter in the chain has settled, and away it goes, oscillating. Perhaps your paragraph on jitter is more applicable to the starting mechanism? Or logic threshold variations from gate-to-gate? \$\endgroup\$
    – glen_geek
    Commented Jan 5, 2017 at 2:02
  • \$\begingroup\$ Each inverter will provide a phase shift of 180 degree plus and additional amount dphi by output resistance and capacitance. So after k inveters the total phase shift will be k*180 mod 360 + k* dphi = 180 + kdphi where k is an odd number. To satisfy oscillation criterion kdphi = 180 or dphi = 180/k. So only after the first pass the total phase shift is already zero. Is there anything wrong in that reasoning above? \$\endgroup\$
    – emnha
    Commented Jan 5, 2017 at 2:48
  • \$\begingroup\$ @glen_geek The first place google pops up for me is here: designers-guide.org/Forum/YaBB.pl?num=1066188298 But perhaps it's just my ignorance (and using a simulator that re-inforced the idea, since the inverter is a class-A amplifier and Spice "likes" to find a nice quiescent point without oscillation, from earlier experience.) \$\endgroup\$
    – jonk
    Commented Jan 5, 2017 at 8:12
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    \$\begingroup\$ @anhnha The output of 3 stages will be inverted \$3\cdot T_d\$ after the input change. In effect, \$3\cdot\left(180^\circ+T_d\right)=540^\circ+3\cdot T_d\equiv 180^\circ+3\cdot T_d\$. Put through twice, this is \$0^\circ+6\cdot T_d\$. Or did I miss something? \$\endgroup\$
    – jonk
    Commented Jan 5, 2017 at 8:24
  • \$\begingroup\$ @jonk: what I am confused is that from 180 + 3*Td, because 3*Td is equivalent to 180 degree shift so it already satisfies oscillating condition of 360 degree phase shift. Is that wrong? \$\endgroup\$
    – emnha
    Commented Jan 5, 2017 at 13:37

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