11.4.2 DLL Simulation Case I => B = 10 bits, k = 0.3016 Hz/bit
188.8.131.52 Input Bit Clock Frequency Swings ±50 ppm
spent a great amount of time designing a DLL with a 1024-bit elastic
store memory and a 10-bit digital error signal, so we will observe the
simulation results of that circuit first.
The simulation plot in
Figure 11.16 shows the DLL synthesized frequency response to a step in
frequency by the input bit clock of +50 ppm or +77.2 Hz above the center
frequency, followed some time later by another frequency step of −100
ppm down to −77.2 Hz below center frequency.
Figure 11.16 B = 10 frequency deviation = ±50 ppm
distance between minor tick marks on the time axis is the equivalent of
106 input bit clock periods. The distance between major tick marks is
the equivalent of 10 . 106 input bit clock periods. The thick line is the plot of the synthesized VCXO clock frequency fS (t), and the thin line is the value of the loop error voltage VO
(t ). The values of the DLL circuit parameters listed to the right of
the plot are frozen at the time the plot is stopped and represent the
latest parameter measurements. These measurements correspond to the last
plotted point, which is the last point on the right-hand side of the
The positive and negative frequency step of the
input bit clock is shown as a dashed line. To avoid clutter on the
following plots, this dashed step will not be shown, but the reader can
rightfully assume that a step has occurred whenever an exponential
response of fS (t) and VO (t) is initiated.
what information can we glean from this graphical simulation? Well, we
can immediately see from this plot that the loop response is indeed
exponential, just as our derived transient response of Equation 11.16
We can also see that after the input bit clock
frequency instantaneously experiences a step of +77.2 Hz or +50 ppm, the
DLL exponentially tracks and eventually locks to the input clock
frequency. We can also see that when the input bit clock frequency
experiences a negative step such that it is below the center frequency
by 77.2 Hz, the DLL exponentially tracks and locks to the new input
clock frequency. This time, the negative step was −2 . 77.2 Hz = −154.4 Hz, which corresponds to a negative step of −100 ppm.
time t = 0, the loop error voltage is set to its center value of 1.5
volts. The error voltage exponentially increases to 2.25 volts in
response to the input frequency step of +77.2 Hz or +50 ppm and then
exponentially decays to 0.75 volts following the negative tributary
clock frequency step of –154.4 Hz or −100 ppm down to 1.544 MHz − 77.2
It is interesting to note that the plot of both the synthesized bit clock frequency fS (t) curve and the VCXO control voltage VO
(t) curve pass through the horizontal axis at the same point. This is
as it should be, since the time axis is positioned at both the loop
center frequency of 1.544 MHz and at the midpoint of the control voltage
of 1.5 volts. Remember that when a 1.5 volt error signal is applied to
the VCXO, it will synthesize the loop center frequency. So it is good to
know that the hardware bit-level simulation behaves correctly in this
We can also see from the parameter measurement list on
the right-hand side of the plot that the value of the loop time constant
was measured to be k = 0.3018 Hz/bit. As shown in Table 11.4, this
value for k is well within 0.1% of the calculated value of k = 0.3016
Hz/bit. The value of k is continuously computed during the 10% to 90%
rise/fall time of the response waveform. The accuracy of the computation
for k depends on the number of points in the rise or fall time of the
curve. A shorter transient response curve will produce less accurate
The parameter measurement list also shows that when
the simulation was stopped, the frequency of the input bit clock (trib
offset) is equal to the center frequency –50 ppm and the input bit clock
frequency is 1, 543, 922.8 Hz, both of which are expected. In addition
we can see from the parameter list that the digital error signal εD
is equal to 256. The ideal VCXO frequency versus voltage curve of
Figure 11.13 says that when the value of the error signal reaches 256,
the VCXO output frequency should be 1.544 MHz – 50 ppm or 1, 543, 922.8
Hz and that the VCXO input error voltage should be 0.75 volts. A quick
check of the plot measurement list shows that the synthesized frequency FS does indeed equal 1, 543, 922.8 Hz and the error voltage VO does indeed equal 0.75 volts.
parameter list also shows that the difference between the input bit
clock and the synthesized bit clock at this particular frequency is clk
diff = 0 Hz. This tells us that the loop was able to exactly synthesize
this particular frequency. This will only be true when the input
frequency exactly equals one of the 1024 discrete frequencies that can
be synthesized by this 10-bit DLL.
And finally we do not pay any
attention to the “num clks” value in the plot parameter list. That is
the number of clocks that have occurred since t = 0. This number is only
meaningful during the loop transient response computations. Since the
loop is already reached the steady state condition, this number is
Now that you are familiar with the format of the
simulation plot, let’s view some additional plots to see how the DLL
responds to input bit clock frequency steps of different magnitudes.
184.108.40.206 Input Bit Clock Frequency Deviation of +40 ppm
the DLL was locked to the input bit clock frequency with a deviation of
±50 ppm, we saw that the DLL was able to synthesize a clock frequency
that perfectly matched the frequency of the input bit clock. The DLL was
able to do this because the digital error signal for those two extreme
frequencies just happened to be integer values of 768 and 256. What
happens when our DLL cannot synthesize a bit clock frequency that
exactly matches the frequency of the input bit clock? Let’s take a look
at the case where the deviation of the input bit clock frequency is +40
ppm or +61.76 Hz. The digital error signal that would drive the DLL to
synthesize this frequency is calculated by
Since εD must be an integer, the closest error values that the loop can compute are the two that bracket 716.7109, which are εD = 716 and εD = 717. And that is exactly what the DLL will try to do. It will generate an error signal of εD = 716 and produce an output frequency fS1 that is slightly less than the frequency of the input bit clock FI.
Slowly over time the small positive difference between the input bit
clock frequency and the synthesized bit clock frequency (FI − fS1) will accumulate until the error signal eventually increases by 1 bit to εD = 717. The extra error bit will cause the DLL to synthesize a higher output frequency fS2 that is 0.3016 Hz greater than fS1 and is a slightly greater than the frequency of the input bit clock FI.
Slowly over time the small negative difference between the input bit
clock frequency and the synthesized bit clock frequency (F I − fS2) will cause the error accumulation to decay until the error signal eventually decreases by 1 bit back to εD =716. This process will repeat itself forever.
switching back and forth between two synthesized frequencies that
closely bracket the target frequency gives rise to synthesized clock
jitter. The magnitude and the frequency of the jitter are dependent on
the frequency resolution of the loop. The smaller the time constant k,
the better the resolution and the smaller the jitter, both in magnitude
and frequency. The jitter magnitude is smaller because the difference
between the synthesized and tar-get frequency is smaller, and the jitter
frequency is smaller because it takes longer for the digital error to
accumulate or decay by 1 LSB. The opposite is true if the value of k
gets larger. In this case, the resolution decreases and the magnitude
and frequency of the synthesized frequency jitter increases.
verify this by looking at the simulator plot for our design for the
case where B equals 10 and the calculated value for our loop time
constant k = 0.3016. The plot of the synthesized frequency is
illustrated in Figure 11.17. The parameter measurements on the right
side of the plot show a post transient response clock difference of (F I − fS1 ) equal to –0.06 Hz. We can also see that the value for the digital error is εD
= 717 . The frequency resolution for our B = 10 bit design is good
enough that we can barely see any jitter in the plot of the output
frequency. In the plot window, an arrow points to a box that encloses
the portion of the plot where the synthesized frequency is actually 0.06
Hz greater than the target input bit clock frequency.
Now take a look at Figure 11.18. Here the error has decayed to the point where the digital error signal εD = 716. Now the difference between the input bit clock frequency and the synthesized clock frequency (FI − fS2)
is equal to 0.24 Hz. Once again there is an arrow in the plot window
that points to the box that encloses the portion of the plot where this
0.24 Hz discrepancy occurs. This oscillation between FI − 0.24 Hz ≤ fS ≤ FI
= 0.06 Hz will continue for forever and is the jitter about the center
of the desired synthesized frequency. Figure 11.17 shows that the jitter
period is measured to be approximately 20.59 seconds. The jitter
frequency is 0.05 Hz and the jitter magnitude is equal to (0.06 +
0.24)Hz = 0.30 Hz. This also is the value for our loop time constant k.
It shouldn’t be a surprise that k is also the value of the loop jitter.
The loop error signal oscillating between two adjacent levels of εD and k is equal to the frequency difference between these adjacent values.
Figure 11.17 B = 10; frequency deviation = +40 ppm (1)
this little constant k has just taken on another huge role in the
behavior of the DLL. Let’s see if we can count all the fun things k
contributes to the DLL performance:
- k is the frequency resolution of the DLL.
- k is the circuit time constant that determines the attack rate of the loop.
- k is the magnitude of the jitter in the loop synthesized frequency.
turns out that this constant k is an extremely important player in the
performance of our DLL. The jitter we see for the B = 10 bit case
doesn’t sound like a big deal here, but we will show some examples in
the next section where this jitter for smaller values of B becomes very
One other thing we need to mention here is the rise
time of the exponential response. The rise time is defined to be the
time required for the curve to rise from 10% to 90% of its final value.
The rise time is dependent on the value of k. We will look at the
computation of the rise time figure in greater detail when we compare
the simulation results with the DLL transient response results we
calculate from either of the two expressions presented in Equation
11.16. For now, all we need to know is that the computed rise time for
this 10-bit DLL is 7.28 seconds.
Figure 11.18 B = 10; frequency deviation = +40 ppm (2)
mentioned several times that the clock jitter becomes more pronounced
and the transient time of the circuit decreases as the value of the
circuit time constant k increases. This makes sense since the units of k
are Hz/ bit. As the time constant k gets larger, the frequency
resolution decreases, which allows the loop to acquire a step in input
frequency more quickly, but because its frequency resolution is
decreased, once locked to in input frequency, the loop cannot track with
as much precision. The opposite is true when the value of k decreases.
The next few simulation cases will demonstrate this very clearly as we
take a look at the simulation examples for B = 8, 9, 11, and 12 bits.
Once we acquire an understanding of these examples, we will look at some
examples of how loop performance degrades when we set B to lower values
such as 6 and 7. These two examples are shown primarily to allow the
reader to see what problems can arise if loop parameters are not
selected with care.
The following are a few things to note about the 10-bit DLL driven by an input clock of 1.544 MHz ± 50 ppm and its response to a ±40 ppm input frequency step:
- The calculated value of the circuit time constant k is 0.3016.
- The jitter period is approximately 20.59 seconds.
- The jitter frequency is approximately 0.0486 Hz.
- The jitter magnitude is 0.06 + 0.24 = 0.3 Hz, which is equal to k.
- The exponential rise time is 7.28 seconds.
- The digital error switches between 716 and 717 bits when tracking the high deviation +50 ppm input.
- The digital error switches between 307 and 308 bits when tracking the low deviation −50 ppm input.