Innovative RoboClock Family Application
2
Cascading two RoboClocks in series will also dramatically
increase the output possibilities. In this case, one of the out-
puts of the fir st stag e wil l serve as the REF input for the sec-
ond stage. Multiple feedback configurations are possible,
which can result in an innovative set of outputs.
RoboClock Configuration Methodologies
Using One Small Table
The entire set of programmable skew configurations is sum-
marized in a single small table shown in
Tabl e 2
. Every pos-
sibl e combination can be driven from thi s sm all tabl e. For ex-
ample, if +2tU is required from 3Qx (3Q0 or 3Q1) outputs,
based on
Table 2
, th e correspondi ng 3Fx input s should be set
as 3F1= MID and 3F0=HIGH. Any one of 1Qx, 2Qx, or 4Qx
outputs may be used as FB input, by leaving its corresponding
1Fx, 2Fx, or 4Fx inputs floating (i.e., 1F1= MID, 1F0= MID).
Note that
Table 2
represents only the cases where the feed-
back is an output with no skew, divide, or invert function. Ba-
si ca lly, a 0tU output is used for FB input.
Now, to generate additional output functions, if the feedback
output is programmed to skew, divide or invert, then output
func tions of other o utp uts m ay n ot b e dire ctly rea d from
Table
2
. In this ca se , to figur e out the final output f uncti on obs erv ed
on the output, simply subtract whatever the feedback term is
programmed to, from t he outp ut func ti on progr amm ed on th e
corresponding output. Therefore, by using only
Table 2
and
the following simple algorithm, every single combination of
RoboClock can be figured out .
Final Output Function = O utput Function – FB Function
If t here is any am bigui ty, the following example should cl arify
the use of this method. Let’s say +7tU of delay is required.
Ob vious ly, +7tU is not a choice , av ai lab le in
Table 2
. Ho wever ,
any two functions from two different outputs of
Tab l e 2
may
be com bined t o achi eve a desired fu nction . For this ex am ple,
there are several sol uti ons, and only one of them wil l be pre-
sented. One way to achieve +7tU is to subtract –3tU from +4tU.
+7t
U
= +4t
U
– (–3t
U
)
Therefore, if 1Qx out put i s progr am med to have –3t U of skew
(1F1=LOW, 1F0=MID), and used as the FB input, and if the
3Qx is programmed to have +4tU of skew (3F1=HIGH,
3F0=LOW), the final output f unction observ ed on 3Qx wil l be
+7tU.
One exception to this simple rule is that if a divided output is
used as the FB inpu t, the n the othe r output s will be m ult iplied
by t he sa me f acto r (2 or 4). Th e r eason f o r thi s is t hat th e PLL
will force the FB to align with the REF both in phase and
frequency. Therefore , if the FB term is programm ed to di vide
by 2, the PLL wil l speed up t wic e to f orce t he FB term to ali gn
with the REF frequency . As an example, if advance by 6tU and
multi ply by 4 function is requi red (–6tU and f*4) , then
Final Function = –6t
U
– (divi de by 4) => (– 6t
U
and f*4)
The solution for this example is to program 3Qx to divide by
4 (3F1=HIGH, 3F0=HIGH) and use it as FB, and program 4Qx
to hav e – 6tU of skew (4F1=LOW , 4F0=MID). The final function
observed on 4Qx will be REF frequency multiplied by 4 and
advanced by 6tU (−6tU and f*4).
By this metho d, one can easi ly d ete rmine if a de sired func tion
can be implemented by RoboClock or not. RoboClock can
generate a wav eform composed of any two functions from two
different outputs of
Table 2
.
Using Three Tables for Multiple Outputs
If multiple outputs with var ious functions are required, using
the previous method could be a little cumbersome. All the
possible combinations of RoboClock outputs are in three ta-
bles , i llu s tra ted in
Tabl es 3
through
5
. Each table represents
all th e possib le outpu t combinat ions with a given output co n-
nected to FB input. For example, once 3Qx output is used as
FB input, then all the possible output combinations could be
found in
Tab l e 4
. These three tables are extremely valuable
tools in dete rmining wh at FB term to u se and how to confi gure
the RoboClock, when mul tiple outputs with var ious functions
are required.
Once the req uired m ult iple funct i ons are det ermi ned in te rms
of tU, an effort shou ld be m ade to l ocate one r ow in one o f the
three tables that contains the required functions. For exam-
ple, if one of t he de sired funct ions i s divid e by 2 an d dela y 4tU
(+4tU an d f/2), t hen by observation, that c hoice can be l ocated
in row 1 of
Table 3
, row 3 of
Table 4
, and row 3 of
Table 5
.
Now, the one to be selected as a solution would depend on
what the other required functions are, because once an out-
put, which is programmed to perform a certain function, is
selected as FB input, all the outputs of a RoboClock are lim-
ited to a single row found in
Tables 3
through
5
. If in the pre-
vious example, the second required function happens to be
invert and skew b y 4tU (+4tU and INV), then the only solution
is ro w 1 of
Table 3
. In this case 1Qx could b e used as the FB
input with its inputs hardwired to GND (1F1=LOW,
1F0=LOW), and 3F1=HIGH, 3F0=HIGH to generate (+4tU
and f/2) function on 3Qx outputs, and set 4F1=HIGH,
4F0=HIGH to generate (+4tu and INV) function on 4Qx out-
puts. In this configuration, the 2Qx outputs could be pro-
grammed to have any one of 0tU through +8tU sk ew. For ex-
ample, if +7tU is another required output, then 2F1=High,
2F0=Mid will generate +7tU skew on 2Qx. Note that even
though the 1Qx outputs are programmed to have –4tU skew,
they a re forced by PL L to al ign wi th t he REF fr equency, t here-
fore 1Qx out put coul d be used as a 0tU output.
The Table method is recommended for multiple outputs with
various function requirements. If the exact required outputs
cannot al l be f ound i n one row, then the designe r can use the
Table 2. Output Adjustment Configurations
Function Selects Output Functi ons
1F1, 2F1,
3F1, 4F1 1F0, 2F0,
3F0, 4F0 1Q0, 1Q1,
2Q0, 2Q1 3Q0, 3Q 1 4Q0, 4Q 1
LOW LOW –4tUDivide by 2 Divide by 2
LOW MID –3tU–6tU–6tU
LOW HIGH –2tU–4tU–4tU
MID LOW –1tU–2tU–2tU
MID MID 0tU0tU0tU
MID HIGH +1tU+2tU+2tU
HIGH LOW +2tU+4tU+4tU
HIGH MID +3tU+6tU+6tU
HIGH HIGH +4tUDivide by 4 Inverted