SX Family FPGAs
1-18 v3.2
Step 3: Calculate DC Power Dissipation
DC Power Dissipation
P
DC
= (I
standby
) × V
CCA
+ (I
standby
) × V
CCR
+ (I
standby
) ×
V
CCI
+ X × V
OL
× I
OL
+ Y(V
CCI
– V
OH
) × V
OH
EQ 1-12
For a rough estimate of DC Power Dissipation, only use
P
DC
=(I
standby
) × V
CCA
. The rest of the formula provides a
very small number that can be considered negligible.
P
DC
= (I
standby
) × V
CCA
P
DC
= .55 mA × 3.3 V
P
DC
= 0.001815 W
Step 4: Calculate Total Power Consumption
P
Tot a l
= P
AC
+ P
DC
P
Tot a l
= 1.461 + 0.001815
P
Tot a l
= 1.4628 W
Step 5: Compare Estimated Power Consumption
against Characterized Power Consumption
The estimated total power consumption for this design is
1.46 W. The characterized power consumption for this
design at 200 MHz is 1.0164 W.
Step 1: Define Terms Used in Formula
V
CCA
3.3
Module
Number of logic modules switching
at f
m
(Used 50%)
m 264
Average logic modules switching rate
f
m
(MHz) (Guidelines: f/10)
f
m
20
Module capacitance C
EQM
(pF) C
EQM
4.0
Input Buffer
Number of input buffers switching at f
n
n1
Average input switching rate f
n
(MHz)
(Guidelines: f/5)
f
n
40
Input buffer capacitance C
EQI
(pF) C
EQI
3.4
Output Buffer
Number of output buffers switching at f
p
p1
Average output buffers switching rate
f
p
(MHz) (Guidelines: f/10)
f
p
20
Output buffers buffer capacitance
C
EQO
(pF)
C
EQO
4.7
Output Load capacitance C
L
(pF) C
L
35
RCLKA
Number of Clock loads q
1
q
1
528
Capacitance of routed array clock (pF) C
EQCR
1.6
Average clock rate (MHz) f
q1
200
Fixed capacitance (pF) r
1
138
RCLKB
Number of Clock loads q
2
q
2
0
Capacitance of routed array clock (pF) C
EQCR
1.6
Average clock rate (MHz) f
q2
0
Fixed capacitance (pF) r
2
138
HCLK
Number of Clock loads s
1
0
Variable capacitance of dedicated
array clock (pF)
C
EQHV
0.61
5
Fixed capacitance of dedicated
array clock (pF)
C
EQHF
96
Average clock rate (MHz) f
s1
0
Step 2: Calculate Dynamic Power Consumption
V
CCA
× V
CCA
10.89
m × f
m
× C
EQM
0.02112
n × f
n
× C
EQI
0.000136
p × f
p
× (C
EQO
+C
L
) 0.000794
0.5 (q
1
× C
EQCR
× f
q1
) + (r
1
× f
q1
) 0.11208
0.5(q
2
× C
EQCR
× f
q2
) + (r
2
× f
q2
)0
0.5 (s
1
× C
EQHV
× f
s1
) + (C
EQHF
× f
s1
)0
P
AC
= 1.461 W