SA8281 Mitel Networks Corporation, SA8281 Datasheet - Page 8

• Manufacturer Part Number: SA8281
• Description: Three-Phase PWM Waveform Generator
• Manufacturer: Mitel Networks Corporation

Amplitude selection
the amplitude of the waveform samples stored in the ROM by
the value of the 8-bit amplitude select word (AMP).
where A = decimal value of AMP.
POWER-UP C0NDITIONS
up in an unidentified state. Holding RST low or using the SET
TRIP input will ensure that the PWM outputs remain inactive
(i.e., low) until the device is initialised.
SA828 PROGRAMMING EXAMPLE
12·288 MHz is used (12·288 MHz crystals are readily available).
This clock frequency will allow a maximum carrier frequency of
24 kHz and a maximum power frequency of 4 kHz.
Initialisation Register Programming Example
carrier frequency of 6kHz, a pulse deletion time of 10 s and an
underlap of 5 s.
1. Setting the carrier frequency
frequency, pulse deletion time and pulse delay time are all
defined relative to the carrier frequency.
carrier frequency:
010 in temporary register R1.
2. Setting the power frequency range
power frequency:
100 in temporary register R1.
3. Setting the pulse delay time
seen at the PWM outputs, it is sensible to set the pulse delay time
before the pulse deletion time, so that the effect of the pulse delay
time can be allowed for when setting the pulse deletion time.
8
SA828
The power waveform amplitude is determined by scaling
The percentage amplitude control is given by:
All bits in both the Initialisation and Control registers power-
The following example assumes that a master clock of
A power waveform range of up to 250Hz is required with a
The carrier frequency should be set first as the power
We must calculate the value of n that will give the required
From Table 4, n = 4 corresponds to a 3-bit CFS word of
We must calculate the value of m that will give the required
From Table 5, m = 16 corresponds to a 3-bit FRS word of
As the pulse delay time affects the actual minimum pulse width
AMP
Power Amplitude, A
7
AMP
n =
Fig.14 Temporary register R2
m =
6
AMP
512 x f
f
RANGE
f
f
5
RANGE
SELECT WORD
CARR
k
AMP
AMP
AMPLITUDE
AMP 0
f
CARR
CARR
x 384
=
POWER
7
4
=
= MSB
= LSB
=
AMP
f
512 x n
CARR
384
512 x 6 x 10
12·288 x 10
=
k
=
3
x m
AMP
250 x 384
255
6 x 10
A
2
AMP
x 100%
3
6
3
1
= 4
AMP
= 16
0
purpose of the pulse delay is to prevent ‘shoot-through’ (where
both top and bottom arms of the inverter are on simultaneously),
it is sensible to round the pulse delay time up to a higher, rather
than a lower figure.
of 5·2 s. From Table 6, pdy = 16 corresponds to a 6-bit PDY
word of 110000 in temporary register R2.
4. Setting the pulse deletion time
width) account must be taken of the pulse delay time, as the
actual minimum pulse width seen at the PWM outputs is equal
to t
instance, be set 5·2 s longer than the minimum pulse length
required
Now,
up or down – the choice of which will depend on the application.
Assuming we choose in this case the value 46 for pdt , this gives
a value of t
– 5·2 s = 9·8 s.
7-bit word in temporary register R0 of 1010010.
registers R0, R1 and R2 (for transfer into the initialisation
register) in order to achieve the parameters in the example
given, is shown in Fig. 15.
pulse delay time:
However, the value of pdy must be an integer. As the
Thus, if we assign the value 16 to pdy this gives a delay time
In setting the pulse deletion time (i.e., the minimum pulse
Therefore, the value of the pulse deletion time must, in this
Again, pdt must be an integer and so must be either rounded
From Table 7, pdt = 46 corresponds to a value of PDT, the
The data which must be programmed into the three temporary
We must calculate the value of pdy that will give the required
pd
– t
FRS
CR PDT
pdy
1
1
X
X
.
pd
2
Minimum pulse length required = 10 s
, of 15 s and an actual minimum pulse width of 15
FRS
= 15·2 x 10
t
PD
= 5 x 10
X
1
0
X
to be set to 10 s + 5·2 s = 15·2 s
6
1
PDT
FRS
PDY
Temporary Register R0
Temporary Register R1
Temporary Register R2
pdy = t
t
pdt = f
0
0
1
pdy
–6
t
5
0
pd
5
–6
=
x 6 x 10
PDT
PDY
=
x 6 x 10
X
Fig. 15
1
X
1
pdy
pd
f
CARR
f
4
4
CARR
x f
x f
PDT
pdy
PDY
CARR
3
CARR
pdt
X
x 512
0
X
0
x 512 = 15·4
3
x 512
3
x 512 = 46·7
3
x 512
PDT
x 512
CFS
PDY
0
0
0
2
2
2
PDT
CFS
PDY
1
1
0
1
2
1
PDT
CFS
PDY
0
0
0
0
2
0