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SA8281IG Datasheet(PDF) 8 Page - Mitel Networks Corporation |
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SA8281IG Datasheet(HTML) 8 Page - Mitel Networks Corporation |
8 / 14 page SA828 8 SA828 PROGRAMMING EXAMPLE The following example assumes that a master clock of 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 A power waveform range of up to 250Hz is required with a carrier frequency of 6kHz, a pulse deletion time of 10 µs and an underlap of 5 µs. 1. Setting the carrier frequency The carrier frequency should be set first as the power frequency, pulse deletion time and pulse delay time are all defined relative to the carrier frequency. We must calculate the value of n that will give the required carrier frequency: From Table 4, n = 4 corresponds to a 3-bit CFS word of 010 in temporary register R1. 2. Setting the power frequency range We must calculate the value of m that will give the required power frequency: From Table 5, m = 16 corresponds to a 3-bit FRS word of 100 in temporary register R1. 3. Setting the pulse delay time As the pulse delay time affects the actual minimum pulse width 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. 12·288 x 106 512 x 6 x 103 ⇒ n = = = 4 AMP7 AMP6 AMP5 AMP4 AMP3 AMP2 AMP1 AMP0 AMPLITUDE SELECT WORD AMP7 = MSB AMP0 = LSB Fig.14 Temporary register R2 Amplitude selection The power waveform amplitude is determined by scaling the amplitude of the waveform samples stored in the ROM by the value of the 8-bit amplitude select word (AMP). The percentage amplitude control is given by: where A = decimal value of AMP. POWER-UP C0NDITIONS All bits in both the Initialisation and Control registers power- 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. Power Amplitude, APOWER = x 100% fCARR = k 512 x fCARR fRANGE = x m ⇒ m = = = 16 fCARR 384 fRANGE x 384 fCARR 250 x 384 6 x 103 A 255 k 512 x n However, the value of pdy must be an integer. As the 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. Thus, if we assign the value 16 to pdy this gives a delay time of 5·2 µs. From Table 6, pdy = 16 corresponds to a 6-bit PDY word of 110000 in temporary register R2. 4. Setting the pulse deletion time In setting the pulse deletion time (i.e., the minimum pulse width) account must be taken of the pulse delay time, as the actual minimum pulse width seen at the PWM outputs is equal to tpd – tpdy. Therefore, the value of the pulse deletion time must, in this instance, be set 5·2 µs longer than the minimum pulse length required Minimum pulse length required = 10 µs ∴ t PD to be set to 10µs + 5·2µs = 15·2µs Now, ⇒ pdt = f pd x fCARR x 512 = 15·2 x 10–6 x 6 x 103 x 512 = 46·7 tpd = Again, pdt must be an integer and so must be either rounded 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 tpd, of 15 µs and an actual minimum pulse width of 15 – 5·2 µs = 9·8µs. From Table 7, pdt = 46 corresponds to a value of PDT, the 7-bit word in temporary register R0 of 1010010. The data which must be programmed into the three temporary 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. Fig. 15 1 1 0 1 0 0 1 0 CR PDT6 PDT5 PDT4 PDT3 PDT2 PDT1 PDT0 Temporary Register R0 1 0 0 X X 0 1 0 Temporary Register R1 FRS2 FRS1 FRS0 X X CFS2 CFS2 CFS2 X X 1 1 0 0 0 0 Temporary Register R2 X X PDY5 PDY4 PDY3 PDY2 PDY1 PDY0 We must calculate the value of pdy that will give the required pulse delay time: ⇒ pdy = t pdy x fCARR x 512 = 5 x 10–6 x 6 x 103 x 512 = 15·4 pdy fCARR x 512 tpdy = pdt fCARR x 512 |
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