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USING THE ST7 SPI TO EMULATE A 16-BIT SLAVE

In the case of concurrent interrupts, the SPI interrupt must always be the first served. The only

way to ensure this happens is to have only the SPI interrupt enabled. Otherwise, if another in-

terrupt is served between two received words, an overrun condition will occur.

In the case of nested interrupts, the SPI interrupt only needs to have the highest priority, to be

always immediately served.

CONCLUSION: The maximum baud rate, which can be reached is 2 MBaud.

4 MEASUREMENTS

The goal of the measurement is to verify the above theory and to see whether the software

correctly receives both bytes. What is especially important is the time needed to read the data

register, which releases the shift register for the reception of the 2nd byte. To measure this

time an I/O port pin is used and cleared after reading the data register. The interrupt routine is

Figure 6. Measurement Software

time between both is the time needed by “ld _SPI_Interrupt$L-1,a“and “clr PBDR”, which is

4cy+5cy=1,125us. So to Ax is 1,125us before Bx.

The measurement shows that Ax is always before the deadline. This proves that the 8-bit SPI

can be used to receive 16-bit SPI messages at speeds up to 2MBaud.

_SPI_Interrupt:

btjf _SPISR,#7,L522

ld a,_SPIDR

ld _SPI_Interrupt$L-1,a

clr _PBDR

L332:

btjf _SPISR,#7,L332

ld a,_SPIDR

ld _SPI_Interrupt$L-2,a

ld a,#255

ld _PBDR,a

iret

L522:

ld a,#196

ld_SPICR,a

iret

if (ValBit(SPISR,SPIF))

{

twobytesDR.b_form.low=SPIDR;

PBDR=0x00;

while(!ValBit(SPISR,SPIF));

twobytesDR.b_form.high=SPIDR;

PBDR=0xFF;

}

else

{

SPICR=0xC4;

}

C code

Assembly code