I have a problem where my code locks up on receiving a RX interrupt on
UART1. I have been transmitting data to a serial display with no
problems, although with interrupts disabled. Anybody see any
omissions or problems with this? I didn't include code such as PLL
setup, because a lot of other features work. But if anyone needs it,
I'll post it. Thanks.
Sutton - dodge55
Notes on code operation:
- DUART_Write_xxx - I have moved my display to utilize a Dual UART
chip instead of UART1 (which printed fine without interrupts ON)
- I have an infinite loop (main) printing out U1IIR which is always
0xc1 as soon as everything is setup. I don't think this is correct
(pending interrupt).
- I have a dumb terminal (WYSE) connected directly to UART1, running
at 19200. As soon as I press any key on the WYSE, the code stops (and
'INT' is never printed.
- I have verified that loopback works fine on the WYSE.
- If I print something out of UART1 to the WYSE, it works fine
(without interrupts).
void uart1_read(void) __irq
{
DUART_Write_Str("INT");
if (U1IIR & 0x04 != 0)
{
DUART_Write_Char(U1RBR);
}
VICVectAddr = 0; // Acknowledge interrupt
}
void set_up_uart1(unsigned int baudrate)
{
PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
PCONP |= 0x00000010;
U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
switch(baudrate)
{
case 9600:
U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 = 0x25
U1DLL = 0x25; // 9601.71 baud, at 12Mhz
break;
case 19200:
U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 = 0x17
U1DLL = 0x17; // 19181.585 baud, at 12Mhz
break;
}
U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
U1FCR = 0x07;
// set up UART1 as a vectored interrupt
VICVectAddr1 = (unsigned long) uart1_read;
VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
U1IER = 7; // RBR, THRE, RLS interrupt enable
}
main()
{
set_up_uart1(19200);
for(;;)
{
char tmp[3];
sprintf(tmp,"%x ",U1IIR);
DUART_Write_Str(tmp); // other supporting code exists
delay(SECS,0.25); // other supporting code exists
}
}
UART1 RX LPC2148
Started by ●January 9, 2007
Reply by ●January 9, 20072007-01-09
On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
>I have a problem where my code locks up on receiving a RX interrupt on
>UART1. I have been transmitting data to a serial display with no
>problems, although with interrupts disabled. Anybody see any
>omissions or problems with this? I didn't include code such as PLL
>setup, because a lot of other features work. But if anyone needs it,
>I'll post it. Thanks.
>
Don't know if it applies to the 2148, but I had a problem on a 2103, where setting the fifo control
reg to an invalid value (fifo off) caused a continuous RX interrupt even with no data.
Reason was the fifo control register is a write-only register but I was using a bitfield-defined
SFR to set it up, so a subsequent read-modify-write operation overwrote the intended value.
>Sutton - dodge55
>Notes on code operation:
>- DUART_Write_xxx - I have moved my display to utilize a Dual UART
>chip instead of UART1 (which printed fine without interrupts ON)
>- I have an infinite loop (main) printing out U1IIR which is always
>0xc1 as soon as everything is setup. I don't think this is correct
>(pending interrupt).
>- I have a dumb terminal (WYSE) connected directly to UART1, running
>at 19200. As soon as I press any key on the WYSE, the code stops (and
>'INT' is never printed.
>- I have verified that loopback works fine on the WYSE.
>- If I print something out of UART1 to the WYSE, it works fine
>(without interrupts).
>void uart1_read(void) __irq
>{
> DUART_Write_Str("INT");
>
> if (U1IIR & 0x04 != 0)
> {
> DUART_Write_Char(U1RBR);
> }
> VICVectAddr = 0; // Acknowledge interrupt
>}
>void set_up_uart1(unsigned int baudrate)
>{
> PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> PCONP |= 0x00000010;
>
> U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
>
> switch(baudrate)
> {
> case 9600:
> U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 = 0x25
> U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> break;
>
> case 19200:
> U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 = 0x17
> U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> break;
> }
>
> U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> U1FCR = 0x07;
>
> // set up UART1 as a vectored interrupt
>
> VICVectAddr1 = (unsigned long) uart1_read;
> VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
>
> U1IER = 7; // RBR, THRE, RLS interrupt enable
>}
>main()
>{
> set_up_uart1(19200);
>
> for(;;)
> {
> char tmp[3];
>
> sprintf(tmp,"%x ",U1IIR);
> DUART_Write_Str(tmp); // other supporting code exists
> delay(SECS,0.25); // other supporting code exists
> }
>}
>Yahoo! Groups Links
>
>I have a problem where my code locks up on receiving a RX interrupt on
>UART1. I have been transmitting data to a serial display with no
>problems, although with interrupts disabled. Anybody see any
>omissions or problems with this? I didn't include code such as PLL
>setup, because a lot of other features work. But if anyone needs it,
>I'll post it. Thanks.
>
Don't know if it applies to the 2148, but I had a problem on a 2103, where setting the fifo control
reg to an invalid value (fifo off) caused a continuous RX interrupt even with no data.
Reason was the fifo control register is a write-only register but I was using a bitfield-defined
SFR to set it up, so a subsequent read-modify-write operation overwrote the intended value.
>Sutton - dodge55
>Notes on code operation:
>- DUART_Write_xxx - I have moved my display to utilize a Dual UART
>chip instead of UART1 (which printed fine without interrupts ON)
>- I have an infinite loop (main) printing out U1IIR which is always
>0xc1 as soon as everything is setup. I don't think this is correct
>(pending interrupt).
>- I have a dumb terminal (WYSE) connected directly to UART1, running
>at 19200. As soon as I press any key on the WYSE, the code stops (and
>'INT' is never printed.
>- I have verified that loopback works fine on the WYSE.
>- If I print something out of UART1 to the WYSE, it works fine
>(without interrupts).
>void uart1_read(void) __irq
>{
> DUART_Write_Str("INT");
>
> if (U1IIR & 0x04 != 0)
> {
> DUART_Write_Char(U1RBR);
> }
> VICVectAddr = 0; // Acknowledge interrupt
>}
>void set_up_uart1(unsigned int baudrate)
>{
> PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> PCONP |= 0x00000010;
>
> U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
>
> switch(baudrate)
> {
> case 9600:
> U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 = 0x25
> U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> break;
>
> case 19200:
> U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 = 0x17
> U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> break;
> }
>
> U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> U1FCR = 0x07;
>
> // set up UART1 as a vectored interrupt
>
> VICVectAddr1 = (unsigned long) uart1_read;
> VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
>
> U1IER = 7; // RBR, THRE, RLS interrupt enable
>}
>main()
>{
> set_up_uart1(19200);
>
> for(;;)
> {
> char tmp[3];
>
> sprintf(tmp,"%x ",U1IIR);
> DUART_Write_Str(tmp); // other supporting code exists
> delay(SECS,0.25); // other supporting code exists
> }
>}
>Yahoo! Groups Links
>
Reply by ●January 9, 20072007-01-09
> -----Original Message-----
> From: l...
> [mailto:l...]On Behalf
> Of Sutton Mehaffey
> Sent: Tuesday, January 09, 2007 12:31 PM
> To: l...
> Subject: [lpc2000] UART1 RX LPC2148
> I have a problem where my code locks up on receiving a RX interrupt on
> UART1. I have been transmitting data to a serial display with no
> problems, although with interrupts disabled. Anybody see any
> omissions or problems with this? I didn't include code such as PLL
> setup, because a lot of other features work. But if anyone needs it,
> I'll post it. Thanks.
>
> Sutton - dodge55
>
Do you have any other interrupts working? Have you enabled interrupts
in the CSPR?
Mike
> From: l...
> [mailto:l...]On Behalf
> Of Sutton Mehaffey
> Sent: Tuesday, January 09, 2007 12:31 PM
> To: l...
> Subject: [lpc2000] UART1 RX LPC2148
> I have a problem where my code locks up on receiving a RX interrupt on
> UART1. I have been transmitting data to a serial display with no
> problems, although with interrupts disabled. Anybody see any
> omissions or problems with this? I didn't include code such as PLL
> setup, because a lot of other features work. But if anyone needs it,
> I'll post it. Thanks.
>
> Sutton - dodge55
>
Do you have any other interrupts working? Have you enabled interrupts
in the CSPR?
Mike
Reply by ●January 9, 20072007-01-09
The function DUART_Write_Str() is been called from the main loop and
the interrupt. Is the function written to handle the calls from int?
You may not enable the Transmitter Interrupt as you are not checking
for it in your IRQ.
suvidh
--- In l..., Mike Harrison wrote:
>
> On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
>
> >I have a problem where my code locks up on receiving a RX interrupt on
> >UART1. I have been transmitting data to a serial display with no
> >problems, although with interrupts disabled. Anybody see any
> >omissions or problems with this? I didn't include code such as PLL
> >setup, because a lot of other features work. But if anyone needs it,
> >I'll post it. Thanks.
> > Don't know if it applies to the 2148, but I had a problem on a 2103,
where setting the fifo control
> reg to an invalid value (fifo off) caused a continuous RX interrupt
even with no data.
> Reason was the fifo control register is a write-only register but I
was using a bitfield-defined
> SFR to set it up, so a subsequent read-modify-write operation
overwrote the intended value.
> >Sutton - dodge55
> >
> >
> >Notes on code operation:
> >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> >chip instead of UART1 (which printed fine without interrupts ON)
> >- I have an infinite loop (main) printing out U1IIR which is always
> >0xc1 as soon as everything is setup. I don't think this is correct
> >(pending interrupt).
> >- I have a dumb terminal (WYSE) connected directly to UART1, running
> >at 19200. As soon as I press any key on the WYSE, the code stops (and
> >'INT' is never printed.
> >- I have verified that loopback works fine on the WYSE.
> >- If I print something out of UART1 to the WYSE, it works fine
> >(without interrupts).
> >
> >
> >void uart1_read(void) __irq
> >{
> > DUART_Write_Str("INT");
> >
> > if (U1IIR & 0x04 != 0)
> > {
> > DUART_Write_Char(U1RBR);
> > }
> > VICVectAddr = 0; // Acknowledge interrupt
> >}
> >
> >
> >void set_up_uart1(unsigned int baudrate)
> >{
> > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > PCONP |= 0x00000010;
> >
> > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> >
> > switch(baudrate)
> > {
> > case 9600:
> > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 = 0x25
> > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > break;
> >
> > case 19200:
> > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 = 0x17
> > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > break;
> > }
> >
> > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > U1FCR = 0x07;
> >
> > // set up UART1 as a vectored interrupt
> >
> > VICVectAddr1 = (unsigned long) uart1_read;
> > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> >
> > U1IER = 7; // RBR, THRE, RLS interrupt enable
> >}
> >
> >
> >main()
> >{
> > set_up_uart1(19200);
> >
> > for(;;)
> > {
> > char tmp[3];
> >
> > sprintf(tmp,"%x ",U1IIR);
> > DUART_Write_Str(tmp); // other supporting code exists
> > delay(SECS,0.25); // other supporting code exists
> > }
> >}
> >
> >
> >
> >
> >Yahoo! Groups Links
> >
> >
>
the interrupt. Is the function written to handle the calls from int?
You may not enable the Transmitter Interrupt as you are not checking
for it in your IRQ.
suvidh
--- In l..., Mike Harrison wrote:
>
> On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
>
> >I have a problem where my code locks up on receiving a RX interrupt on
> >UART1. I have been transmitting data to a serial display with no
> >problems, although with interrupts disabled. Anybody see any
> >omissions or problems with this? I didn't include code such as PLL
> >setup, because a lot of other features work. But if anyone needs it,
> >I'll post it. Thanks.
> > Don't know if it applies to the 2148, but I had a problem on a 2103,
where setting the fifo control
> reg to an invalid value (fifo off) caused a continuous RX interrupt
even with no data.
> Reason was the fifo control register is a write-only register but I
was using a bitfield-defined
> SFR to set it up, so a subsequent read-modify-write operation
overwrote the intended value.
> >Sutton - dodge55
> >
> >
> >Notes on code operation:
> >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> >chip instead of UART1 (which printed fine without interrupts ON)
> >- I have an infinite loop (main) printing out U1IIR which is always
> >0xc1 as soon as everything is setup. I don't think this is correct
> >(pending interrupt).
> >- I have a dumb terminal (WYSE) connected directly to UART1, running
> >at 19200. As soon as I press any key on the WYSE, the code stops (and
> >'INT' is never printed.
> >- I have verified that loopback works fine on the WYSE.
> >- If I print something out of UART1 to the WYSE, it works fine
> >(without interrupts).
> >
> >
> >void uart1_read(void) __irq
> >{
> > DUART_Write_Str("INT");
> >
> > if (U1IIR & 0x04 != 0)
> > {
> > DUART_Write_Char(U1RBR);
> > }
> > VICVectAddr = 0; // Acknowledge interrupt
> >}
> >
> >
> >void set_up_uart1(unsigned int baudrate)
> >{
> > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > PCONP |= 0x00000010;
> >
> > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> >
> > switch(baudrate)
> > {
> > case 9600:
> > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 = 0x25
> > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > break;
> >
> > case 19200:
> > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 = 0x17
> > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > break;
> > }
> >
> > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > U1FCR = 0x07;
> >
> > // set up UART1 as a vectored interrupt
> >
> > VICVectAddr1 = (unsigned long) uart1_read;
> > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> >
> > U1IER = 7; // RBR, THRE, RLS interrupt enable
> >}
> >
> >
> >main()
> >{
> > set_up_uart1(19200);
> >
> > for(;;)
> > {
> > char tmp[3];
> >
> > sprintf(tmp,"%x ",U1IIR);
> > DUART_Write_Str(tmp); // other supporting code exists
> > delay(SECS,0.25); // other supporting code exists
> > }
> >}
> >
> >
> >
> >
> >Yahoo! Groups Links
> >
> >
>
Reply by ●January 9, 20072007-01-09
Yes. I have Timer 0 working as an interrupt.
I have been trying to look at the LPC2148 UART example in the Files
section to see if I could find the source of my problem. I don't see
any mention of CSPR in that example. No, I'm not doing anything with
that register. I don't see what that has anything to do with this
UART1 interrupt, as it is a Code Security register. If it does,
educate me please.
Sutton
--- In l..., "Michael Anton" wrote:
>
> > -----Original Message-----
> > From: l...
> > [mailto:l...]On Behalf
> > Of Sutton Mehaffey
> > Sent: Tuesday, January 09, 2007 12:31 PM
> > To: l...
> > Subject: [lpc2000] UART1 RX LPC2148
> >
> >
> > I have a problem where my code locks up on receiving a RX interrupt on
> > UART1. I have been transmitting data to a serial display with no
> > problems, although with interrupts disabled. Anybody see any
> > omissions or problems with this? I didn't include code such as PLL
> > setup, because a lot of other features work. But if anyone needs it,
> > I'll post it. Thanks.
> >
> > Sutton - dodge55
> > Do you have any other interrupts working? Have you enabled interrupts
> in the CSPR?
>
> Mike
>
I have been trying to look at the LPC2148 UART example in the Files
section to see if I could find the source of my problem. I don't see
any mention of CSPR in that example. No, I'm not doing anything with
that register. I don't see what that has anything to do with this
UART1 interrupt, as it is a Code Security register. If it does,
educate me please.
Sutton
--- In l..., "Michael Anton" wrote:
>
> > -----Original Message-----
> > From: l...
> > [mailto:l...]On Behalf
> > Of Sutton Mehaffey
> > Sent: Tuesday, January 09, 2007 12:31 PM
> > To: l...
> > Subject: [lpc2000] UART1 RX LPC2148
> >
> >
> > I have a problem where my code locks up on receiving a RX interrupt on
> > UART1. I have been transmitting data to a serial display with no
> > problems, although with interrupts disabled. Anybody see any
> > omissions or problems with this? I didn't include code such as PLL
> > setup, because a lot of other features work. But if anyone needs it,
> > I'll post it. Thanks.
> >
> > Sutton - dodge55
> > Do you have any other interrupts working? Have you enabled interrupts
> in the CSPR?
>
> Mike
>
Reply by ●January 9, 20072007-01-09
I have UART1 disabled in DUART_Write_Str for this test and then reenabled.
Can't I have the TX interrupt enabled and just ignore it? Even if it
was causing continuous TX interrupts, that's not a problem for this
test. Because, I'm not even getting one interrupt. Besides, I'm not
tranmitting any data, only receiving for this test.
Basically, what I am doing is async typing in chars from a WYSE
terminal (trying to receive them over UART1 RX) and 'echo' them out to
my display (connected via DUAL UART chip). So, I'm not transmitting
anything out over UART1 (yet).
Sutton
--- In l..., "suvidhk" wrote:
> The function DUART_Write_Str() is been called from the main loop and
> the interrupt. Is the function written to handle the calls from int?
>
> You may not enable the Transmitter Interrupt as you are not checking
> for it in your IRQ.
>
> suvidh
>
> --- In l..., Mike Harrison wrote:
> >
> > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> >
> > >I have a problem where my code locks up on receiving a RX
interrupt on
> > >UART1. I have been transmitting data to a serial display with no
> > >problems, although with interrupts disabled. Anybody see any
> > >omissions or problems with this? I didn't include code such as PLL
> > >setup, because a lot of other features work. But if anyone needs it,
> > >I'll post it. Thanks.
> > >
> >
> > Don't know if it applies to the 2148, but I had a problem on a 2103,
> where setting the fifo control
> > reg to an invalid value (fifo off) caused a continuous RX interrupt
> even with no data.
> > Reason was the fifo control register is a write-only register but I
> was using a bitfield-defined
> > SFR to set it up, so a subsequent read-modify-write operation
> overwrote the intended value.
> >
> >
> > >Sutton - dodge55
> > >
> > >
> > >Notes on code operation:
> > >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> > >chip instead of UART1 (which printed fine without interrupts ON)
> > >- I have an infinite loop (main) printing out U1IIR which is always
> > >0xc1 as soon as everything is setup. I don't think this is correct
> > >(pending interrupt).
> > >- I have a dumb terminal (WYSE) connected directly to UART1, running
> > >at 19200. As soon as I press any key on the WYSE, the code stops
(and
> > >'INT' is never printed.
> > >- I have verified that loopback works fine on the WYSE.
> > >- If I print something out of UART1 to the WYSE, it works fine
> > >(without interrupts).
> > >
> > >
> > >void uart1_read(void) __irq
> > >{
> > > DUART_Write_Str("INT");
> > >
> > > if (U1IIR & 0x04 != 0)
> > > {
> > > DUART_Write_Char(U1RBR);
> > > }
> > > VICVectAddr = 0; // Acknowledge interrupt
> > >}
> > >
> > >
> > >void set_up_uart1(unsigned int baudrate)
> > >{
> > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > PCONP |= 0x00000010;
> > >
> > > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> > >
> > > switch(baudrate)
> > > {
> > > case 9600:
> > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 0x25
> > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > break;
> > >
> > > case 19200:
> > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 0x17
> > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > break;
> > > }
> > >
> > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > U1FCR = 0x07;
> > >
> > > // set up UART1 as a vectored interrupt
> > >
> > > VICVectAddr1 = (unsigned long) uart1_read;
> > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > >
> > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > >}
> > >
> > >
> > >main()
> > >{
> > > set_up_uart1(19200);
> > >
> > > for(;;)
> > > {
> > > char tmp[3];
> > >
> > > sprintf(tmp,"%x ",U1IIR);
> > > DUART_Write_Str(tmp); // other supporting code exists
> > > delay(SECS,0.25); // other supporting code exists
> > > }
> > >}
> > >
> > >
> > >
> > >
> > >Yahoo! Groups Links
> > >
> > >
> > >
>
Can't I have the TX interrupt enabled and just ignore it? Even if it
was causing continuous TX interrupts, that's not a problem for this
test. Because, I'm not even getting one interrupt. Besides, I'm not
tranmitting any data, only receiving for this test.
Basically, what I am doing is async typing in chars from a WYSE
terminal (trying to receive them over UART1 RX) and 'echo' them out to
my display (connected via DUAL UART chip). So, I'm not transmitting
anything out over UART1 (yet).
Sutton
--- In l..., "suvidhk" wrote:
> The function DUART_Write_Str() is been called from the main loop and
> the interrupt. Is the function written to handle the calls from int?
>
> You may not enable the Transmitter Interrupt as you are not checking
> for it in your IRQ.
>
> suvidh
>
> --- In l..., Mike Harrison wrote:
> >
> > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> >
> > >I have a problem where my code locks up on receiving a RX
interrupt on
> > >UART1. I have been transmitting data to a serial display with no
> > >problems, although with interrupts disabled. Anybody see any
> > >omissions or problems with this? I didn't include code such as PLL
> > >setup, because a lot of other features work. But if anyone needs it,
> > >I'll post it. Thanks.
> > >
> >
> > Don't know if it applies to the 2148, but I had a problem on a 2103,
> where setting the fifo control
> > reg to an invalid value (fifo off) caused a continuous RX interrupt
> even with no data.
> > Reason was the fifo control register is a write-only register but I
> was using a bitfield-defined
> > SFR to set it up, so a subsequent read-modify-write operation
> overwrote the intended value.
> >
> >
> > >Sutton - dodge55
> > >
> > >
> > >Notes on code operation:
> > >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> > >chip instead of UART1 (which printed fine without interrupts ON)
> > >- I have an infinite loop (main) printing out U1IIR which is always
> > >0xc1 as soon as everything is setup. I don't think this is correct
> > >(pending interrupt).
> > >- I have a dumb terminal (WYSE) connected directly to UART1, running
> > >at 19200. As soon as I press any key on the WYSE, the code stops
(and
> > >'INT' is never printed.
> > >- I have verified that loopback works fine on the WYSE.
> > >- If I print something out of UART1 to the WYSE, it works fine
> > >(without interrupts).
> > >
> > >
> > >void uart1_read(void) __irq
> > >{
> > > DUART_Write_Str("INT");
> > >
> > > if (U1IIR & 0x04 != 0)
> > > {
> > > DUART_Write_Char(U1RBR);
> > > }
> > > VICVectAddr = 0; // Acknowledge interrupt
> > >}
> > >
> > >
> > >void set_up_uart1(unsigned int baudrate)
> > >{
> > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > PCONP |= 0x00000010;
> > >
> > > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> > >
> > > switch(baudrate)
> > > {
> > > case 9600:
> > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 0x25
> > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > break;
> > >
> > > case 19200:
> > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 0x17
> > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > break;
> > > }
> > >
> > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > U1FCR = 0x07;
> > >
> > > // set up UART1 as a vectored interrupt
> > >
> > > VICVectAddr1 = (unsigned long) uart1_read;
> > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > >
> > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > >}
> > >
> > >
> > >main()
> > >{
> > > set_up_uart1(19200);
> > >
> > > for(;;)
> > > {
> > > char tmp[3];
> > >
> > > sprintf(tmp,"%x ",U1IIR);
> > > DUART_Write_Str(tmp); // other supporting code exists
> > > delay(SECS,0.25); // other supporting code exists
> > > }
> > >}
> > >
> > >
> > >
> > >
> > >Yahoo! Groups Links
> > >
> > >
> > >
>
Reply by ●January 9, 20072007-01-09
> -----Original Message-----
> From: l...
> [mailto:l...]On Behalf
> Of Sutton Mehaffey
> Sent: Tuesday, January 09, 2007 1:40 PM
> To: l...
> Subject: [lpc2000] Re: UART1 RX LPC2148
> Yes. I have Timer 0 working as an interrupt.
>
> I have been trying to look at the LPC2148 UART example in the Files
> section to see if I could find the source of my problem. I don't see
> any mention of CSPR in that example. No, I'm not doing anything with
> that register. I don't see what that has anything to do with this
> UART1 interrupt, as it is a Code Security register. If it does,
> educate me please.
>
> Sutton
>
CPSR contains the global IRQ enable mask. It may be set by your
startup code, but interrupts need to be enabled here for them to
work.
I have had difficulties in calling functions from inside an interrupt
handler. I'm not exactly sure what my issue is, but I understand that
there may be a special assembler stub required to get this to work. So,
you might want to verify that calling functions inside your handler
works correctly.
As near as I can tell you have everything set appropriatly, so it
should work as written, but I haven't inspected in detail to make
sure all of the bits you have set are correct. The comments in
your code seem to indicate that you are doing things correctly.
Mike
> From: l...
> [mailto:l...]On Behalf
> Of Sutton Mehaffey
> Sent: Tuesday, January 09, 2007 1:40 PM
> To: l...
> Subject: [lpc2000] Re: UART1 RX LPC2148
> Yes. I have Timer 0 working as an interrupt.
>
> I have been trying to look at the LPC2148 UART example in the Files
> section to see if I could find the source of my problem. I don't see
> any mention of CSPR in that example. No, I'm not doing anything with
> that register. I don't see what that has anything to do with this
> UART1 interrupt, as it is a Code Security register. If it does,
> educate me please.
>
> Sutton
>
CPSR contains the global IRQ enable mask. It may be set by your
startup code, but interrupts need to be enabled here for them to
work.
I have had difficulties in calling functions from inside an interrupt
handler. I'm not exactly sure what my issue is, but I understand that
there may be a special assembler stub required to get this to work. So,
you might want to verify that calling functions inside your handler
works correctly.
As near as I can tell you have everything set appropriatly, so it
should work as written, but I haven't inspected in detail to make
sure all of the bits you have set are correct. The comments in
your code seem to indicate that you are doing things correctly.
Mike
Reply by ●January 9, 20072007-01-09
The Thing is that u will be getting an xmit interrupt you wont be
looking at it and it wont be cleared unless u write data to xmitter
FIFO.Once your data is transmitted the transmitter int. will be
generated again and again as u are not handling it. You need to
disable it in the interrupt if their is no data to write and reenable
when you have some data to write in FIFO.
In your case u can continuously disable xmit interrupt and check.
~~~~~~~ The transmit int. is cleared when u write data to xmitter fifo
~~~~~~
Suvidh
--- In l..., "Sutton Mehaffey" wrote:
>
> I have UART1 disabled in DUART_Write_Str for this test and then
reenabled.
>
> Can't I have the TX interrupt enabled and just ignore it? Even if it
> was causing continuous TX interrupts, that's not a problem for this
> test. Because, I'm not even getting one interrupt. Besides, I'm not
> tranmitting any data, only receiving for this test.
>
> Basically, what I am doing is async typing in chars from a WYSE
> terminal (trying to receive them over UART1 RX) and 'echo' them out to
> my display (connected via DUAL UART chip). So, I'm not transmitting
> anything out over UART1 (yet).
>
> Sutton
> --- In l..., "suvidhk" wrote:
> >
> >
> > The function DUART_Write_Str() is been called from the main loop and
> > the interrupt. Is the function written to handle the calls from int?
> >
> > You may not enable the Transmitter Interrupt as you are not checking
> > for it in your IRQ.
> >
> > suvidh
> >
> > --- In l..., Mike Harrison wrote:
> > >
> > > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> > >
> > > >I have a problem where my code locks up on receiving a RX
> interrupt on
> > > >UART1. I have been transmitting data to a serial display with no
> > > >problems, although with interrupts disabled. Anybody see any
> > > >omissions or problems with this? I didn't include code such as PLL
> > > >setup, because a lot of other features work. But if anyone
needs it,
> > > >I'll post it. Thanks.
> > > >
> > >
> > > Don't know if it applies to the 2148, but I had a problem on a 2103,
> > where setting the fifo control
> > > reg to an invalid value (fifo off) caused a continuous RX interrupt
> > even with no data.
> > > Reason was the fifo control register is a write-only register but I
> > was using a bitfield-defined
> > > SFR to set it up, so a subsequent read-modify-write operation
> > overwrote the intended value.
> > >
> > >
> > > >Sutton - dodge55
> > > >
> > > >
> > > >Notes on code operation:
> > > >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> > > >chip instead of UART1 (which printed fine without interrupts ON)
> > > >- I have an infinite loop (main) printing out U1IIR which is always
> > > >0xc1 as soon as everything is setup. I don't think this is correct
> > > >(pending interrupt).
> > > >- I have a dumb terminal (WYSE) connected directly to UART1,
running
> > > >at 19200. As soon as I press any key on the WYSE, the code stops
> (and
> > > >'INT' is never printed.
> > > >- I have verified that loopback works fine on the WYSE.
> > > >- If I print something out of UART1 to the WYSE, it works fine
> > > >(without interrupts).
> > > >
> > > >
> > > >void uart1_read(void) __irq
> > > >{
> > > > DUART_Write_Str("INT");
> > > >
> > > > if (U1IIR & 0x04 != 0)
> > > > {
> > > > DUART_Write_Char(U1RBR);
> > > > }
> > > > VICVectAddr = 0; // Acknowledge interrupt
> > > >}
> > > >
> > > >
> > > >void set_up_uart1(unsigned int baudrate)
> > > >{
> > > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > > PCONP |= 0x00000010;
> > > >
> > > > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> > > >
> > > > switch(baudrate)
> > > > {
> > > > case 9600:
> > > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 > 0x25
> > > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > > break;
> > > >
> > > > case 19200:
> > > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 > 0x17
> > > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > > break;
> > > > }
> > > >
> > > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > > U1FCR = 0x07;
> > > >
> > > > // set up UART1 as a vectored interrupt
> > > >
> > > > VICVectAddr1 = (unsigned long) uart1_read;
> > > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > > >
> > > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > > >}
> > > >
> > > >
> > > >main()
> > > >{
> > > > set_up_uart1(19200);
> > > >
> > > > for(;;)
> > > > {
> > > > char tmp[3];
> > > >
> > > > sprintf(tmp,"%x ",U1IIR);
> > > > DUART_Write_Str(tmp); // other supporting code exists
> > > > delay(SECS,0.25); // other supporting code exists
> > > > }
> > > >}
> > > >
> > > >
> > > >
> > > >
> > > >Yahoo! Groups Links
> > > >
> > > >
> > > >
> > >
>
looking at it and it wont be cleared unless u write data to xmitter
FIFO.Once your data is transmitted the transmitter int. will be
generated again and again as u are not handling it. You need to
disable it in the interrupt if their is no data to write and reenable
when you have some data to write in FIFO.
In your case u can continuously disable xmit interrupt and check.
~~~~~~~ The transmit int. is cleared when u write data to xmitter fifo
~~~~~~
Suvidh
--- In l..., "Sutton Mehaffey" wrote:
>
> I have UART1 disabled in DUART_Write_Str for this test and then
reenabled.
>
> Can't I have the TX interrupt enabled and just ignore it? Even if it
> was causing continuous TX interrupts, that's not a problem for this
> test. Because, I'm not even getting one interrupt. Besides, I'm not
> tranmitting any data, only receiving for this test.
>
> Basically, what I am doing is async typing in chars from a WYSE
> terminal (trying to receive them over UART1 RX) and 'echo' them out to
> my display (connected via DUAL UART chip). So, I'm not transmitting
> anything out over UART1 (yet).
>
> Sutton
> --- In l..., "suvidhk" wrote:
> >
> >
> > The function DUART_Write_Str() is been called from the main loop and
> > the interrupt. Is the function written to handle the calls from int?
> >
> > You may not enable the Transmitter Interrupt as you are not checking
> > for it in your IRQ.
> >
> > suvidh
> >
> > --- In l..., Mike Harrison wrote:
> > >
> > > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> > >
> > > >I have a problem where my code locks up on receiving a RX
> interrupt on
> > > >UART1. I have been transmitting data to a serial display with no
> > > >problems, although with interrupts disabled. Anybody see any
> > > >omissions or problems with this? I didn't include code such as PLL
> > > >setup, because a lot of other features work. But if anyone
needs it,
> > > >I'll post it. Thanks.
> > > >
> > >
> > > Don't know if it applies to the 2148, but I had a problem on a 2103,
> > where setting the fifo control
> > > reg to an invalid value (fifo off) caused a continuous RX interrupt
> > even with no data.
> > > Reason was the fifo control register is a write-only register but I
> > was using a bitfield-defined
> > > SFR to set it up, so a subsequent read-modify-write operation
> > overwrote the intended value.
> > >
> > >
> > > >Sutton - dodge55
> > > >
> > > >
> > > >Notes on code operation:
> > > >- DUART_Write_xxx - I have moved my display to utilize a Dual UART
> > > >chip instead of UART1 (which printed fine without interrupts ON)
> > > >- I have an infinite loop (main) printing out U1IIR which is always
> > > >0xc1 as soon as everything is setup. I don't think this is correct
> > > >(pending interrupt).
> > > >- I have a dumb terminal (WYSE) connected directly to UART1,
running
> > > >at 19200. As soon as I press any key on the WYSE, the code stops
> (and
> > > >'INT' is never printed.
> > > >- I have verified that loopback works fine on the WYSE.
> > > >- If I print something out of UART1 to the WYSE, it works fine
> > > >(without interrupts).
> > > >
> > > >
> > > >void uart1_read(void) __irq
> > > >{
> > > > DUART_Write_Str("INT");
> > > >
> > > > if (U1IIR & 0x04 != 0)
> > > > {
> > > > DUART_Write_Char(U1RBR);
> > > > }
> > > > VICVectAddr = 0; // Acknowledge interrupt
> > > >}
> > > >
> > > >
> > > >void set_up_uart1(unsigned int baudrate)
> > > >{
> > > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > > PCONP |= 0x00000010;
> > > >
> > > > U1LCR |= 0x80; // enable access to Divisor Latches, DLAB = 1
> > > >
> > > > switch(baudrate)
> > > > {
> > > > case 9600:
> > > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 > 0x25
> > > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > > break;
> > > >
> > > > case 19200:
> > > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 > 0x17
> > > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > > break;
> > > > }
> > > >
> > > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > > U1FCR = 0x07;
> > > >
> > > > // set up UART1 as a vectored interrupt
> > > >
> > > > VICVectAddr1 = (unsigned long) uart1_read;
> > > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
> > > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > > >
> > > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > > >}
> > > >
> > > >
> > > >main()
> > > >{
> > > > set_up_uart1(19200);
> > > >
> > > > for(;;)
> > > > {
> > > > char tmp[3];
> > > >
> > > > sprintf(tmp,"%x ",U1IIR);
> > > > DUART_Write_Str(tmp); // other supporting code exists
> > > > delay(SECS,0.25); // other supporting code exists
> > > > }
> > > >}
> > > >
> > > >
> > > >
> > > >
> > > >Yahoo! Groups Links
> > > >
> > > >
> > > >
> > >
>
Reply by ●January 9, 20072007-01-09
I understand your answer. However, there is another underlying
problem, because I am not getting any interrupts to fire. Under your
scenario, I should be getting TX interrupts continuous. I wish I
were, but I'm not getting any interrupts whatsoever.
I've even typed in line for line the UART0 example from the Files
section and changed UART0 to UART1. Still no interrupts. I can still
TX out using no interrupts.
Sutton
--- In l..., "suvidhk" wrote:
>
> The Thing is that u will be getting an xmit interrupt you wont be
> looking at it and it wont be cleared unless u write data to xmitter
> FIFO.Once your data is transmitted the transmitter int. will be
> generated again and again as u are not handling it. You need to
> disable it in the interrupt if their is no data to write and reenable
> when you have some data to write in FIFO.
> In your case u can continuously disable xmit interrupt and check.
> ~~~~~~~ The transmit int. is cleared when u write data to xmitter fifo
> ~~~~~~
> Suvidh
> --- In l..., "Sutton Mehaffey" wrote:
> >
> > I have UART1 disabled in DUART_Write_Str for this test and then
> reenabled.
> >
> > Can't I have the TX interrupt enabled and just ignore it? Even if it
> > was causing continuous TX interrupts, that's not a problem for this
> > test. Because, I'm not even getting one interrupt. Besides, I'm not
> > tranmitting any data, only receiving for this test.
> >
> > Basically, what I am doing is async typing in chars from a WYSE
> > terminal (trying to receive them over UART1 RX) and 'echo' them out to
> > my display (connected via DUAL UART chip). So, I'm not transmitting
> > anything out over UART1 (yet).
> >
> > Sutton
> >
> >
> >
> >
> > --- In l..., "suvidhk" wrote:
> > >
> > >
> > > The function DUART_Write_Str() is been called from the main loop and
> > > the interrupt. Is the function written to handle the calls from int?
> > >
> > > You may not enable the Transmitter Interrupt as you are not
checking
> > > for it in your IRQ.
> > >
> > > suvidh
> > >
> > > --- In l..., Mike Harrison wrote:
> > > >
> > > > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> > > >
> > > > >I have a problem where my code locks up on receiving a RX
> > interrupt on
> > > > >UART1. I have been transmitting data to a serial display with no
> > > > >problems, although with interrupts disabled. Anybody see any
> > > > >omissions or problems with this? I didn't include code such
as PLL
> > > > >setup, because a lot of other features work. But if anyone
> needs it,
> > > > >I'll post it. Thanks.
> > > > >
> > > >
> > > > Don't know if it applies to the 2148, but I had a problem on a
2103,
> > > where setting the fifo control
> > > > reg to an invalid value (fifo off) caused a continuous RX
interrupt
> > > even with no data.
> > > > Reason was the fifo control register is a write-only register
but I
> > > was using a bitfield-defined
> > > > SFR to set it up, so a subsequent read-modify-write operation
> > > overwrote the intended value.
> > > >
> > > >
> > > > >Sutton - dodge55
> > > > >
> > > > >
> > > > >Notes on code operation:
> > > > >- DUART_Write_xxx - I have moved my display to utilize a Dual
UART
> > > > >chip instead of UART1 (which printed fine without interrupts ON)
> > > > >- I have an infinite loop (main) printing out U1IIR which is
always
> > > > >0xc1 as soon as everything is setup. I don't think this is
correct
> > > > >(pending interrupt).
> > > > >- I have a dumb terminal (WYSE) connected directly to UART1,
> running
> > > > >at 19200. As soon as I press any key on the WYSE, the code stops
> > (and
> > > > >'INT' is never printed.
> > > > >- I have verified that loopback works fine on the WYSE.
> > > > >- If I print something out of UART1 to the WYSE, it works fine
> > > > >(without interrupts).
> > > > >
> > > > >
> > > > >void uart1_read(void) __irq
> > > > >{
> > > > > DUART_Write_Str("INT");
> > > > >
> > > > > if (U1IIR & 0x04 != 0)
> > > > > {
> > > > > DUART_Write_Char(U1RBR);
> > > > > }
> > > > > VICVectAddr = 0; // Acknowledge interrupt
> > > > >}
> > > > >
> > > > >
> > > > >void set_up_uart1(unsigned int baudrate)
> > > > >{
> > > > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > > > PCONP |= 0x00000010;
> > > > >
> > > > > U1LCR |= 0x80; // enable access to Divisor Latches,
DLAB = 1
> > > > >
> > > > > switch(baudrate)
> > > > > {
> > > > > case 9600:
> > > > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 > > 0x25
> > > > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > > > break;
> > > > >
> > > > > case 19200:
> > > > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 > > 0x17
> > > > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > > > break;
> > > > > }
> > > > >
> > > > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > > > U1FCR = 0x07;
> > > > >
> > > > > // set up UART1 as a vectored interrupt
> > > > >
> > > > > VICVectAddr1 = (unsigned long) uart1_read;
> > > > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
>
> > > > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > > > >
> > > > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > > > >}
> > > > >
> > > > >
> > > > >main()
> > > > >{
> > > > > set_up_uart1(19200);
> > > > >
> > > > > for(;;)
> > > > > {
> > > > > char tmp[3];
> > > > >
> > > > > sprintf(tmp,"%x ",U1IIR);
> > > > > DUART_Write_Str(tmp); // other supporting code exists
> > > > > delay(SECS,0.25); // other supporting code exists
> > > > > }
> > > > >}
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >Yahoo! Groups Links
> > > > >
> > > > >
> > > > >
> > > >
> > >
>
problem, because I am not getting any interrupts to fire. Under your
scenario, I should be getting TX interrupts continuous. I wish I
were, but I'm not getting any interrupts whatsoever.
I've even typed in line for line the UART0 example from the Files
section and changed UART0 to UART1. Still no interrupts. I can still
TX out using no interrupts.
Sutton
--- In l..., "suvidhk" wrote:
>
> The Thing is that u will be getting an xmit interrupt you wont be
> looking at it and it wont be cleared unless u write data to xmitter
> FIFO.Once your data is transmitted the transmitter int. will be
> generated again and again as u are not handling it. You need to
> disable it in the interrupt if their is no data to write and reenable
> when you have some data to write in FIFO.
> In your case u can continuously disable xmit interrupt and check.
> ~~~~~~~ The transmit int. is cleared when u write data to xmitter fifo
> ~~~~~~
> Suvidh
> --- In l..., "Sutton Mehaffey" wrote:
> >
> > I have UART1 disabled in DUART_Write_Str for this test and then
> reenabled.
> >
> > Can't I have the TX interrupt enabled and just ignore it? Even if it
> > was causing continuous TX interrupts, that's not a problem for this
> > test. Because, I'm not even getting one interrupt. Besides, I'm not
> > tranmitting any data, only receiving for this test.
> >
> > Basically, what I am doing is async typing in chars from a WYSE
> > terminal (trying to receive them over UART1 RX) and 'echo' them out to
> > my display (connected via DUAL UART chip). So, I'm not transmitting
> > anything out over UART1 (yet).
> >
> > Sutton
> >
> >
> >
> >
> > --- In l..., "suvidhk" wrote:
> > >
> > >
> > > The function DUART_Write_Str() is been called from the main loop and
> > > the interrupt. Is the function written to handle the calls from int?
> > >
> > > You may not enable the Transmitter Interrupt as you are not
checking
> > > for it in your IRQ.
> > >
> > > suvidh
> > >
> > > --- In l..., Mike Harrison wrote:
> > > >
> > > > On Tue, 09 Jan 2007 19:30:47 -0000, you wrote:
> > > >
> > > > >I have a problem where my code locks up on receiving a RX
> > interrupt on
> > > > >UART1. I have been transmitting data to a serial display with no
> > > > >problems, although with interrupts disabled. Anybody see any
> > > > >omissions or problems with this? I didn't include code such
as PLL
> > > > >setup, because a lot of other features work. But if anyone
> needs it,
> > > > >I'll post it. Thanks.
> > > > >
> > > >
> > > > Don't know if it applies to the 2148, but I had a problem on a
2103,
> > > where setting the fifo control
> > > > reg to an invalid value (fifo off) caused a continuous RX
interrupt
> > > even with no data.
> > > > Reason was the fifo control register is a write-only register
but I
> > > was using a bitfield-defined
> > > > SFR to set it up, so a subsequent read-modify-write operation
> > > overwrote the intended value.
> > > >
> > > >
> > > > >Sutton - dodge55
> > > > >
> > > > >
> > > > >Notes on code operation:
> > > > >- DUART_Write_xxx - I have moved my display to utilize a Dual
UART
> > > > >chip instead of UART1 (which printed fine without interrupts ON)
> > > > >- I have an infinite loop (main) printing out U1IIR which is
always
> > > > >0xc1 as soon as everything is setup. I don't think this is
correct
> > > > >(pending interrupt).
> > > > >- I have a dumb terminal (WYSE) connected directly to UART1,
> running
> > > > >at 19200. As soon as I press any key on the WYSE, the code stops
> > (and
> > > > >'INT' is never printed.
> > > > >- I have verified that loopback works fine on the WYSE.
> > > > >- If I print something out of UART1 to the WYSE, it works fine
> > > > >(without interrupts).
> > > > >
> > > > >
> > > > >void uart1_read(void) __irq
> > > > >{
> > > > > DUART_Write_Str("INT");
> > > > >
> > > > > if (U1IIR & 0x04 != 0)
> > > > > {
> > > > > DUART_Write_Char(U1RBR);
> > > > > }
> > > > > VICVectAddr = 0; // Acknowledge interrupt
> > > > >}
> > > > >
> > > > >
> > > > >void set_up_uart1(unsigned int baudrate)
> > > > >{
> > > > > PINSEL0 |= 0x00050000; // TX1 and RX1 on P0.8 and P0.9
> > > > > PCONP |= 0x00000010;
> > > > >
> > > > > U1LCR |= 0x80; // enable access to Divisor Latches,
DLAB = 1
> > > > >
> > > > > switch(baudrate)
> > > > > {
> > > > > case 9600:
> > > > > U1FDR = 0x0000009A; // MULVAL = 9, DIVVAL = 10
> > > > > U1DLM = 0; // (12000000 / 9600 / 16) * (9/19) = 37 > > 0x25
> > > > > U1DLL = 0x25; // 9601.71 baud, at 12Mhz
> > > > > break;
> > > > >
> > > > > case 19200:
> > > > > U1FDR = 0x000000A7; // MULVAL = 10, DIVVAL = 7
> > > > > U1DLM = 0; // (12000000 / 19200 / 16) * (10/17) = 23 > > 0x17
> > > > > U1DLL = 0x17; // 19181.585 baud, at 12Mhz
> > > > > break;
> > > > > }
> > > > >
> > > > > U1LCR = 0x03; // 8 bits, 1 stop, no parity, DLAB = 0
> > > > > U1FCR = 0x07;
> > > > >
> > > > > // set up UART1 as a vectored interrupt
> > > > >
> > > > > VICVectAddr1 = (unsigned long) uart1_read;
> > > > > VICVectCntl1 = 0x27; // IRQ 7 plus enable bit
>
> > > > > VICIntEnable |= 0x00000080; // enables the UART1 Interrupt
> > > > >
> > > > > U1IER = 7; // RBR, THRE, RLS interrupt enable
> > > > >}
> > > > >
> > > > >
> > > > >main()
> > > > >{
> > > > > set_up_uart1(19200);
> > > > >
> > > > > for(;;)
> > > > > {
> > > > > char tmp[3];
> > > > >
> > > > > sprintf(tmp,"%x ",U1IIR);
> > > > > DUART_Write_Str(tmp); // other supporting code exists
> > > > > delay(SECS,0.25); // other supporting code exists
> > > > > }
> > > > >}
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >Yahoo! Groups Links
> > > > >
> > > > >
> > > > >
> > > >
> > >
>
Reply by ●January 9, 20072007-01-09
Anybody know where the decription of the CPSR register (for global
interrupts) is. I can't seem to find it in the user's manual. I
think it is being set in startup, because my other interrupts work,
and there is no mention in any of my code. I apologize if it is plain
daylight, but I can't find it. :(
Sutton
--- In l..., "Michael Anton" wrote:
>
> > -----Original Message-----
> > From: l...
> > [mailto:l...]On Behalf
> > Of Sutton Mehaffey
> > Sent: Tuesday, January 09, 2007 1:40 PM
> > To: l...
> > Subject: [lpc2000] Re: UART1 RX LPC2148
> >
> >
> > Yes. I have Timer 0 working as an interrupt.
> >
> > I have been trying to look at the LPC2148 UART example in the Files
> > section to see if I could find the source of my problem. I don't see
> > any mention of CSPR in that example. No, I'm not doing anything with
> > that register. I don't see what that has anything to do with this
> > UART1 interrupt, as it is a Code Security register. If it does,
> > educate me please.
> >
> > Sutton
> > CPSR contains the global IRQ enable mask. It may be set by your
> startup code, but interrupts need to be enabled here for them to
> work.
>
> I have had difficulties in calling functions from inside an interrupt
> handler. I'm not exactly sure what my issue is, but I understand that
> there may be a special assembler stub required to get this to work. So,
> you might want to verify that calling functions inside your handler
> works correctly.
>
> As near as I can tell you have everything set appropriatly, so it
> should work as written, but I haven't inspected in detail to make
> sure all of the bits you have set are correct. The comments in
> your code seem to indicate that you are doing things correctly.
>
> Mike
>
interrupts) is. I can't seem to find it in the user's manual. I
think it is being set in startup, because my other interrupts work,
and there is no mention in any of my code. I apologize if it is plain
daylight, but I can't find it. :(
Sutton
--- In l..., "Michael Anton" wrote:
>
> > -----Original Message-----
> > From: l...
> > [mailto:l...]On Behalf
> > Of Sutton Mehaffey
> > Sent: Tuesday, January 09, 2007 1:40 PM
> > To: l...
> > Subject: [lpc2000] Re: UART1 RX LPC2148
> >
> >
> > Yes. I have Timer 0 working as an interrupt.
> >
> > I have been trying to look at the LPC2148 UART example in the Files
> > section to see if I could find the source of my problem. I don't see
> > any mention of CSPR in that example. No, I'm not doing anything with
> > that register. I don't see what that has anything to do with this
> > UART1 interrupt, as it is a Code Security register. If it does,
> > educate me please.
> >
> > Sutton
> > CPSR contains the global IRQ enable mask. It may be set by your
> startup code, but interrupts need to be enabled here for them to
> work.
>
> I have had difficulties in calling functions from inside an interrupt
> handler. I'm not exactly sure what my issue is, but I understand that
> there may be a special assembler stub required to get this to work. So,
> you might want to verify that calling functions inside your handler
> works correctly.
>
> As near as I can tell you have everything set appropriatly, so it
> should work as written, but I haven't inspected in detail to make
> sure all of the bits you have set are correct. The comments in
> your code seem to indicate that you are doing things correctly.
>
> Mike
>
