The base address for THR and RHR registers are same. So is it possible to transmit and receive at the same time?
It is specific to your particular UART hardware implementation, but it is unlikely that they are in fact the same register. They are two registers that have the same address - one is read-only (RHR), the other write-only (THR), so they do not need separate addresses.
In the hardware logic the correct register will be selected depending on the state of the read/write logic state as if that were an additional address line.
So yes, full duplex operation will be supported. You should read the user manual and/or data sheet for your particular part.
Related
When writing to device registers on a Cortex M0 (in my case, on an STM32L073), a question arises as to how careful one should be in a) ordering accesses to device memory and b) deciding that a change to a peripheral configuration has actually completed to the point that any dependencies become valid.
Taking a specific example to change the internal voltage regulator to a different voltage. You write the change to PWR->CR and read the status from PWR->CSR. I see code that does something like this:
Write to PWR->CR to set the voltage range
Spin until (PWR->CSR & voltage flag) becomes zero
In my mind there are three issues here:
Access ordering. This is Device Memory so transaction order is preserved relative to other Device access transactions. I would assume this means a DSB is not required between the write to CR and the read from CSR. A linked question and the answer to this is: [ARM CortexA]Difference between Strongly-ordered and Device Memory Type
Device memory can be buffered. Is there a possibility that a write to CR could still be in process when the read from CSR occurs. This would mean that the voltage flag would be clear and the code would proceed. In actual fact the flag hasn't gone high yet!
Hardware response time. Is there a latency between the write and the effects becoming final? In actuality this should always be documented - for the STM32 the docs definitively say that the flag is set when the CR register changes.
Are there any race condition possibilities here? It's really the buffering that worries me - that a peripheral write is still in progress when a peripheral read takes place.
Access ordering.
Accesses are strongly ordered and you do not need barrier instructions to read back the same register.
Device memory can be buffered. Is there a possibility that a write to CR
Yes, it is possible. But it is not because of buffering but because of the bus propagation time. It may take several clocks before a particular operation will go through all bridges.
Hardware response time. Is there a latency between the write and the
effects becoming final
Even if there is a latency it is not important from your point of view. If you set bit in the CR register and wait for the result in the status register. Simply wait for the status bit to have the expected value.
I am reading manuals regarding controlling a device with C,and in general its just playing with addresses; however when we are connected through UART we have the BAUDRATE present.
So how putting a value into some address have to do with baud-rate?
Is it necessary in embedded programming?
Those addresses are not memory. They are memory-mapped I/O registers.
The address for your UART's baud rate divisor register is a hardware register. The value in hardware registers directly control the hardware. The value written to baud rate divisor register is typically a counter reload value, and one bit period is the time it takes to count up-to (or down from) the value in the divisor given a specific peripheral clock source. So for exaample if the UART peripheral clock were say 12MHz, and you wanted a baud rate of 19200, you would set the divisor register to 12x106/19200 = 625.
Although you can read and write hardware registers as if they were memory, they do not necessarily behave like memory. Some registers may be read-only, others write only, and some writing may have a different effect than reading, such that if you write a value, the value read back will not be what was written. This often works at the bit level, so that each bit in a register may exhibit different behaviour.
For example on many UART implementations the register to which your write data to be sent, is the same address you read for received data - however they are not the same register, but rather a read-only register and a write-only register mapped to the same address.
It is not specifically an embedded programming thing, but rather an I/O hardware thing; it is simply that outside of embedded systems you are not typically writing directly to the hardware unless you happen to be writing a kernel device driver, where you will encounter the same thing.
As well as the device manuals which necessarily assume existing knowledge and expertise, perhaps you should consult a more general reference. Now you know the key term: "memory-mapped I/O" or MMIO, you are in a better position to Google it. Examples:
http://www.cs.uwm.edu/classes/cs315/Bacon/Lecture/HTML/ch14s03.html
https://en.wikipedia.org/wiki/Memory-mapped_I/O
In ARM there is a concept of Banked Register. While reading many questions and their answer and various other resources about what is Banked mean here. Then I got this definition:
Register banking refers to providing multiple copies of a register at the same address. Not all registers can be seen at once.
But my query here is that How multiple copies of registers are created. Because we have single register file in our core. And if there is another mode then it will get the new copy of the banked register that will not contain any data and not going to access the data of another mode register.
Then how this copy of register is created?
Register banking refers to providing multiple copies of a register at the same address. Not all registers can be seen at once.
This is some what correct. However, the register does not have a 'traditional address'. The majority of the arm instructions or 'binary encodings' have register as source or destination arguments. There are sixteen base register so four bits are needed for each register in a binary instruction. A typical instruction takes 12 bits (out of 32bits) to describe the three registers (two source and one destination). These bits in the instruction are the 'address' in the definition above.
But my query here is that How multiple copies of registers are created. Because we have single register file in our core. And if there is another mode then it will get the new copy of the banked register that will not contain any data and not going to access the data of another mode register. Then how this copy of register is created?
They are not 'created' dynamically. The banked registers are part of the 'register file' of the core and always exist. The issue is that the typical instructions can not access some banked registers unless a 'mode switch' occurs. This maybe from user to IRQ mode or from normal to secure world with trustzone.
So running the same code in different modes may end up accessing different (banked) registers. In this way, user code never affects the IRQ stack and vice-versa. Perhaps more importantly, the IRQ code could corrupt non-banked user registers if careful context saving is not performed at the start and end of an IRQ.
See: Accessing banked registers on ARM for information on how you might access these different registers.
The newer ARMv7 instruction mrs r2,sp_svc breaks this banked register rule and allows access to the banked registers directly without switching a mode. the intent is to allow context switching code to easily access the banked registers for saving and restoring without a mode switch.
The traditional instruction ldm rN, {sp,lr}^ allows saving of user stack pointer and link register without switching modes. Again, this has a special encoding (or addressing as per your definition).
Banking is also done with CP15 system registers in trustzone. TrustZone monitor mode and banked IFSR... maybe interesting for anyone looking at that ARM 'banking' which is conceptually the same as the register banking.
I count 31 registers needed to support the traditional arm. Several r13s and r14s a bunch for FIQ mode. First and foremost you are confusing tasks and modes. At the application level the tasks will all share the same set of registers, there is no banking there, when you switch tasks you have to save registers, the operating system allocates memory for this and for each task switch saves the old tasks registers and restores the next tasks registers.
As far as register banking there are multiple r13s for example. For each access to the register there is more than a simple offset into the register file it also has other inputs, for example
unsigned int get_r13 ( unsigned int mode )
{
switch(mode)
{
case SYS: return r13_sys;
case SVC: return r13_svc;
case ABT: return r13_abt;
case UND: return r13_und;
case IRQ: return r13_irq;
case FIQ: return r13_fiq;
}
}
but in logic although the logic could very well look about the same, that or they take the mode bits and logically convert them into some of the address bits into the register file or a combination thereof.
A single register file does not mean there are only 16 registers (r0-r15, not counting cpsr, etc) in the register file, there are 31 or more depending on whether or not it contains *PSR registers.
I want to understand how the registers of various peripherals/IPs are mapped to the ARM processor memory map in a microcontroller.
Say, I have a CONTROL register for UART block. When I do a write access to address (40005008), this register gets configured. Where does this mapping happens: Within the peripheral block code itself or while integrating this peripheral to the SoC/microcontroller.
For a simple peripheral like a UART it's straightforward - taking the ARM PL011 UART as an example (since I know where its documentation lives):
The programmer's model defines a bunch of registers at word-aligned offsets in a 4k block.
In terms of the actual hardware, we see the bus interface matches what the programmer's model suggests - PADDR[11:2] means only bits 11:2 of the address are connected, meaning it can only understand word-aligned addresses from 0x000 to 0xffc (similarly, note that only 16 bits of read/write data are connected, since no register is wider than that).
The memory-mapping between the UART's 12-bit address and the full 32-bit address that the CPU core spits out happens in the interconnect hardware between them. At design time, the interconnect address map will be configured to say "this 4k region at 0x40005000 is assigned to the UART0 block", etc., and the resulting bus circuitry will be generated for that.
More complex things like e.g. DMA-capable devices typically have separate interfaces for configuration and data access, so the registers can be mapped in a small relocatable block on a low-speed peripheral bus much like the UART.
Most significant bits are defined by your ASIC design, least significant bits are defined by the IP design. Your IP has several registers. The number of register, their order, is defined by the IP design. Here, your register is at address 8. Then when designing the ASIC, the peripherals are connected to the memory bus, and the way they are connected define their address. Your UART is at 40005000. You may have an other instance of the same IP at (for instance) 40006000. The two UART would be strictly identical, and you would be able to access CONTROL register of your second UART at address 40006008.
I'm looking for ways to interface a slow device (an ADC - ~1.2MHz) to my mcf5270 microcontroller, which is a rather fast device (~150MHz). I am using ADC0817, and have tried the following: GPIO read off the data lines, memory mapped access (since the adc is uP compatible).
When mapped as a location in memory, I am using the maximum wait states (15ws, which is about 107ns). The mcf5270 does not support any further waiting without using their bus model (which the ADCs do not support).
Using GPIO, only junk values are read.
Note: The ADC is actually working, and other than reading data the interface to it is functional. I am using C/C++ and assembly.
In short, I am looking for suggestions for ways in which to try to read the data off the ADC0817.
Comments and responses are greatly appreciated.
You could trigger the ADC through some GPIO, do an appropriate number of NOPs and then read the value. Also, you'd need to disable interrupts while doing this.
I think memory mapped should work - normally I would wait for the interrupt from the ADC, then read the byte at the specified address. Since you get an interrupt, there is no wait state needed. Am I missing something? GPIO should also work. How do you know the ADC is working? Can you put a logic analyzer on the data & interrupt to prove the ADC is producing the correct output?
I think from what he's saying MMIO wont' work because he would need more than the maximum number of wait states on the bus to get the glue logic to respond - ie, the A/D won't set its bus pins soon enough for the uC to be happy. That's a problem sometimes.
But the GPIO approach should work. If junk values are being read I assume that you aren't following the timing diagram provided. You'll have to be able to wait certain amounts of time after you signal a read before the data is valid. Check out the datasheet for specifics.