There are 7+ mainstream STM32F0 MCUs with 8 USART interfaces. There are several high performance STM32F4/F7 chips with 4 USART + 6 UART (for 10 total) interfaces, and three dozen chips with 4 USART + 4 UART (for 8 total).
I am pretty sure you can find similar options from other manufacturers if you stop looking for DIP package. The adapters from LQFP to DIP ...
The STM32F413 looks like it has 10 UARTS! https://www.st.com/en/microcontrollers-microprocessors/stm32f4-series.html
Even better for you, it looks like the STM32F413 is available on a Discovery Board, https://www.st.com/content/st_com/en/products/evaluation-tools/product-evaluation-tools/mcu-mpu-eval-tools/stm32-mcu-mpu-eval-tools/stm32-discovery-kits/...
Solved problem. There are bridge chips that can be controlled over I2C or SPI and have a UART.
Examples for the phrase "i2c uart ic":
XR20M1172 (double UART)
Example breakout: http://www.aliexpress.com/item/32772833676.html
The Teensy will be fine with 5 V on the inputs; that's what tolerant means. The main question is whether the 3.3 V output from the Teensy will be recognised as a logic HIGH by the RockBLOCK. The answer is in the datasheet:
T/Rx High Minimum: 3.0 V
As long as the Teensy will drive the load to a voltage >3.0 V, it will be fine.
However, given the cost of ...
What are you connecting your UART/serial comms to?
The MAX3323 uses external capacitors to generate +5V and -5V, and uses those supplies to convert signals from the low-voltage side into signals which transition between +5V and -5V. It only requires one low-voltage supply to do this.
The shield devices you link to require an external supply ( +5V ) to be ...
yes, for such slow signals either kind would work.
Or you could use resistors for voltage drop and a TTL compatible buffer for the upwards shift.
but "rock7" says
There are various FTDI USB cables, the precise model used with our products is the TTL-232R-3V3, and it's available to purchase from our shop or from various electronics outlets ...
You can also bit-bang a UART interface in software. This does not require dedicated UART hardware. Note however, that transmitting on a bit-banged interface takes more time than a hardware interface.
If you use Arduino, this might be of interest to you:
Especially if you have a crystal-controlled UART at the other end, you are reasonably safe with as much as +/-5% error.
However that error should include initial oscillator error, drift error (temperature, Vcc, time etc.) and any error in divider digital ratio.
No, because the UART stands for "asynchronous". You have somwhere in the datasheet a ratio baudrate vs sampling clock, so it is oversampled, then filtered (decimated). The asynchronous clock is reset once the start bit is detected, so there is plenty room for clock inequalities.
For the standard oversampling ratio of 16 , the clocks can be off by +/-5.11% (...
UART is very robust when it comes to clock accuracy requirements. This is a consequence of the fact that each byte is processed independently and any time difference associated to clock mismatch doesn't propagate for more than 10 bits (start bit + 8 data bits + stop bit) in time.
This analysis from Maxim concludes that 2% in clock frequency mismatch is ...
You can't connect the TX lines from two modules together. If you want to use your "TTL module" to monitor the communication then you can only connect the RX line. So, you can only monitor one direction of data transmission at a time.
Of course, you must also have compatible voltage levels for the serial interfaces in all three devices. You haven't provided ...
There is also a chance that the system may not detect it as an error. It will simply further sample the idle line and assume stop condition and stop.
The first byte if reported as error, can be cleared in software and the rest of the bytes will be received properly. Hence, in general there will be CRC or error checking mechanism to make sure that the ...
If you turn the receiver on after the start of a data word, then of course, the receiver will not interpret it correctly. You could regard this as misusing, rather than using a link.
If the word is immediately followed by a further word, then the misalignment can continue. As long as bit 2 is 1 and bit 3 is zero, the RX will 'see' a start edge at the time ...
Usually in the PLC communication world terms cyclic means data that are exchanged each scan continuously, for example IO variables, meanwhile acyclic means you send something on demand also you receive the response, usually this means the exchange of parameter data, configuration, diagnosis,..etc.
What are you looking on particular page of linked spec. is ...
The micro controllers in embedded systems share common clock baud rate generator clocked from the same cliyck source(TI, Microchip, ST, I have seen again). This won't allow the system to operate the same UART port with different baud rate.
To get around the problem:.
Use two UART ports, one for transmission and one for reception..both at different ...
In a typical situation between two communicating systems A and B there are two signals:
One from A's transmitting output to B's receiving input
One from B's transmitting output to A's receiving input
Same speed in different directions?
Other than simplicity, there's no particular reason the signal should go at the same rate in the two directions. Many ...
Synchronous communication will have a clock line and a data line. That makes it very easy:
The TX sets the data on the data line and then strobes the clock line.
The RX waits for the clock strobe and then reads the data.
It doesn't matter what the baud rate is and it could even vary during the transmission of a byte, for example.
In Asynchronous data ...
"Synchronous" means (in this context) that there's a separate line that just carries a clock strobe that tells you "when this signal goes high, measure the voltage on your RX line to get a value".
In asynchronous communications, there's no such signal. Instead, the receiver just watches the RX signal for a beginning-of-frame marker, and then simply measures ...
CSCTL1 |= (DCORSEL | DCOFSEL_3);//DCO 8Mhz 0x0046
The comment is wrong. This sets the FSEL field to 7, which is invalid and results in 24 MHz, which does not work without more FRAM wait states.
If you do not want to keep any bits that happen to be set in the register, do not use the | operator.
And you should clear the password after you have initialized ...
As mentioned above, setting it to output push-pull mode will solve the problem, however you may also want to check the "Maximum output speed" is set to 'very high', and if your system tolerates it, add the internal pull-up resistors; I know there's various schools of thought on whether USART should have pullups or not, so try it and see if it does what you ...
Do you have a schematic for your circuit?
In addition to the misconfigured GPIO pin, another possibility is that you have a capacitor on your UART line, acting as a low-pass filter, and preventing the signal from rising quickly.
Most likely the malloc and free are not thread safe so you can't allocate in interrupt and free in main loop. In general, there should be no dynamic memory allocation in interrupts. At least you should disable receive interrupts when popping a byte from the linked list. And in general rarely there should be need for linked lists using dynamic allocation in ...