designpinoutpulldownpullupstm32
I'm designing STM32f446 chip in PCB in a project. It will be programmed through SWD.
but I'm stuck with (BOOT0) pin. I want it to be (0) as recommended.
The thing is that there is no enough info. about it.
I've looked at the data sheets to find the rating (voltage and current), I just found the voltage to be 9V !
I need to select the pull-down resistor, what should I do?
thanks,
The cynic in me says that everyone uses a 10kΩ resistor because under the technical data section of the FSR above, a chart is given with output voltages specified with 5V supply and a 10k pull-down resistor. I have copied the chart below for reference: \$\begin{array}{} Force (lb)& Force (N)& \begin{matrix}FSR\\ Resistance\end{matrix} &(FSR + R) Ω&\begin{matrix}\text{Current thru}\\ FSR+R\end{matrix} &\begin{matrix}Voltage\\ \text{across }R\end{matrix} \\ \hline None& None&\infty&\infty&0 mA&0V\\ 0.04 lb &0.2 N &30kΩ&40 kΩ&0.13 mA&1.3 V\\ 0.22 lb &1 N &6 kΩ&16 kΩ&0.31 mA&3.1 V\\ 2.2 lb &10 N &1 kΩ&11 kΩ&0.45 mA&4.5 V\\ 22 lb &100 N &250 Ω&10.25 kΩ&0.49 mA&4.9 V \end{array}\$
So we can see that a 10kΩ gives us a reasonable amount of range over the operation of the sensor. If we wanted to have better resolution with the smaller force ranges, we could increase the pull-down resistance. We could also decrease the pull-down resistance to improve the resolution at higher force ranges, although that would also come at a cost of greater current draw.
From a TI datasheet, this is the equivalent circuit for a 7407 buffer:
Note that the input is relatively low impedance (having a 6k pullup to Vcc through the emitter-base junction (which effectively looks like a diode) of the transistor shown). The datasheet says it requires 1.6mA to pull the input low (parameter IIL). This is a lot of current compared with the few microamps, nanoamps, or picoamps typical of CMOS or other high-impedance inputs.
Q: Since the pull-down resistor is supposed to create a low-level at the buffer's input, why doesn't it show something closer to 0V at A?
A 5.1k pulldown resistor is too large to pull the low impedance input down. With 1.6mA flowing through 5.1k, the voltage drop (calculated via Ohm's law) would be 8.16 volts. To pull the input down below 0.8 volts, you would need a 500 Ohm or smaller resistor. I would suggest trying 330 Ohms.
Q: If a buffer's output is supposed to mirror its input, why does it not read the same 1.9V as A?
The output is floating. The voltage you are measuring is just stray charge. Unless the output is actively pulled down by the 7407 or pulled up by an external circuit, it will float.
Also, it should only "mirror" its input in terms of logic state, not voltage. When the input is pulled to a logic low level (below 0.8V), the output transistor is fully switched on. When the input is at a high logic level (above 2V), the output transistor is fully switched off. At input voltages between 0.8V and 2V, the output transistor may be fully on or fully off or partially switched on.
Best Answer
Section 6.2 in the STM32F446 datasheet shows that the absolute maximum voltage allowed on the
BOOT0pin is 9V. However, a voltage above VDD is only used in one, rare, situation. In all other (i.e. normal) cases, you treat theBOOT0pin as a normal CMOS input (input voltage between VDD and VSS).The one situation where that 9V maximum voltage is relevant, is because the
BOOT0pin is also the VPP pin. One option for programming the internal Flash memory of the MCU (which is rarely used outside of mass programming on a production line) involves supplying an external VPP voltage on that pin.That is the only time when VPP is applied to that pin, so that is the only time when the 9V maximum applies. See section 3.5.2 "Program/erase parallelism" in Reference manual RM0390 for your MCU.
For each family of STM32 MCUs, ST produces a "Getting started with [...] MCU hardware development" application note. For your MCU, you need "Application note AN4488 - Getting started with STM32F4xxxx MCU hardware development". That shows you how to connect the
BOOT0pin for normal use.If you don't ever want to change this
BOOT0setting, you could tie that pin low (i.e. to your 0V rail), to force the MCU to always boot from your program in its Flash. However that will prevent you booting the MCU using its built-in bootloader (or from SRAM) in future, if your plans change. Therefore leaving theBOOT0pin as a logic high/low selectable option (e.g. via jumper pins or solder jumper etc.) may be sensible, as shown in the ST "Getting started" application note linked above.