Is it possible to change the timing of SPI slave select line on a microcontroller? How is the timing of slave select line determined?
I am using SPI interface of nrf52DK microcontroller board. For 125kHz SPI clock, the slave select line has a low level of 165 us but the high level is quite long (appears for a long time before transitioning to next low level) and I want to change this. I want to reduce the length of high level on the slave select line. I checked this on an oscilloscope.
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I had written a code on atmel studio for blinking a led on pin 13. After uploading the code with xloader mega's builtin led was blinking.
I uploaded fade code on my mega and the builtin led was blinking instead of led. What should i do?
I am using arduino mega 2560.
int main(void)
{
DDRB=0b00000000;
while (1)
{
PORTB=0b10000000;
_delay_ms(1000);
PORTB=0b00000000;
}
}
What you should do? Read the manual.
Please refer to https://ww1.microchip.com/downloads/en/devicedoc/atmel-2549-8-bit-avr-microcontroller-atmega640-1280-1281-2560-2561_datasheet.pdf
Chapter 13.2.
The DDxn bit in the DDRx Register selects the direction of this pin.
If DDxn is written logic one, Pxn is configured as an output pin. If
DDxn is written logic zero, Pxn is configured as an input pin.
Working with registers doesn't make sense if you don't know what they do.
DDRB=0b00000000;
Gives you inputs only.
why would you use Arduino and try to program it without its conventional macros and functions?
If you are trying to blink an led or make it breath then use the Arduino IDE and its built-in functions analogWrite() to generate a pwm pulse for your led or any led on suitable pins which support analogwrite(). You shouldn't try to do any direct modifications on registers if you have no suitable knowledge, because your risk destroying your development kit and maybe burning some other stuff around. Please use your kit's schematics to spot the pins which support analogwrite() and then use the code in examples.
That way you will achieve your goal faster and without any issues.
TL/DR: you have to set 7th bit in DDRB to one.
In AVR ports are configured by bits in two registers: DDRx and PORTx.
When the corresponding bit in the DDRx register is set to one, the port is configured as output. And the corresponding bit in the PORTx register chooses which electrical level is output on the pin. If it is 0 then internal MOSFET shorts the pin to "ground" lane, and sinks current from external source. When the bit of the PORTx is one, then the pin is connected to "VCC", sourcing big amount of current enough to lit up a LED.
But if the pin is connected to something, what consumes too much of current, or the pin is shorted to GND or VCC (let's say you have a button connected and pressed), then output MOSFETS might be overloaded and damaged.
If the bit in DDRx is set to zero, then the pin is configured as input. If the corresponding bit in the PORTx is zero, then the pin has no internal connection to power lines, it is called "Hi-impedance" state, or Tri-state. It does not source or sink any current. So, if no external source of current is connected, then pin level is floating, influenced by electrical interference. Logical level is not detectable and can change occasionally. If you want to connect, for example, a button (between the pin and GND), then logical level will be defined only when button is pressed. When it is released, the logical level will be undefined.
But! If the bit in the PORTx is set to one, then internal MOSFET connects the pin thru a resistor (about 35 kOhm) to VCC line. This make the pin to source a little amount of current, setting its logical level to high. Therefore, if a button is connected, when it is released, then pin will have defined high level. This is called "pull-up resistor". When button is pressed, it will not short and damage the MCU, because current flowing thru the button is limited by the resistor, but the logical level will be defined low.
What if instead of button you have a LED connected to the pin? Very small amount of current will flow thru the LED, makes it barely glow.
Read more in the datasheet (chapter 13. I/O-Ports)
Im currently trying to get an electric signal from arduino, its 5v and 1amp that i get from a powersupply.
I want to input that signal into an arduino pin, lets say pin 4.
The main powersource from my arduino is via usb, but the 5v signal is from an external device.
I just want to know the number of time that signal became active, like a switch.
As far as i know arduino can take only .04amp from 5v.
Is there anyway i can reduce the current?
Anyway to obtain the value of a resistor to make it less dangerous for my arduino?
Your question is a very common application for Arduino!
You can give your Arduino some additional protection by placing a 10kOhm resistor between the Arduino analog pin you wish to use and the positive voltage output of the power supply.
If you're worried that the voltage could increase above 5V, you can protect your arduino with a simple voltage divider using two resistors. There's a detailed tutorial for this approach here: https://startingelectronics.org/articles/arduino/measuring-voltage-with-arduino/ Here's a simplified circuit diagram with a voltage divider that reduces voltage 11 fold - making voltages up to 55V safe to measure (where the battery could be replaced by your power supply):
For your code, you can use analogread() to read the voltage of the pin. If you wired it correctly, it should return near 0 when the powersupply is at 0, and 1026 or thereabouts if it is at 5v (or whatever the maximum value your voltage divider is designed for). Here is an example to get you started :
https://www.arduino.cc/reference/en/language/functions/analog-io/analogread/
If you need any support with your code to count the number of times the voltage goes high, post that as a separate question along with the code you have so far.
I want to write a stepper motor driver with a STM32L152RE.
I would like to control the position of the stepper motor with a potentiometer.
As I am new, I do not know how to communicate with (DIR, STEP).
Can someone give me a light or show me a way?
I'm using an A4988 - DMOS Microstepping Driver with Translator And Overcurrent Protection
I tried to use STM32 tim, but I could not.
Actually I have wrote whole driver for a4988, it is irq based. But I can't uncover it.
I can describe a path how to start. Anyway you should have some kind of hardware, because A4988 needs extra components for current control (resistors), and some capacitors..
You could try POLOLU HW.
If you have some kind of custom board, there may be some flaws.. So recheck pins.
Especially ROSC pin, SENSE1, SENSE2 pins as those may cause that motor wont work even if other pins are ok.
ROSC pin is for low power mode, so here you should calculate, if you just ignore it, be sure to connect at least to 10k resistor. Don't let it float. SENSE1, SENSE2 pins can be connected to 0.25omh resistors. You should check it.
Also from power pins very decisive VREG pin. It should get from 0 to 2000mV if I remember. Actually it controls current for your motor. So it depends on your stepper motor. Here also may appear nasty flaws. For example you have small stepper motor, and setting too high VREG value, than A4988 will sink too much current and your motor will glitch. Anyway you should read A4988 data sheet very accurately.
DIR PIN is simply for direction, push-pull pin configuration and HIGH/LOW values controls direction, clock wise, anti clock wise.
RESET INPUT PIN A4988 must get HIGH from your MCU.
ENABLE INPUT PIN A4988 must get LOW from your MCU.
SLEEP INPUT PIN A4988 must get HIGH from your MCU, also it is very useful to control it when your stepper job done, else if you leave it always HIGH, stepper motor will eat current and will heat up at idle state.
Also there are 3 MICROSTEPPING PINS, those are for controlling stepping.
As you just starting to play, it will be enough connect those pins to GND, you'll get full stepping regime.If you'll control those pins you can get other regimes like 1/2 stepping, 1/4,1/8,1/16...
And general pin is STEP pin, it should be driven with TIMER as PWM output with constant pulse width and alternating period.
Here is an example of STEP PIN control:
RCC_AHBPeriphClockCmd(RCC_AHBPeriph_GPIOA,ENABLE);
RCC_APB2PeriphClockCmd(RCC_AHBPeriph_GPIOA,ENABLE);
GPIO_InitTypeDef init;
init.GPIO_Mode = GPIO_Mode_AF;
init.GPIO_OType = GPIO_OType_PP;
init.GPIO_Pin = GPIO_Pin_9;
init.GPIO_PuPd = GPIO_PuPd_UP;
init.GPIO_Speed = GPIO_Speed_Level_2;
GPIO_Init(GPIOA,&init);
GPIO_PinAFConfig(GPIOA,GPIO_PinSource9,GPIO_AF_2);
for example I connect STEP output from MCU PA9 to A4988 STEP input. Which can be driven from timer as PWM. Check your concrete MCU datasheet.
Firstly output pin should be configured as AF, with push-pull and resistor is UP.
Also setup line for alternating pin.
Now configuring timer:
RCC_APB2PeriphClockCmd(RCC_APB2Periph_TIM1, ENABLE);
TIM_TimeBaseInitTypeDef timerInitStructure;
timerInitStructure.TIM_Prescaler = 48;
timerInitStructure.TIM_CounterMode = TIM_CounterMode_Up;
timerInitStructure.TIM_Period = 0;
timerInitStructure.TIM_ClockDivision = 0;
timerInitStructure.TIM_RepetitionCounter = 0;
TIM_TimeBaseInit(TIM1, &timerInitStructure);
TIM_OCInitTypeDef osc;
osc.TIM_OCMode = TIM_OCMode_PWM1;
osc.TIM_OutputState = TIM_OutputState_Enable;
osc.TIM_OutputNState = TIM_OutputNState_Disable;
osc.TIM_Pulse = 1000;
osc.TIM_OCPolarity = TIM_OCNPolarity_High;
osc.TIM_OCNPolarity = TIM_OCNPolarity_Low;
osc.TIM_OCIdleState =TIM_OCIdleState_Reset;
osc.TIM_OCNIdleState =TIM_OCNIdleState_Set;
TIM_OC2Init(TIM1, &osc);
TIM_Cmd(TIM1, ENABLE);
Here I configure 1us timer, as my MCU frequency is 48MHz.
Also you have configure that timer would drive PWM output.
TIM1->CCR = 10; with this register I can control pulse width, in this example it is 10us.
TIM1->ARR = 30; with ARR register I can control period, so it means STEP pulse frequency which is equal to stepper motor speed. In this case 30us.
If you are using HAL and CUBEMX you can get those configurations pretty fast.
im trying to turn on the light (bulb) with my arduino UNO and one module with 4 relays. I can do it with one LED but with an bulb i can't. I have connected the wires like in the photo:
http://i.stack.imgur.com/GUuAS.jpg
I need a 1k ohm resistor or the module include it??
Here the bulb that I have:
http://i57.tinypic.com/10dbp90.jpg
Thanks!!!
In your wiring picture it appears that the only source of power is what the Arduino is getting from the USB cable. The purpose of a relay is typically to control the flow of a higher voltage source (such as multiple small batteries in series, a larger battery, or a wall outlet) using a lower voltage control signal (e.g. one of your Arduino's GPIO pins). The maximum current from VCC to ground that can be draw without damaging your Arduino is 200 mA (source). Power = voltage * current (p = i * v) and VCC is 5V. This means the total amount of power your Arduino can supply is 1 watt. This is likely significantly less than the amount of power required to turn on your light bulb.
The purpose of the resistor in the LED circuit is to limit the current going through the LED. This is more commonly done when the LED is connected directly to a GPIO pin in order to prevent more current from being drawn from a pin than the amount that will damage that pin. From the same source as the current limit from VCC, the limit for a GPIO pin is only 40 mA. I would recommend seeing if you can power your light bulb with a battery. You could then use this same size battery as the power source for your relay board.
I know this sounds a bit funny :). But I am trying to eliminate possibilities:
On the Arduino Uno I have attached an interrupt triggered on HIGH to a routine which only increments a volatile defined long counter. This counter is displayed on an LCD screen.
If I connect a pulse generator with a frequency of 1 Hz at TTL levels, I would expect the counter to increase with about 1 per second. However this is not the case.
As the frequency is 1 Hz (duty cycle 50%) could it be possible that once the counter is incremented the IRS is exited (and clears the interrupt flag) BUT: the INT0 level is still HIGH so the ISR would be called again? At 1 Hz 50% duty, the HIGH would stay for 500 ms and at 16 mHz...
The processor at the heart of any Arduino has two different kinds of interrupts: “external”, and “pin change”. There are only two external interrupt pins on the ATmega168/328 (ie, in the Arduino Uno/Nano/Duemilanove), INT0 and INT1, and they are mapped to Arduino pins 2 and 3. These interrupts can be set to trigger on RISING or FALLING signal edges, or on low level. The triggers are interpreted by hardware, and the interrupt is very fast. The Arduino Mega has a few more external interrupt pins available.
So as commented: It triggers on an edge!
See more details on the Arduino Playground web page.
Two electrical reasons can explain why interrupt does not function as you need.
1- The pulse generator output and MCU input can have an impedance mismatch, which can cause ringing on the waveform edges. For example, if your function generator has a 50 ohm output capable of generating high frequencies you might see a problem driving a high impedance input like the Arduino at low frequency.
The name "pulse generator" makes me think this is a 50 ohm out device intended to make very short pulses with sharp edges. In such a case, you add a terminating resistor at the destination (load) to match the impedance of the source (pulse generator). For a 50 ohm output, 47 ohm would be close enough. If the output is 100 kohm, then place a matching resistor at the Arduino.
2- Just the opposite, the generator waveform edges may be so slow that the voltage passes through TTL 0 to 1 transition multiple times. If you have noise on your signal input, a slow edge could be causing multiple triggers. For example, if you are picking up some 60 Hz ripple from a power supply and grounding issues, your square wave edges won't be as square as you think.
In such cases hysteresis is a solution. There are many ways to de-glitch (debounce) in code. There is no answer that is right for all problems. A simple example would be that the ISR you require that the input reads high twice in a row for the edge to be accepted.