I am designing 8*8 reed switch matrix which will be connected to Arduino. I am using Arduino Duemilanove. I have completed the matrix portion but don't know how to connect and where to connect the other elements.
My Matrix circuit
I decided to use the following items:
Arduino (Duemilanove)
64x Reed Switch
64x Diode (1N4148)
20x Resistors (10K)
1x 74HC595
1x 74HC165
I dont know where to place the Resistors and shift Registers and connection from shift registers to arduino. I need help in the circuit diagram. Please help me out with the circuit diagram.
Reference: https://mtifall10.wordpress.com/2010/12/10/magnetic-sensing-chessboard/
I'd use something like that:
Where LOAD should be in HIGH state by default, LOW level loads inputs on 74HC165, and LOW to HIGH change loads outputs.
So you just send short LOW pulse to the LOAD, and then send next row to scan (will be loaded in next round) to the SPI. You have to remember the previous row or you have to connect QH* output from 595 to SER on 165 and shift out two bytes, one dummy and second will be the new row to be scanned.
And the matrix scanning is Active LOW. So you have to send the byte with one bit cleared (selected row) and rest bits set. And it's similar for reads (but you can use inverted output from 165 as well to get inverted values directly).
I'm not an expert but I'll try my best. I can't assure you that this is 100% correct.
Because of the limited amount of digital ports you will also have to use the analog ports.. In your case, the a, b, c, d... pins are "positive"(current goes from plus to minus). The pins 1, 2, 3... are your grounds/negative side.
The resistors should be connected to each positive pin, in your case pin a, b, c and so on.
EDIT: Sorry, I did bit consider the shift register. You will have to read the datasheet. You'll probably use a port with pulse width modulation(PWM), GND port and VoltageOut port.
Related
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)
I am fairly new to programming MCU. I have an ATtiny2313 and the datasheet came with a layout of how the pins are configured; however, finding pin 1, since the mcu is a square, is tough for me. I share the images of the diagram and the actual unit. Maybe I am overthinking the issue? Further reading materials are also greatly appreciated.
This is the datasheets diagram
This is the actual unit
Check the Packaging information section at the end of datasheet. There are physical properties of packages described including any marking to identify pin 1.
In some situations (some really small ICs) it can be hard to identify right pin 1 marking. In production, one simply rely on orientation of chip in the tape, but for populating boards by hand, little trick can come handy anytime you are not sure, you have identified pin 1 correctly.
In virtually any IC, there are ESD protection diodes from GND to any IO pin and prom IO pin to power rail. So, if you put a multimeter in diode measuring mode, you should see a drop about 700 mV (actual value can vary) when the positive probe is touching a ground pin and the negative probe any data pin, or when the negative probe is touching a positive voltage rail pin and the positive probe any IO pin. It is not 100 % fail-proof way, but really good and useful hint.
I'm working through a set of beginner exercises with the Arduino Uno microcontroller. (A generic one, though, as this is what I've been supplied with.)
The program I'm running, which alternates between sending 1's and 0's to serial output depending on the state of a momentary switch, has set pin 2 to be the input for the switch. But. Whilst wiring up, I accidentally plugged the jumper cable in to pin 3 initially, and found it still mostly sent the 1's when the button was pushed. Some 0's, yet mostly 1's.
Initially I thought maybe it was just the board was a bit dodgy, but thought I'd experiment a bit. Plugging into pin 3 instead of pin 2 still fairly consistently sent 1's when the button was pushed, though the 1's flowed a little bit less consistently than when it was in pin 2. In pin 2 it was completely consistent by comparison. So I tried pin 4, but with that one there's no response at all.
Am I right in presuming the program's readings seems to get a little bit less responsive the further away I move the cable from the pin that I've programmed to act as input? Can anyone help me understand why this happens?
It's probably quite obvious that I'm new to electronics. :)
The program I've got uploaded to the board is as follows:
// digital pin 2 has a pushbutton attached to it. Give it a name:
int pushButton = 2;
// the setup routine runs once when you press reset:
void setup() {
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
// make the pushbutton's pin an input:
pinMode(pushButton, INPUT);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input pin:
int buttonState = digitalRead(pushButton);
// print out the state of the button:
Serial.println(buttonState);
delay(1); // delay in between reads for stability
}
Floating pins are prone to noise. If you are not actually connecting anything to pin 2, you will be reading noise. Any wire connected to pin 2 (even connections on the board) will act like an antenna and pick up noise. You should always use the pin number that you are physically connecting in situations like these.
Leaving input pins open makes the microcontroller read a floating value which swings between 0 to 1. Also when wiring a switch to any pin, make sure to hook some pull-down resistor to make the input 0.
These are common for many electronics and proper notice to be taken while designing circuits of your own.
You need to look into datasheet where described functions of the pins.
The pins of MCU can be assigned various functions through special registers.
Two most common functions of the pins are input and output. MCUs provide internal pull-up and pull-down resistors which when used properly significantly simplifies electronic schemas.
If the input activated as input without any pull-??? then it's state is not defined and can be used as initiator of random number generator. Due this reason it is better to define what is default state of the input pin by connecting pull-??? resistors.
In Arduino IDE you are not limited to functions provided -- you still can use register manipulation directly, you just need to learn internals of the MCU.
If you do it properly then 2kbit program very often can be made as small as a few hundred bytes and it will work hundred times faster.
Operating registers in C is not much different from assembly, in C++ you get right away significant overhead -- although some benefits of registers still can be significant.
Libraries hide from programmers internals of MCU what is nice as it simplifies the programming and does not require to understand how MCU works, what registers are changed in what sequence.
But when you know hardware in and out -- you can squeeze from small MCU what is not possible with use of libraries (code will just not fit into the chip). MCUs are not that complicated (Atmel) to learn about it's internals -- benefits are significant.
Knowledge is a power which many avoid.
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.
I am a CS guy getting started with Arduino. This is probably a very basic electronics question but from going over the arduino tutorials everything is connected to the arduino with a resistor.
Well since i am following the tutorials i know what type of resistor i should use but what i do not know is why i should use one? and What type of resistor to pick i am to do something which is not covered in a tutorial.
The resistor simply serves to limit the current into or out of a pin in case something goes awry. If your AVR decides to output high on a pin that something else wants low (or vice-versa), large, damaging currents can occur if not limited by some resistance. The current limit for AVRs is about 20 milliamps, and given that the voltages are usually 5V, something larger than 250 ohms "would work".
To give a margin of safety, 1-10k is a great choice; for digital signals it seldom matters unless you're into very high-speed applications (beyond the AVRs capability anyways). For analog inputs, a similar resistor would also be advisable, as the amount of current the ADC takes to sample is negligible when your resistor is in the few kilo-ohm range.
The underlying principle that you want to learn is Ohm's Law, which describes the relationship between voltage, resistance, and current in a circuit.
Resistors are used to
limit current,
devide voltage
protect against over voltage
pull-up, pull down
current to voltage conversion
etc ...
1) limit output current, the absolute max current per IO is 40mA, a typical LED works on ±2V 20mA.
the resistance value can by calculated by (5V - 2V)/(0.02A)=150Ω usually a 220Ω resistor is used, because: it consumes less power, there doesn't flow 20 milliamps, and there is no notable difference in emitted light.
2) if you have a analog voltage that variates between 0 and 10 Volts, you 'll need a voltage divider of 1/2. pick by example z2 10k and calculate z1 by 10k*(Vin,max/5V -1). take a value of resistance higher than the original calculated. and recalculate the new Vout.
3) place a resistor of 10k in series between the analog input of the arduino and the 'to measure voltage'
3) if you have to measure a analog current, you place a resistor to ground and the analog input, calculate the resistor by Z=5V/amps.
4) if you connect a button to the arduino, you 'll need to place a pull up or a pull down resistor. if you 're not using a resistor, the input is floating and can take any value (high or low). or you can enable the internal weak pull up. by pinMode(xx,INPUT); digitalWrite(xx,HIGH);. and disabeling by digitalWrite(xx,LOW); by default the pull-up is disabled.