I am currently using analog pin 3 on my Arduino Uno to send out voltage from 0 to 5V.
I am using that voltage to control the motor and currently I am using the function:
analogWrite(pin, PWM_PULSE);
I am using 255 pwm for 5V and 127 for 2.5V.
The problem is that PWM is sending full cycle at 255pwm(5V), but on 127V the cycle is at 50% which causes my motor to twitch a little bit.
How can I solve this? I am searching for a way to send full PWM cycle even at lower volts. Is it possible?
First I believe you mean D3 not A3, since PWM does not exist on A3.
Assuming you are driving a DC motor, and not something like a servo or stepper
You have two problems.
1st. you may need a smoothing capacitor. Where your formula would be F=L*C Noting that analogWrite uses a F=490Hz. The concept is simple, in short the cap average out the high and lows of the PWM, based on the duty cycle. And the capacitance needed is based on the frequency and impedance. This will provide the analog voltage.
2nd. And bigger problem is the output of the Arduino can not supply sufficient current to drive the motor correctly. It will max out at about 20ma, and the motor likely needs more. So at low speeds the pulses which were week, stall out during their low periods.
You should have your PWM output drive a transistor, which in turn will ON-OFF the motor directly from the power supply. Now your motor may have enough inertia as not to need the cap.
see adafruit-arduino-lesson-13-dc-motors/breadboard-layout
and here for a discussion about the smoothing cap
Related
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.
Currently I have this and a 12v power supply: http://www.ebay.com.au/itm/High-Power-10W-LED-Driver-MBI6651-PWM-DIM-
I want to dim a 12v 10w LED with PWM. Is there any way I can do this with an Arduino?
The pin description says this:
PWM terminal. When applied with +5v or suspended, full amount of current will be output and when connected with ground, output current will be 0.
So, as the Arduino runs off 5v, does that mean I can use the Arduino PWM to tell this board to DIM? Or I am I getting that totally wrong?
I'm a bit of an electronics noob, so forgive me if these questions are simple.
You can generate an PWM signal in the Arduino and link it to the PWM pin in the power supply, it should work as specified in the link.
PWM terminal. When applied with +5v or suspended, full amount of
current will be output and when connected with ground, output current
will be 0.
So if you generate a full signal, you will have the full power in the power supply, 100% light, 50% PWM will generate 50% power to the LED and so on...
You can check Arduino documentation for more information about how to use the PWM using analogWrite()
https://www.arduino.cc/en/Tutorial/PWM
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.
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.