The design assumes USB provides a regulated 5v so no further regulation is necessary. The power connecter is meant to accept a wider range of voltage and regulate it to 5v, and the kind of regulators used on the Arduino board need that much higher voltage to provide regulation. If you already have a regulated +5v source you can supply it to the 5v pin. You'...
No. You can draw power from the 5v and VIN pins on the arduino. VIN is the completely unaltered input power before the regulator (it will be useless if regulated 5v is supplied directly). Outputting power is what the 5v pin is intended to be used for, not as a power input.
From the arduino website:
5V. This pin outputs a regulated 5V from the regulator ...
Consideration (go down for the actual solution)
If you have a lot of different China-Arduino-clones, the easiest way will be to just test it. The cloned Arduinos use a lot of different voltage regulators and often you won't even be able to find datasheets for them on Google.
Someone here provided an Arduino sketch which helps you identify the frequency ...
The regulator should be marked K850(5.0V) or K833(3.3V).
A 5 volt part has a 16MHz resonator may be marked with "A1" or "A'N"
A 3.3 volt part has a 8MHz resonator may be marked with "80'0"
As others have indicated, you can apply up to 12V at the RAW pin, and measure the output of the regulator.
The genuine Pro Mini's use a MIC5205 regulator which should accept up to 16V at it's input normally, with an absolute maximum of 20V. It's unlikely that the regulator would be damaged by 15.1V.
However, the component you have indicated that has blown is a capacitor. SMD capacitors are available in different voltage ratings, typically 4V, 6.3V, 10V, 16V, 25V,...
The ADC inside the Arduino does not measure voltage, but rather a
voltage ratio. Namely the ratio from the voltage at the analog input
to the voltage at the Vref pin.
In the default configuration, the Vref pin is internally tied to the
+5 V line. You can select to use instead an internal reference as
This reference ...
To drive such high currents, you may have to cascade several transistors (you can also use a Darlington transistor). There are arrays of Darlingtons mounted in a chip (e.g. the ULN2803A has 8 darlington transistors, but is limited to 500mA).
You probably will have to deal with higher power transistors; as an example I have found STMicroelectronics TIP110 ...
Well, it depends on what you have attached to the Arduino. If you are using the pins to do things you will definitely require more current.
If you ONLY want to power the Arduino, then I calculated what I think it should be below. (You should also check with someone who owns this device for an experimental value, and not just a calculated value. )
Based on ...
You basically have three options:
Switch to an Arduino Due which has a built-in DAC which outputs a real voltage.
Add an external DAC chip (such as the MCP4821/2) to create the voltage for you
Use a low-pass filter (R-C network) on a PWM pin.
Of the three options I usually use an MCP4822 since it gives the best results and doesn't cost as much as using a ...
I have good news, and I have bad news.
The good news is that only two changes to the board are required to change it from a 5V board to a 3.3V (or any voltage within spec) board.
The bad news is that there's no way in hell you'll be able to do them.
So, we work around them and get the board up and running.
The first is obvious: a 3.3V board should be ...
The Vin pin serves the same purpose as the barrel jack. The only difference is (usually) that it doesn't go through the schottky protection diode that causes a small voltage drop.
Normally its voltage is at around 0.3-0.4V below whatever you feed into the barrel jack.
If you power from USB then the USB voltage (~5V) is "back-fed" through the 5V regulator ...
Unfortunately there is no one "clear cut" answer for all you ask.
There are some hard limits, though, which you can get from the main chip's datasheet:
Operating Voltage: 1.8 - 5.5V
But that's not the whole story. The minimum voltage depends on the clock speed of the chip, as shown in this graph:
But wait, there's more. The Arduino also contains a 3.3V ...
A USB-powered Arduino Nano will have an ADC voltage reference which can't be relied on, due to the +/- 5% tolerance of the incoming USB voltage. On top of that, the Nano has an MBR0520 Schottky diode (D1) that will drop between 0.1 and 0.5 V depending on its own manufacturing tolerances, its temperature, and the current draw of your board.
What can you ...
For a 5V logic output to a 3V3 logic input, you can use a resistive divider to lower the voltage.
When unloaded, a 3V3 logic output is just enough to drive a 5V logic input. Check the AVR datasheet for the exact voltages (0.6 × Vcc = 3V, found under DC Characteristics in the datasheet).
In other words, with a little bit of special care it may just work.
The board has a voltage regulator which converts 9-12V to the 5V that the chip uses.
Why does it need 9V if the board works at 5V? Well, it doesn't need 9 exactly, but it needs something greater than 5. The reason for this is that the regulator has whats called a dropout voltage.
Looking at the schematic of the Uno we see it uses a NCP1117, which has a ...
Since the ATTiny can easily run at 5V I don't see any problem with programming it at 5V levels and then run it at 3.3V. Is that correct?
Yes. But if there are any components on the board that can't run at 5V then you must provide some way of protecting them from the excessive voltage.
If I change the fuses on Tiny to run it at 3.3V can I still use the 5V ...
You would use a transistor. Something like this would work.
In this case the Arduino controls a transistor to drive a brushless motor.
Source: Arduino Cookbook: (https://www.inkling.com/read/arduino-cookbook-michael-margolis-2nd/chapter-8/figure-8-9)
For the sake of clarity I should point out that the TIP102 is the device to use for the 1A ...
It's... mostly wrong. For three independent reasons.
The LEDs themselves, like all diodes, absorb some of the electrovoltaic potential of the electricity passing through them, resulting in a voltage decrease in their circuit. This means that although 5V is being used to power the resistor and LED in a circuit, only some of that voltage is passing through ...
There are lots of ways to switch higher loads, and jfpoilpret has described some good options. I'll summarise a couple of relay-based solutions, which are mainly appropriate for comparatively slow switching speeds (i.e. not usually suitable for PWM).
Solid State Relays
Solid State Relays (SSRs) are effectively semiconductor-based switches. They come in a ...
And the third part of your question: It is possible to bypass the voltage regulator and feed REGULATED 5 volts directly to the 5V pin.
That's how the Arduino is powered using USB. The USB power is already regulated 5V, and is fed directly to the 5V pin.
Beware of doing this yourself however. You better be certain the 5V you're feeding in is a clean, ...
The logic level is the same as the voltage you supply the chip.
You can think of the IO pins as tiny switches that either connect the output pin to the supply voltage or to ground (when in output mode) or compare the incoming voltage to an input pin to a percentage of the supply voltage (typically <0.3×Vcc for low and >0.6Vcc for high)
So yes, if you ...
Using the MAX3232 data sheet , which is what the TTL Converter has on it's board :
The main difference between using 3.3v or 5v is on the receiver data line "R" (or Rout as it is called on the MAX3232).
Simply put, you want to power the TTL converter from the same voltage as your Arduino is using (3.3v or 5v). Meaning the voltage that your I/O pins ...
At 5V, a 1 ohm resistor will try to sink 1A and far exceed the 40mA specs. Please use at least a 5/0.040=125 ohm resistor to protect your pin. And if you put the a capacitor between your resistor and ground, the RC circuit of the capacitor will smooth out the PWM into an analog voltage.
Please try the suggested @russell answer with a 47K resistor and ...
and an external 5V DC power supply
OK, assuming you are putting 5V into the barrel jack this is exactly what you expect to happen.
From the Arduino Uno specifications you are supposed to supply a recommended 7V to 12V. Your supply of 5V is too low.
The barrel jack first goes through a M7 protection diode which would have about a 0.7V forward voltage drop, ...
Not only can you share the grounds - sharing the grounds is required for there to be any form of meaningful circuit for signals to get around the place.
To copy-and-paste a blog post I wrote some time back:
A lot of the time on the Arduino forums we get questions regarding wiring things together. One common format is:
I want to connect my 12V powered ...
Firstly, if I understand your schematic right, you're not "dividing" anything there. If the ringed red symbol is where you are connecting the ADC then either you will see near 4V on there (if it's only connected to the ADC), or 0V (if it's connected to ground).
You should be connecting the ADC between the resistors to get a reading.
Secondly, you should be ...
If I understand your writeup correctly, this is what you have built:
simulate this circuit – Schematic created using CircuitLab
What you didn't mention in your question is that you made a connection from Arduino ground to the base of the transistor and the seconds circuit's ground connection. I drew the extra lead 'loose' for illustration only, but ...
Use the analog comparator in the ATmega328P to trigger an interrupt once the input voltage rises enough to indicate a button press.
Connect the analog network to both the analog input and D6.
If your lowest analog voltage is not greater than 1.2V then apply a voltage greater than 0V but more than 40mV less than your lowest analog voltage to D7. If you have ...
You could try not digitalRead, but analogReadVoltage. Analog Read Voltage
From Arduino examples:
Reads an analog input on pin 0, converts it to voltage, and prints the result to the serial monitor.
Attach the center pin of a potentiometer to pin A0, and the outside pins to +5V and ground.
This example code is in the public ...