You should search for this data yourself, but since you are saying you are a newbie I'll tell you how to find this.
First of all you should search for a document called "datasheet". On a datasheet the producer writes all the relevant data about his product.
In your case, you should search for the electrical characteristics of the ESP8266. For instance here ...
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 ...
The answers are in the notes:
Although no output capacitor is need for stability, it does improve transient response.
Required if regulator is locate an appreciable distance from power supply filter.
So the values aren't critical to the operation, but you should have them there. Bigger capacitors give more energy to the regulator and more energy ...
There are some suggestions at Protecting Inputs in Digital Electronics .
Testing of my earlier answer involving clamping diodes gave unsatisfactory results. Due to the small current flow the diodes allowed a considerable voltage to reach the input pin (like, 9 V).
The amended circuit below, tested with 12V input, satisfactorily keeps the ...
Wouldn't this be simpler?
Note: amended answer increases R1 to 10 k.
The 10 k resistor limits current to 0.17 mA. The 3.3 V zener diode clamps the input voltage to 3.3 V (I measured 2.3 V on mine so that is well within spec). The other end of the zener diode goes to the Arduino ground pin.
I originally had R1 as 1 k however if you had a high voltage input ...
In this data sheet it says:
When the output-enable (OE) input is high,the outputs are in the
However on page 3 you will notice that the *OE pin controls all outputs at the same time. So if you connect 2 output pins and one is high and the other is low and you hold *OE low ... you will likely burn out the driver for one of the 2 ...
Different color LEDs are not the same. The are made with different impurities. And some colors (blue) were really hard to figure out. So much so that the blue LED inventor was awarded the Nobel Prize!
Consequently, it should not be assume all LEDs operate at the same voltage and current. As such, when driving an LED with a constant voltage source, ...
Some good results you can get with optocouplers.
It's electrically disconnects one part of your circuit from another.
For example, here is my IoT-ish device, which is detects water leaks in my bathroom:
I use the transformer (T1), which have two separate output windings, so, that windings does not connected electrically, only magnetically. The block in ...
I don't know what that adapter is (a link would be nice, or at least the chip number it uses) but in general the 3.3v on a USB TTL adapter can only provide a small current. For instance the FT232RL can only provide up to 50mA at 3.3v. That may not be enough current to drive the RF circuitry of the ESP. The early Arduinos used the FT232RL chip for the USB ...
As you have already noticed, it may cause the 5V regulator to get hotter than would be normally considered reasonable. However, how hot depends entirely on how much current your board is drawing (not including the 12V devices)- the more current you draw the hotter it will get.
Even at 12V it can get uncomfortably hot.
For running off voltages greater than ...
wouldnt that then send current through the Vin pin and back to the coin cell battery? Is that what would happen in this instance?
Yep, that's right.
shouldn't there be a design feature to prevent users from damaging it like that? Fool/fail safe?
No. VIN can also be VOUT. You can use it to power other things that want a higher voltage.
The VIN port is connected to the barrel jack. It must only be used if you are powering the board through that jack.
To access the USB power just use the 5V pin - it is directly connected to the USB.
However, the USB, although the socket it is plugged into can supply more, is limited to 500mA. This is because of a 500mA auto-resetting polyfuse on the USB ...
An optoisolator is good for digital signals, but won't transfer an analog signal accurately.
Instead, you could put ADCs on the different circuits and optoisolate their communication with a master Arduino.
No, you cannot safely power an LED with 5V without a resistor. The resistor is absolutely 100% required.
The resistor isn't put there purely on a whim, it's required to set the current based on the supply voltage minus the LED forward voltage and the resistance of the resistor.
The 3V input on the DS1307 is for a 3V battery, usually a CR2032 or similar. The supply voltage of the DS1307 is 5V, which means that it won't work with a 3.3V Arduino without a separate supply and level shifter.
They are using the same SOT223 LDO as every other product. These have a very low limit of output current when run from an input voltage such as 12V. If you replace the LDO with a switching regulator, you can eliminate the waste heat regardless of the output current (Except for the 5V to 3.3V secondary LDO which is a constant). The PDF linked below shows a ...
All the IO pins contain a pair of clamping diodes to protect against over-voltage and negative-voltage situations:
As long as the current flowing through one of those diodes doesn't exceed its maximum forward current all will be well and the Arduino won't be damaged. If too much current flows then you will damage the pin.
Since you have 10KΩ resistors, at -...
Could someone explain why the examples use 10uF and the data sheet specifies 0.33uF and 0.1uF?
One obvious reason is that those two examples are non compliant with the datasheet.
But generally speaking the ordinary 78xx regulators use emitter output and are quite tolerant to the output capacitance. 10if is not a bad number.
However, there are also ...
For this I would recommend isolating the power from the batteries with P-channel MOSFETs and only switching them on when you want to sample the voltage. This means that there is no power consumption at all (aside from an absolutely minuscule leakage current) when turned off.
The whole setup would require 4 MOSFETs - two P-channel and two N-channel. They don'...
The 6.5V is too much. I have had an ATtiny chip which seems to run fine after applying 8V to it. But how can I be sure that it is okay ? There are so many parts in the chip die that can be damaged.
There has been a few broken voltage regulators reported, when applying 5V to the 5V pin of an Arduino Uno. The reverse diode inside the voltage regulator might ...
If you connect 11.5V directly to the Arduino then you might have killed you Arduino, BUT you say the LED comes up when you hook up the USB lead. So take things one step at a time.
Get an Arduino that you know is working and try it in the PC with the same cable. Does it work? (No -> Cable or PC is broken - Get a new cable/PC)
Remove all wires, shields, ...
Considering that you used resistors to lower the 11.5V to an cough - cough "very normal and regulated 5V", that most certainly is the problem.
The "right and cheap" way to lower such high voltage and connect them directly to your 5V (just assuming, since you said that you used resistors to protect the Arduino) is actually to use a voltage regulator. Any ...
Current through the clamping diodes shouldn't be more than 1mA according to Atmel's Zero Cross Detection Application Note (AVR182).
EDIT: As Edgar Bonet pointed out you have current about (-5+0.5)/(10000/2) = 0.9mA (tested) in the worst case, there is not much space for mistakes.
How would I hook up the arduino to the 12v system? As in, what should I do to protect the io of the arduino?
First you're going to need to do some investigative work. You're going to need to beg, steel, borrow (or use your own if you have one) an oscilloscope. You can then use that to examine the 12V power coming from the cigarette lighter. Armed with that ...
Looking towards the high end of recommended (vs absolute maximum) current and considering Ohm's law at a 5v difference if two outputs are trying to drive each other to opposite values, you would probably get away with as little as 250 ohms. But in most cases that would needlessly waste power, and in the case of multiple instance you could start totaling ...
According to the link you supply you might either receive the WS2812 or the SK6812, datasheet on Ws2812 claims you can supply up to 7V , but this is still less than 9V , the SK6812 is lacking in max voltage but says 5v typical.
If you still want to use the 9V supply you will need a extra voltage regulator.
A little math: each pixel can max draw 60mA, if ...