Nowadays, most smartphones come with a charger that is linked to the USB plug of the phone.
The charger for my HTC says: 5V, 1A, I guess this voltage must be regulated (but I haven't checked it has the charger is sealed).
The charger itself has a female USB socket on which you can plug any USB cord to link to your Arduino.
IMPORTANT! I have never tried it ...
The last resort option is to open a 5v phone charger and replace the metal prongs that go in a socket with two wires, and replace the usb socket with another pair of wires that go the arduino board, but still, an Arduino Pro Mini or an Arduino Nano is smaller than the phone charger.
You will probably have to do something like that.
Because of the nature of ...
Should the Resistor be Placed Before or After the LED?
The important thing to remember is that the direct current electricity must flow through the circuit to be useful. Since the resistor is part of the circuit (meaning that there isn't another path for the current to take) it resists the overall flow, thereby helping to protect the LED. Whether before or ...
If you are talking about running at 5V, then I doubt there is any safety issue. After all, 9V batteries (the sort you put in your smoke detector) just have bare terminals and no-one tells you not to touch them.
There is quite a bit of related information at How much voltage is “dangerous”?. That link seems to have enough information that there isn't a lot ...
I have connected the Relay and IR receiver directly to the Arduino
mini. Pin 10-14 are used for the relays Ins and PWD 3 for IR output.
I hope you don't mean that you have connected the coils of the relays directly to the Arduino pins. If that's the case, the surprise won't be the regulator blowing up, but the MCU surviving. Please edit the post and add ...
Welcome! I used bidirectional optocouplers. The input has a resistor I believe 27K and the output is a NPN transistor with a pull up to +5. The emitter goes to electronic ground. I connected these across the relay coils. They have not given me any problems since installed about 15 years ago. Nice thing they are not polarity sensitive.
The return signal is probably the other wire's servo. This only works when the control pulses to both servos are not in sync. Pull one signal wire off the breadboard and both servos will stop working.
You need to interconnect grounds or you will get strange effects sooner or later.
Well, if you put 5V into your barrel jack, your UNO's onboard 5V regulator is going to drop that down to 3.5V or so. You must have more than 5V into the regulator to get 5V out of it. This is referred to as the regulator's "drop out" voltage.
The specs for an UNO say to power it from at least 7V for that reason.
As an experiment you ...
wanted to do a similar thing and I found and just ordered these:
A fully sealed power supply, disk-shaped, 5cm diameter and 11mm thick. THey are available in 3.3V, 5V, and 12V. They are designed to be put inside walls or behind switches.
An obvious answer has been omitted which is: transformerless power supply.
If your average power requirements are low (<10mA) then this is the most compact way to achieve the desired voltage.
For an in-depth study see: http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf
Another solution: jeelabs.org /2011/11/27/ultra-low-power-supply/
Measuring the peak-to-peak current and scaling the result will get you an answer which will at least go up and down with the magnitude of your average AC current, so yes it is NOT incorrect. As you suggested it will be sensitive to noise from the sensor - in fact as you are taking the highest and lowest single readings the noise will always cause the ...
An input pin can only sense the voltage that is applied to it - whether it is above or below a pair of thresholds.
In your existing circuit when the button is not pressed the voltage sensed by the pin is 0V because it is connected to 0V (also known as Ground) through the resistor. When you press the button the input pin is connected directly to +5V, so the ...
That circuit is fine as long as the input signals don't go above 12.5V. You can protect it further by adding a 5.2V zener diode across each of the signals as they enter the Arduino:
simulate this circuit – Schematic created using CircuitLab
Why is this circuit not going to burn my Arduino if it is using 12V?
Because it uses a voltage divider to ...
Should I power the Arduino with the buck converter parallel to the same PSU (the 12V 20A Solar Charge Controller) I'm using for the pumps
Yes. However I wouldn't use a 9V buck regulator, but a 5V one and feed it to the 5V pin. That way you bypass the wasteful LDO on the Arduino, and when running from solar every bit of power saving you can muster helps.
A low cost option is the TCRT5000 infrared proximity sensor. While this is designed to sense objects coming close to the sensor, by reflection of the infrared light emitted by the integrated IR LED, it also does very well in detecting black versus reflective surfaces placed near it.
There are two parts to making this work:
Reduce the drive current for the ...
You can connect digital pin to analog input directly. That's not very interesting though, since you'll only see two values from the analog input (at least theoretically).
The digital pin gives out 0 or 5 Volts. The analog pin expects 0 to 5 Volts input. The only thing that matters to the analog input is the voltage with respect to ground. That can be ...
The arduino website states that the operating voltage of an arduino is 7-12V.
That's when it goes through the voltage regulator. It's a linear regulator, meaning it wastes a lot of power as heat. The regulator outputs 5V that's given to the MCU. The USB voltage bypasses the regulator and goes to a fuse before going to the common 5V for the board. If less ...
Try an LED driver. Their typical output is 11.8V DC, which is in perfect specs for the Arduino. You can find them nowadays in all shapes and sizes, round or square, low profile or boxy. A 500ma version (6W) is more than enough for your needs.
Since your project is to be battery powered, you want to make your system as efficient as is reasonable.
So I am going to take an approach that ends up improving the efficiency of your project.
The button you describe in your link states :
This button has a domed red plastic cover with a black retaining ring around it. There's an optional LED ...
The 1k (pulldown-)resistor must be placed at the other side of the button (the pin8 side, not the 5v side).
The led needs a resistor, to limit the current.
There is no need for the buttonstate variable, as you can just check the brightness to see if the PWM value must be changed.
int ledpin = 9;
int pushbutton = 8;
int brightness = 0;
int fadeamount = 1;
Using a 220k Ohm resistor will lead to approximatively 20µA to light the LED. This is not enough to light the LED, you should replace the resitor by one of value 330Ohm (220 to 470 could be OK depending of the LED)
The two pins aren't +5V power and ground.
One of them will be what you think, the other will be a signal wire with a pull up/down.
So if the ground wire really is ground then the other wire is the button input and is pulled up with a resistor, probably in the 10k region.
You can't power an arduino with a ~10k resistor in series with the power, it's just ...
What you want is a P-channel MOSFET to control the power. Something with a nice low "logic level" gate threshold.
simulate this circuit – Schematic created using CircuitLab
R1 keeps it switched off. Set the output to LOW to pull the gate down and enable the MOSFET.
Be sure to "stop" your Serial connection and set both TX and RX pins to INPUT to ...
There is no "You need X resistor for Y transformer". The resistor you choose depends on what current range you are working with and what voltage range you want that to be expressed as.
That transformer is a 100:5 (or more properly a 20:1) ratio. That means for every 20A going through it you get 1A out.
That output current is then passed through a burden ...
So you want to use the 5V voltage regulator on the Arduino serve as the supply for your 5V components, and feed the 5V regulator from the 12V supply that powers your solenoid?
The problem with that plan is heat. The voltage regulator on the Arduino is a poorly heat-sinked (heat-sunk?) "linear" regulator. A linear voltage regulator is basically a solid state ...