arduino
I am still a beginner, and I wanted to do this project
which will be fairly easy if I follow the code and the pic, but there is something I don't understand about the relay connection: Why is there a transistor, and why is there a diode, can't I just connect the output pin to a resistor then to the coil pin on the relay?
I think you may not fully understand how the servo works. Most servos have a power, ground, and PWM line. The PWM line will pull very little power. This line is what the microcontroller will actually be controlling. The Power and Ground lines are where the servo actually pulls its power from.
The power line can be connected to anything as long as the ground lines are at the same potential as the Arduino. If you are using an Arduino board, the voltage regulator that is on the board is limiting your power to the servo and not the microcontroller itself. If you are using an Arduino in an IC form (ie: breadboard) then you will need to make sure what ever you are using as your voltage source (power supply, battery, voltage regulator...) is at the right Voltage level and can handle the current of the servo.
What are those "larger components"? The only larger thing is the relay, and most relays will fit on a breadboard.
This is how you control the relay (the coil is shown next to the diode), it assumes you can connect the 12V's ground to the Arduino's. Resistor, transistor and diode are normal, small components. This relay is just a few cm long, wide and high. It can switch 10A and 230V. If you tell us more about what you want to switch I can give you more directed advice.
edit re your shopping
The relay requires 90mA from your 5V power supply. That will add a couple of hundreds of mW in the Arduino's voltage regulator. At 12V in that would be 630mW, which is a pity. If you have 12V in it would have been better to use that for a 12V relay.
The TIP31 transistor is overkill. It's a power transistor, and they don't have very high \$H_{FE}\$ (the current gain). Next time go for a TO-92 general purpose transistor like the BC547. The BC547B variant has an \$H_{FE}\$ of minimum 200. Go for a high \$H_{FE}\$. This one is still OK at an \$H_{FE}\$ of 100, but I would take a safety factor, and calculate with 40. Then the base current has to be 90mA/40 = 2.25mA. A 1k\$\Omega\$ base resistor will give you 4.3mA, so that's OK.
Best Answer
the transistor is there because the relay operates at 12V and probably wasnt over 100mA of current to operate. the arduino can't provide a signal of that magnitude so the transistor boots its output capability..
The diode is there because the relay contains an electromagnet, and an electromagnet is a type of inductor, and when you try to turn an inductor off it generates a high voltage, which could easily be hundereds of volts. but the
BC547transistor is can only withstand a few tens of volts Vce when the transistor turns off the energy in the inductor forces the switched terminal more positive until it gets to 0.6V above the 12V supply and then theiN4001starts to conduct the current and he stored energy is burned off as heat in the electromagnet's resistance and in the diode's voltage drop. This is called a freewheeling diode or catch diodeAll this complexity could be avoided by using a solid-state relay as they can be triggered directly by the arduino outputs but they can be quite expensive.