very basic question

[unixyperl]

Hello forum

I am looking for an answer to a very fundamental question.. I think!

I have played with eletronics lots over the years, but never have gotten properly into it as a hobby, because I have failed to fully get my head around understanding circuit diagrams.. I guess I am the sort of person who HAS to understand how it works, and they never quite make sense to me so I don't progress

They don't make sense because, I think, I don't understand the concepts of how eletricity moves around the circuit. I will try to explain exactly what I don't get using the attached Arduino diagram:

Two connections go to GND. But when there is more than one path to ground, how does a component such as the resistor in this Arduino diagram have any effect? I could understand if it was connected to the bottom of the Arduino, but the way I am looking at it, I see it connected directly to + live (when button pressed) and therefore I expect the electricity to go into pin 7 on the Arduino, and the resistor is just sort of sitting in the middle.. either that or the electricity goes into pin 7 and through the resistor, eith/both ways still going to ground.

My best answer is that as the electricity has to flow back from GND to +, it does so through this resistor, thus resiting the current through pin 7? But if so, why does it do that? Does it take the path of least resistence, ie through pin 7, pin 7 has a diode (or software equivalent) connected so it, so current must flow through the resitor?

Sorry, pretty long winded there, but if anyone has the kindness and patience to help me, I would greatly appriciate it! Thank you people

 

[Harald Kapp]

Welcome to elctronicspoint.
Unfortunately the attachment is missing.

Current always takes the pat of least resistance. It there is more than one path for current to flow, the current splits. One part takes the route of low resistance, the other part takes the route of high resistance. Of course, more current flows through the low resistance path than through the high resistance part. The voltage across both pathes (resistors) is the same. This is know a s Kirchhoff's laws.

Consider a water tap and two hoses with different diameter as an analogon. The pressure of water at the tap is the same for both hoses (the electrical analogon is the voltage). The amount of water flowing in the two hoses is different (the electrical analogon is current). The flow of water is proportional to the diameter: small diameter, little water, big diameter, much water.
The sum of water flowing in the small and the large hose is the "current" flowing out from the tap. If you make the big hose wide enough (electrical analogon: low resistance), almost no currrent will flow through the small hose (high resistance).

Here are some tutorials that may interest you.

Kind regards,
Harald

 

[unixyperl]

Welcome to elctronicspoint.
Unfortunately the attachment is missing.

This is know a s Kirchhoff's laws.

If you make the big hose wide enough (electrical analogon: low resistance), almost no currrent will flow through the small hose (high resistance).

Harald

Thank you Harald. That helps me somewhat. But I will try again to attach the file if you can look at it for me? I really want to be certain I understand how the resistor in the curcuit functions. Thank you very much

 

[Arouse1973]

Hello

The resistor in your drawing is what we call a pull down. It ensures the port pin is in a determinate state, in this case it is asserting a logic zero or LOW to the port pin. This makes sure the port does not float and cause a false trigger i.e. It thinks the switch is pressed when it is not. Also if the port pin does not have a resistor pull down or pull up (This is the opposite of a pull down) then when the port floats due to say contamination on the PCB it can go into a state where it's neither in a LOW or HIGH state (indeterminate). This can cause excess current to be drawn by the internals of the port, the exception to this is an A to D port which doesn't care about the voltage on the pin as long as it's within the limits of the port pin.

Thanks
Adam

 

[unixyperl]

Hello

The resistor in your drawing is what we call a pull down. It ensures the port pin is in a determinate state

Adam

Thanks Adam. That is good information that I can research

But, I'm trying to grasp how exactly the electricity is moving around the circuit.

Here is the break-down as I can see it:

1) switch is pessed, allows electricity to flow through
2) the electricty chooses the path of lower resitance (pin 7), rather than through the resistor.
3) after causing whatever internal effects to the Arduino curcuits, it exits to GND
4) *Here I am unsure* Does it flow from GND, through the resistor and back to pin 7?

Thanks very much

 

[Harald Kapp]

1) switch is pessed, allows electricity to flow through
2) the electricty chooses the path of lower resitance (pin 7), rather than through the resistor.
3) after causing whatever internal effects to the Arduino curcuits, it exits to GND
4) *Here I am unsure* Does it flow from GND, through the resistor and back to pin 7?

"Electricity" has two main components, namely voltage and current. Let's start with the latter: Current is the flow (movement) of electrons within a conductor (wire). The more electrons move per unit of time, the higher the current.
Imagine voltage as the force that pushes the elecrons through the wire. The higher the voltage, the higher the push on the electrons.
You can have voltage without current , e.g. across the terminals of a battery.
It is difficult to have current without voltage (not impossible but difficult) as you need a force to push the electrons.

1) when you close a switch, you allow current to flow. At the same time, the voltage present on one side of the switch appears on the other side of the switch, as a switch has very low resistance.
2) the current indeed choses the path of lowest resistance, but not solely. Some current will flow through the resistor, too, see my water analogon. In fact, the arduino's input pins have a very high resistance and almost no current flows into them. They are voltage controlled. It suffices to connect either a high voltage or a low voltage to the pin to make the arduino react.
3) as said in 2), almost no current flows into the arduino's input. The arduino acts upon the voltage alone.
But in case you have another circuit with low input resistance, the current will flow either to ground or to Vcc, depending on the magnitude of the input voltage.
4) in "your" circuit, if pin 7 had low resistance, the current would flow from the battery plus through the switch into pin 7 and back to ground.

Please don't take me for being unfriendly, but may I remark that from my point of view you're putting the cart before the horse? The arduino is an advanced piece of semiconductor with hundreds of thousands of transistors. Getting to know basic electronics needs basic circuits. Start with a battery, a switch and a lamp and try to understand how they are connected and what happens within this simple circuit. Then get to know the basic laws (Ohm's law, Kirchhoff's laws etc.). I suggest you get you hands on an elementary hobbyist book or electronics starter kit.