Help understanding power flow and and ground?

[FogSpirit]

Hello,

I have had an Arduino Uno for some time, but just recently found time after graduation to seriously look at it and try to understand it at an intuitive level rather than just powering through as many of the pre-made schematics that came with my booklet to see what it can do.

I've taken Electricity and Magnetism as a Physics II course, and so I understand the basic principles behind the forces at play. Maybe I could figure out maths issues once they arise. However, I'm not sure I'm understanding the flow of power in a simple setup wherein a single LED is lit. See the attachment for an example of this (Hopefully I made this image correctly! It does light up in person!).

First I want to say that I think I understand the general concept of a ground (at least from an elementary standpoint) and the complications between the ground on the arduino uno board and an "earth" such as described at here. Additionally, I understand (although I am not at that point yet) that there are complications that may arise when one introduces additional power sources into the equation such as is discussed here.

What I'm really having trouble is understanding a few concepts regarding to the flow of electricity and how the board itself is managing current.

First of all, in the schematic I've included-- I've uploaded an image of the corresponding board as well -- I don't understand the role of the wire from the 5V to the + strip. I note that the LED still blinks if this is removed, and indeed there is nothing connecting the + strip (Now with 5V input I assume) to the components of interest. I know that there is a standard setup that includes this input-to-(+) and GND-to-(-) paradigm. Is it simply to introduce this idea before it is really necessary?

Secondly, I'm not sure how the ground actually DOES its job on this board. Here's what I imagine generally; I'm not an electrician and have no understanding of how electricity works within a housing unit, so I'm sorry if explanations based on that didn't get through to me!

1. Power is coming through the USB-B (that chonky printer connection to computer).
2. Arduino board recognises this, and uses this to power the 3V and 5V and all the digital pins (although I'm not sure what voltage might be associated with those).
3. The power/electrical current from the pins is transferred from the pin to (in this schematic) Row 10.
4. Then it accesses the LED anode (+), passes through, lighting up the light when the program says to turn it on/allow the pin to be HIGH.
5. The current passes through the resistor.
   5a. The resistor is not providing a solid block that only slows current AT that location, but instead slows the flow rather uniformly, which explains why it can be placed before or after the LED in the circuit (the current is the same before the LED and after the resistor, if I remember/read correctly.
6. The current then passes from the resistor into the - strip.

Here's where I get confused. I know that the current has to go somewhere or it won't move as electrons won't be moving (the conventional flow of current and electron flow discussions set aside). This is why the circuit cannot end there. No flow, no light.

There is the wire allowing the path to travel from the (-) to the GND on the board. However, what happens after this? What allows the current to continue? Where does that current go from there that allows the system to not experience the same issue? I was thinking of two possible answers, but I'm not sure what is correct.

1: The electrical current is somehow sent to a negligible place (sorry for terminology) similarly to what might be there case in a ground literally embedded in the ground (or maybe when connected to a case around the board or something? I don't know what capability the Arduino Uno has to do this, and assuming it's put somewhere, wouldn't that place just "back up" and prevent the flow of current? Or perhaps there's some sort of capacitor or something. I think this whole answer is maybe unlikely?

2: Somehow the electrical current is being sent back into the overall "power circuit" of the board, which is powering the 3V, 5V, digital pins, etc. In which case, the only way for this sort of setup to NOT disrupt the Voltage of those outputs would be to have (at least in the case of the 3V and 5V outputs; I have no idea what the digital pins are doing) the voltage regulators in place AFTER they get their power from the underlying "power circuit" of the Arduino board.

I am ready to move on past lighting up single and multiple LED, etc., but want to fully understand how the overall circuit is actually working. I hope it's obvious that I've been trying to understand and research answers to this, but am missing some knowledge/insight. I keep finding endless information on why common grounds are important for multiple power sources and how the ground is a reference point for the other Voltages in the system.

Thank you, I hope to learn more and truly find out what this device is capable of! I hope this is a suitable location for this elementary question!

[slipstick]

I think you need to go back to the basic idea of an electrical circuit. Current must flow all the way round a complete circuit. That's why it's called a circuit.

Without getting into electrons or holes as the carriers, any current that flows out of the positive power supply/battery terminal will go through whatever the load is, an LED, a motor, lots of LEDs etc and (this is the critical part) then it has to flow back to the negative terminal. In an Arduino and a lot of other electronics the GND terminal is the thing that is connected to the power supply/battery negative.

Steve

[system]

I hate to oversimplify things but all of your questions would be answered if you just study Ohm's Law.

FYI-which direction does current flow ? - Google Search

(the word "convention" is used because everyone simply agreed to that viewpoint. It is essentially irrelevant
what the electrons are doing since we are not physicists. If it weren't for the fact that sometimes we want
to know what is attracted to what and why, we probably wouldn't care which direction it flowed.
We need to have a reference point for everything and for us it is polarity.(+/-)(Vcc/GND) etc...
After that it's Ohm's Law for linear devices and device electrical properties for nonlinear device.
When you buy a new TV you read the user manual. When we shop for ICs we read the datasheet.

[slipstick]

So there you have the gospel according to raschemmel. When you've been an Electronics Engineering Technician for 39 years you'll understand all this. And there's no point asking question about electronics until after you've learnt electronics.

Of course you could just ignore him and feel free to try to learn things from here by asking questions. Many of us are happy to help.

Steve

[FogSpirit]

Hello slipstick and raschemmel,

Thanks for your replies. I think by originally setting out to reply to you each, the process helped me think about things differently.

Is it correct, then, that the power source is not only responsible for establishing the voltage (such as the 3V, 5V, and pins) as well as keeping the GND as a GND (or -)? The power source is connected to both of these, keeping the GND as a GND and mediating the flow of current between the GND and the pins? Maybe now I don't know how the power supply works. I will look more into that.

I think I thought that the power supply was just providing the voltage without interacting to keep the GND. Is this incorrect?

I am sure this seems very simple to you; I never really was taught how power sources work in circuits. :-[

Thanks again for your input!

And for future reference, I understand why there's a convention. My point in mentioning it wasn't at all to have it explained, but the opposite. I didn't want my terminology to be a reason to focus on that point. If my questions do not make sense to you, I appreciate your opinion, but I'm not sure what you truly want me to take from your comment other than that you're much more intelligent than I am on a topic I have not studied in a practical sense. I never said I wanted to you withold the facts, etc. I was asking for help in understanding what I was missing, not to be told that I was missing information. Obviously I knew that.

[system]

@Slipstick,

I think referring the OP to Ohm's Law is apprpriate
in this case, and helpful. You're welcome to address
each and every one of his questions (which, I might
add , you somehow neglected to do.) How exactly
did you help the OP ? It seems to me that we both
said the same thing in different ways. As I recall I went to great trouble to explain current flow
convention and polarity. Did you address his questions ?

Of course you could just ignore him and feel free to try to learn things from here by asking questions. Many of us are happy to help.

So your post is helping and mine is not ? How does that work ? Would I be posting if I didn't want to help ?

I think I thought that the power supply was just providing the voltage without interacting to keep the GND. Is this incorrect?

Think of it like water flowing from a tank through the pipes to your faucet but instead of going into a sewer
it is recirculating back to the tank.

While that is not a best analogy, I'm not the first to use it.
As far as my comments about your questions, do you think it is best not to tell a newbie when their
question does not make sense ? I am interested in your feedback on that .

I'm not sure what you truly want me to take from your comment

I think I made it very clear that I want you to study Ohm's Law. Was that not clear ?

I was asking for help in understanding what I was missing, not to be told that I was missing information

Was it not clear that what you were missing was Ohm's Law ?
I don't know how else to say that. You made no indication that you understood that or plan to
study it. I'm actually confused because I frankly don't understand how you can understand what
you are missing if you don't learn Ohm's Law. How can you be told what you are missing without
telling you what you are missing ? (ie: Ohm's Law) Can you explain that for me ?

[FogSpirit]

Perhaps there is some sort of misunderstanding or language barrier, I am not sure. I think that assuming that someone expects to have to facts hidden/subdued is condescending. You could have simply said what you thought didn't make sense. And if you thought that wasn't worth your time, then I think it wouldn't hurt to wait for someone who doesn't mind engaging in a conversation with someone at my level and leave your talents for those who can ask questions in ways you prefer.

Also, you act like I had 100 questions. I had 2 main questions, which were

1: Why is the wire from the 5V to the + necessary in this circuit, and
2: How does the circuit connect from the GND to the digital pin.

The other questions and paragraphs were my attempts to think it through and show that I was trying to figure it out myself. I said that option 1 for the answer to question 2 was not likely. Thus, the individual questions within that were rhetorical.

I think that option 2 for that second question is essentially what I'm being suggested is the correct answer, but I couldn't phrase it a preferred way because I need to learn about how power sources create voltage differences in the Arduino.

I appreciate your effort in trying to help. I think, however, there was a disconnect between intent to help and execution, which is why it was responded to as such. Just because you say your intent isn't to be hostile and condescending doesn't make it fact... whether I hide that fact or not.

I think the attached image I found (AFTER realizing what my confusion was about) helps me understand how the Arduino does this. Although I'm not familiar with all of the schematics/symbols involved, it shows me how the GND and IN and OUT are connected in a way that keeps the voltage difference between the GND and the pins; I had trouble finding this information prior and didn't know where to look to learn more.

And to think Arduino community is advertised as positive and educational. :')

[system]

I'm glad that schematic helped. You can learn the symbols here:

At some point you will need to learn Ohm's Law, since it is necessary to do calculations of voltage, current and resistance. (and power dissipation of a resistor or other load)

Sorry about the misunderstanding. I apologize if you took offense.

[CrossRoads]

Kirchoff's Law explains it better:
Kirchhoff's Loop Rule Formula
In any "loop" of a closed circuit, there can be any number of circuit elements, such as batteries and resistors. The sum of the voltage differences across all of these circuit elements must be zero. This is known as Kirchhoff's Loop Rule. Voltage differences are measured in Volts (V). When the current I in the loop is given in Amperes (A) and resistance of circuit elements is given in Ohms (Ω), the voltage difference across a resistor can be found using the formula .

In the example with the LED and resistor, what is not pictured is the transistor that connects the LED anode to +5V (from the USB connector) to start the current flow, while Gnd is just the Gnd plane of the board that connects back to the Gnd of the USB connector.
So in this case, the loop is the +5 thru '328P output transistor, thru the LED, thru the resistor, to Gnd.
Mathematically, you can determine the LED current, making some assumptions about the parts: The transistor has a resistance, unstated, but the datasheet also says it has a voltage drop, of about 0.8V, with VCC = 5V and an IO pin load of <=20mA. The LED will have say 2.2V across it when turned on (typical for Red). The current limit resistor is 330 ohm.
So: (5V - 0.8V - 2.2V)/330 ohm = 6mA. With that low of a load, the 0.8V will be less, say 0.4V.
Then (5V - 0.4V - 2.2V)/330 = 7.2mA.

Others have done testing that suggests the Rds of the output transistors is about 40 ohm.
Using that: (5V - 2.2V)/(330 ohm + 40 ohm) = 7.5mA

If there are multiple loads, LED or otherwise, then each circuit can be similarly analysed. The current for each load can then be added up.
There will be a couple of limits - the board has a 500mA current limit polyfuse (resettable after it 'melts' and cools down) between the USB connector and the board, the Absolute max 200mA limit of the '328P VCC and Gnd pins (see the datasheet), the Absolute max current limit per groups of pins (see the datasheet).
And then Time is involved - will all the loads be turned on at the same time, for the same duration?

[MarkT]

FogSpirit:
First of all, in the schematic I've included-- I've uploaded an image of the corresponding board as well -- I don't understand the role of the wire from the 5V to the + strip. I note that the LED still blinks if this is removed, and indeed there is nothing connecting the + strip (Now with 5V input I assume) to the components of interest. I know that there is a standard setup that includes this input-to-(+) and GND-to-(-) paradigm. Is it simply to introduce this idea before it is really necessary?

Don't really know what you mean by this. In modern logic circuits the -ve supply is ground. This is an accident of history, older logic families exist which have +ve rail as ground. Today you can assume the supply is +ve voltage w.r.t. ground. Ground is purely a convention, the electrons know nothing about it.

1. Power is coming through the USB-B (that chonky printer connection to computer).
2. Arduino board recognises this, and uses this to power the 3V and 5V and all the digital pins (although I'm not sure what voltage might be associated with those).
3. The power/electrical current from the pins is transferred from the pin to (in this schematic) Row 10.
4. Then it accesses the LED anode (+), passes through, lighting up the light when the program says to turn it on/allow the pin to be HIGH.
5. The current passes through the resistor.
   5a. The resistor is not providing a solid block that only slows current AT that location, but instead slows the flow rather uniformly, which explains why it can be placed before or after the LED in the circuit (the current is the same before the LED and after the resistor, if I remember/read correctly.
6. The current then passes from the resistor into the - strip.

Yup - current is the same all the way round a circuit because charge is conserved.

Here's where I get confused. I know that the current has to go somewhere or it won't move as electrons won't be moving (the conventional flow of current and electron flow discussions set aside). This is why the circuit cannot end there. No flow, no light.

There is the wire allowing the path to travel from the (-) to the GND on the board.

Yes, without that there is no circuit.

However, what happens after this? What allows the current to continue? Where does that current go from there that allows the system to not experience the same issue? I was thinking of two possible answers, but I'm not sure what is correct.

current returns to the power supply whence it came. There is a loop of current all the way round, in the power supply power is given to the circuit, in the load power is taken from the circuit (turning to light and heat here)

[FogSpirit]

My original reply was eaten by a bad gateway, but I realised through some searching that the Arduino official starter kit is much more thorough than the Vilros kit that I ordered. I found the official kit booklet that I think will help. I will try to find datasheet information about the Uno and specific parts as well...

CrossRoads, thank you for your very patient and polite explanation. Kirchoff's Law is very sensible now, and I think the maths you've outlined will help me think more critically about the components of my circuits as I start to add more components to them.

Don't really know what you mean by this. In modern logic circuits the -ve supply is ground. This is an accident of history, older logic families exist which have +ve rail as ground. Today you can assume the supply is +ve voltage w.r.t. ground. Ground is purely a convention, the electrons know nothing about it.

MarkT, I think the breadboard image on my original post might help you understand what I am trying (but likely failing) to ask regarding this. Namely, there is a wire from the 5V on the left of the board to the (+) strip of the breadbox. This strip isn't even being used in this circuit, and indeed the LED lights up fine without that connection. I got this design from the Vilros kit. I was just confused why it was there since it wasn't being used.

(...) while Gnd is just the Gnd plane of the board that connects back to the Gnd of the USB connector.

current returns to the power supply whence it came. There is a loop of current all the way round, in the power supply power is given to the circuit, in the load power is taken from the circuit (turning to light and heat here)

I think these two statements are answering the heart of what was truly confusing for me, which was how the breadboard/power supply was creating a complete circuit between the GND and the power-providing pins (regardless of whether that is the 5V/3V pins or the digital pins). The complicated schematic I attached in a reply indicated that the power supply was mediating this relationship and maintaining the potential difference between the GND/power supply, but I wasn't able to find/see this information from my kit until I got some replies here; I wasn't even sure what to google/look up because I kept getting answers related to complications from multiple power supplies. I think this may be in a datasheet/more detailed information of the arduino board and I will look for that and be sure to understand the basics of what isn't under my control! I hope that makes sense.

Thanks for your help everyone!

[system]

Namely, there is a wire from the 5V on the left of the board to the (+) strip of the breadbox. This strip isn't even being used in this circuit, and indeed the LED lights up fine without that connection. I got this design from the Vilros kit. I was just confused why it was there since it wasn't being used

The led circuit is connected to a digital output that provides the positive voltage (hence no need for the +54v
strip you are referring to that has nothing connected to it. As already explained , by several posters I think,
the closed loop for the led is from GND strip to which the resistor is connected , through the resistor, through the led and back to the digital output. It is not really relevant whether the current is "going from the 'output' (which suggests something coming 'out' of it) to the led, or whether the current is going from the GND to the output (remember electrons are negatively charged). In general we don't concern ourselves with
this detail because it makes no difference to us for us to use the circuitry. We only care that there is in fact a closed loop, which , in this case , does NOT involve the +5V pin you keep referring to on the left side. As
long as the loop is closed , the current will flow, (which direction ? who cares ?).

If you look at the arduino UNO schematic you'll that Vcc (pin-4) is tied to +5V and if you look up in the upper right corner you'll
see that +5V comes from the onboard 5V regulator chip. if you look at the right side of the ATMega328 chip you'll see that pin you used (D10) is in the upper right with the number "10" next to it. If you
trace it back to the chip you'll see it is label 'PB2'. That pin (16) on the ATMega3287 chip is where the
circuit 'enters/exits' the chip and since that's a positive voltage , it goes through the Vcc pin (4) and on
to the 5V regulator. (GND of course goes from the ATMega328 pin-3 to
the regulator GND pin. (different versions of the UNO use different chips. Some use the 28 pin DIP and
some have the SMD flat pack so pin numbers will vary from version to version).
And that's your complete loop

I hope that helps.

Official UNO R3 Schematic

[FogSpirit]

If you look at the arduino UNO schematic you'll that Vcc (pin-4) is tied to +5V and if you look up in the upper right corner you'll
see that +5V comes from the onboard 5V regulator chip. if you look at the right side of the ATMega328 chip you'll see that pin you used (D10) is in the upper right with the number "10" next to it. If you
trace it back to the chip you'll see it is label 'PB2'. That pin (16) on the ATMega3287 chip is where the
circuit 'enters/exits' the chip and since that's a positive voltage , it goes through the Vcc pin (4) and on
to the 5V regulator. (GND of course goes from the ATMega328 pin-3 to
the regulator GND pin. (different versions of the UNO use different chips. Some use the 28 pin DIP and
some have the SMD flat pack so pin numbers will vary from version to version).
And that's your complete loop

I hope that helps.

Official UNO R3 Schematic

This is perfectly helpful! I took a long time to reply because I really wanted to delve into this and understand this! In terms of what I am looking for to understand this system better, I think this information, paired with this page and this page really help to close the gap between my previous state of "maybe things are connected?" to now "I can see how the power and ground are connected to form the closed circuit!"

Not only was that able to answer my question, it's really satisfying to see the beauty of connections in multi-component technology like this. I definitely benefitted from the second link telling me that Voltage Nodes were connected. Before, I was a bit lost on which 5V was connected to which 5V all over the place, but knowing that they were all connected cleared things up!

This is likely juvenile/naive of me, but it seems like it would be interesting to make a larger-scale version of some of these individual components. For example, make a larger-scale 3.3V regulator and see if it regulates as expected.

At any rate, I think I understand everything well enough to continue experimenting with the parts of the Arduino Uno I can manipulate.

Thanks again to everyone, really, for helping me look at this in a lot more detail than I was able to previously!

[system]

Just for the record, line 1 of Reply#7 is this link I posted. I left it to you pick something on that page.

This is likely juvenile/naive of me, but it seems like it would be interesting to make a larger-scale version of some of these individual components. For example, make a larger-scale 3.3V regulator and see if it regulates as expected.

Have at it:

3.3V Foldback Current Limiting Voltage Regulator

3.3 uH axial inductor

PDS540 SCHOTTKY DIODE

LT3976 40V, 5A, 2MHz Step-Down Switching Regulator with 3.3μA Quiescent Current

Sorry, I couldn't find a UDD Package/24-Lead Plastic QFN (3mm×5mm) breakout board, although they
must exist.

I think you should start simple with an op amp circuit , or just transistors and resistors.

[DVDdoug]

This is likely juvenile/naive of me, but it seems like it would be interesting to make a larger-scale version of some of these individual components. For example, make a larger-scale 3.3V regulator and see if it regulates as expected.

Do you mean "from scratch", using transistors, resistors, and diodes, etc?

You can build a fairly simple regulator around a Zener diode for the voltage reference, but a voltage regulator chip may have hundreds of transistors inside. In the earlier days of integrated circuits the manufactures would publish an actual schematic, but now you usually just get block diagrams or simplified "equivalent circuits", etc.

I assume a regulator chip is similar to an op-amp with high gain and negative feedback (AKA "corrective feedback"). You could build your own regulator around an op-amp with an added transistor or MOSFET to boost the output-current capability. If you understand how op-amps work (including the negative feedback) you'll understand how to do that. (Again, op-amps may contain hundreds of transistors.)

i.e. With an op-amp you can make a fixed-gain amplifier and then connect a Zener (or other voltage reference) to the input to get a known-output. But, "normal" IC op-amps are not designed to put-out significant current.

[MarkT]

Opamps contain a few dozen transistors at most, more transistors than that can't be useful, its just an amplifier,
typically 2, 3 or 4-stage, which some current mirrors to set up the biasing and some protection circuitry. Any more stages and it would be impossible to make the amp stable in heavy-feedback configuration, 3-stage is common, differential pair, voltage-gain, current-buffer.

The simplest opamp circuit is 2-stage, a differential pair driving a follower. Its not great, but it is only 3 or 4 transistors, noone sells anything that crude.

[Paul_B]

raschemmel:
Think of it like water flowing from a tank through the pipes to your faucet but instead of going into a sewer
it is recirculating back to the tank.

It occurs to me that analogy is quite appropriate if you simply consider the sky - the atmosphere - as your battery.

MarkT:
This is an accident of history, older logic families exist which have +ve rail as ground. Today you can assume the supply is +ve voltage w.r.t. ground.

Not so much an accident of history as a consequence of the way dopants function in semiconductor manufacture and earlier, simply that electrons are negatively charged because electrons were the carrier in thermionic valves.

This makes NPN transistors more practical and N-channel FETs which therefore use a negative "ground" reference.

Telephone systems use positive ground.