# conventional flow, electron flow, and the LED flasher

#### ThirtyWest

Joined Jul 15, 2017
150
I understand the difference between conv and elec flow. The part that is confusing, for someone learning to read schematics, is which is which on the paper?

I find many pictures showing a circuit with no notation as to the flow. And by no notation, let me include that spotting the diodes doesn't help me. I see resistors on one side or the other of something (see the LED flasher that is akin to "hello world" in programming") but I can't find a single one that truly describes step by step how things are playing out. This leads me to think it's down to me not understanding the notation they've used.

So, I know in a simple circuit the direction isn't of consequence; however, when diodes are present it is....isn't it?

Am I to just "know" (which I do) that the + and - on the power source are just printed in conventional and should really say - and +. Or, such as the case with the flasher, should I accept that the printings are intentional and the flow really goes the other way?

Last question: In order to have something concrete from which to extrapolate and compare, can someone please explain the need for the "other" resistor in the simple LED flasher and the order of events that occur in the current? I say "other" because if I say 1st or 2nd resistor it sounds like I'm implying a specific direction of flow, and at this point I haven't an ohm of a clue.

patience and thanks as always.

#### dl324

Joined Mar 30, 2015
15,511
So, I know in a simple circuit the direction isn't of consequence; however, when diodes are present it is....isn't it?
Current direction is always important.

When you're not familiar with a circuit, you might assume wrong current directions. If that happens, you just reverse direction.

#### profbuxton

Joined Feb 21, 2014
419
As far as you are concerned current flow is ALWAYS taken to be from positive+ to negative-. Markings on components are meant to reflect that.
If components on drawing have no polarity indication (resistors) then it don't matter.
If you want to follow current flow start from the positive terminal of the supply and follow to the negative terminal.Active components(transistors,diode etc) need to be connected correctly, also polarised capacitors.
PS some times components are deliberately connected in the opposite direction(zener diodes, suppression diodes)..

Joined Jan 15, 2015
7,148
My opinion and just my opinion. It doesn't matter.Initially in my earliest classes I learned Negative to Positive and later learned Positive to Negative. During endless formulas and a 45 plus year career it never mattered. Maybe to a few pedantic types it matters but I never ran into a schematic or drawing or even a design where it mattered. Possibly in advanced physics but other than that I would not worry about it. Diodes? Initially I was taught current flowed into the arrow and later that the arrow pointed in the direction of current flow. If I place a current limiting resistor in series with a LED does it matter if I place it on the anode or cathode side? Show me a schematic where it matters? This subject gets beat to death endlessly in every electrical/electronic forum out there and argued to no end. Learn both theories and then don't worry about it as I doubt you will ever have an instance where it matters. Again, just my opinion on the subject.

Ron

#### WBahn

Joined Mar 31, 2012
28,192
You've got a few points of confusion here. I'll try to deal with them as best as I can separate them.

I understand the difference between conv and elec flow. The part that is confusing, for someone learning to read schematics, is which is which on the paper?
There is almost always no indication of conventional versus electron flow on a schematic. It is a convention used in the analysis of the circuit and thus is assigned by the person performing the analysis. About the only time an indication would be shown on a schematic (that is not annotated for analysis) is if it is important to identify the current in some branch as part of the function of the circuit, such as a current-sense circuit or current-controlled circuit, or as part of some documentation of the circuit.

I find many pictures showing a circuit with no notation as to the flow. And by no notation, let me include that spotting the diodes doesn't help me. I see resistors on one side or the other of something (see the LED flasher that is akin to "hello world" in programming") but I can't find a single one that truly describes step by step how things are playing out. This leads me to think it's down to me not understanding the notation they've used.
Which side of an LED (or other component) a resistor is placed on has no bearing on whether conventional flow or electron flow is being used. These two concepts analysis choices and have no bearing on the physical operation of the circuit. There may or may not be reasons to place series components in a specific order (there often is, but it is sometimes arbitrary) but those reasons have nothing to do with what direction someone performing an analysis chooses to assign their current directions (except, perhaps, by coincidence and convenience based on how the designer visualized things in their mind as they are designing the circuit).

So, I know in a simple circuit the direction isn't of consequence; however, when diodes are present it is....isn't it?
Not which direction we assign to our current designations. The circuit neither knows nor cares about that. You can use either, you just need to be consistent. Unfortunately, electron flow adherents are almost never consistent and, as a result, have to implicitly invoke magical, mystery, minus signs throughout their work.

Am I to just "know" (which I do) that the + and - on the power source are just printed in conventional and should really say - and +. Or, such as the case with the flasher, should I accept that the printings are intentional and the flow really goes the other way?
The + and - on the power source (assuming a voltage source, which is usually the case) have nothing to do with conventional or electron flow. Adherents to both conventions are in universal agreement regarding the polarity of VOLTAGE assignments -- it is CURRENT directions that differ. In fact, this is why the electron flow crowd has to use so many magical, mystery, minus signs because they want to pretend the CHARGE flows in the direction of the electrons but don't realize that this requires that the polarity of the voltages on everything must be inverted in order to be consistent -- hence the liberal use of magical, mystery, minus signs to make things work out.

Last question: In order to have something concrete from which to extrapolate and compare, can someone please explain the need for the "other" resistor in the simple LED flasher and the order of events that occur in the current? I say "other" because if I say 1st or 2nd resistor it sounds like I'm implying a specific direction of flow, and at this point I haven't an ohm of a clue.
There are probably hundreds of LED flasher circuits out there. How can we possibly know which resistor is the "other" resistor unless you provide a schematic?

But, as already noted, it has absolutely nothing to do with whether you choose to use conventional or electron current.

#### WBahn

Joined Mar 31, 2012
28,192
My opinion and just my opinion. It doesn't matter.Initially in my earliest classes I learned Negative to Positive and later learned Positive to Negative. During endless formulas and a 45 plus year career it never mattered. Maybe to a few pedantic types it matters but I never ran into a schematic or drawing or even a design where it mattered. Possibly in advanced physics but other than that I would not worry about it. Diodes? Initially I was taught current flowed into the arrow and later that the arrow pointed in the direction of current flow. If I place a current limiting resistor in series with a LED does it matter if I place it on the anode or cathode side? Show me a schematic where it matters? This subject gets beat to death endlessly in every electrical/electronic forum out there and argued to no end. Learn both theories and then don't worry about it as I doubt you will ever have an instance where it matters. Again, just my opinion on the subject.

Ron
Generally agree. There are extremely few instances that can't be perfectly well analyzed/designed either way. There are instance where the nature of the charge carrier matters in terms of whether the device will operate, such as vacuum tubes and ion beams, but the analysis is perfectly valid either way whether the actual operation is a positive charge carrier moving one direction or a negative charge carrier moving the other.

The only instance I am really aware of where it matters (and I'm sure there are others) are Hall Effect devices. Here the polarity of the charge carrier is fundamental -- a positive charge carrier moving one direction and a negative charge carrier moving the other direction behave fundamentally different due to the magnetic interactions involved.

#### crutschow

Joined Mar 14, 2008
31,580
To summarize, there is absolutely no reason, on a typical schematic, to indicate whether the designer used current flow or electron flow in doing the design because the schematic is drawn exactly the same either way.
All components are drawn the same whether you consider current flow or electron flow, no exceptions.

#### ThirtyWest

Joined Jul 15, 2017
150
I'm still digesting your posts. And sorry, I forgot to include the link to the thing in question.

http://cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html

I'm trying to find the one that had a large arrow pointing from the voltage source, and it was heading in the "-" direction--even though that went against the direction of the diode symbol. It seems like it would matter for an LED though (which side the resistor is put on ) since you could burn one out if the current isn't resisted, correct?

I mean, if you skip the whole flasher discussion and just go with an LED, resistor, and a battery, it's not going to matter?

Understanding the flow and how it 'flows' here is where i was going. Again, thank you for the contributions. I'm home now and slowly re-reading the posts above.

Last edited:

#### bertus

Joined Apr 5, 2008
22,133
Hello,

The posted link is to the website of our moderator @DickCappels .
He might visit this thread and give you an explanation of the circuit.

Bertus

#### ThirtyWest

Joined Jul 15, 2017
150
Hello,

The posted link is to the website of our moderator @DickCappels .
He might visit this thread and give you an explanation of the circuit.

Bertus
Oh that could help. Thanks.

#### crutschow

Joined Mar 14, 2008
31,580
It seems like it would matter for an LED though (which side the resistor is put on )
No.
The LED and the resistor are in series and each will drop the appropriate amount of voltage to equal the applied battery voltage.
It makes absolutely no difference on which side of the LED the resistor voltage drop occurs, it still leaves the same amount of voltage drop across, and current through, the LED.
How could it be different for the two configurations?

#### ThirtyWest

Joined Jul 15, 2017
150
well, if I knew outside of intuition I would be posting it

I assumed it had to do with the flow direction--which was the impetus for this thread. I suppose I was clinging too long to the water analogy--that there was a beginning and end to the flow of current. Not the case here in this DC then; the flow (in the case of the resistor after the LED) going to the LED is already reduced by the resistor prior to the "-" end of the battery.

Last edited:

#### WBahn

Joined Mar 31, 2012
28,192
Understanding the flow and how it 'flows' here is where i was going. Again, thank you for the contributions. I'm home now and slowly re-reading the posts above.[/QUOTE]
well, if I knew outside of intuition I would be posting it

I assumed it had to do with the flow direction--which was the impetus for this thread. I suppose I was clinging to long to the water analogy--that there was a beginning and end to the flow of current. Not the case here in this DC then; the flow (in the case of the resistor after the LED) going to the LED is already reduced by the resistor prior to the "-" end of the battery.
You are trying to think in terms of something happening to the current before it gets to the LED. The current in the resistor and the LED are the same. Neither the resistor nor the LED has the faintest sense regarding what is happening elsewhere in the circuit. Their behavior is characterized by the current through them and the voltage difference across them. The sum of the voltage drops across resistor and the LED is the same regardless of which comes "first", since addition is commutative. Hence the order doesn't matter.

Joined Jan 15, 2015
7,148
Here is Dick's little circuit:

With the circuit Dick provides an explanation oft the circuit's operation:
In this implementation, a common NPN transistor is used. In the circuit, a 1k resistor charged the 330 uf capacitor until the voltage became large enough to get the emitter-base junction to avalanche. In the oscilloscope image, it can be seen that the peak voltage (yellow trace) was a little bit less than 9 volts. At this point transistor turned on quickly and partially discharged the 330 uf capacitor through the LED and the 100 Ohm current limiting resistor. The current waveform, which is the voltage drop across the 100 Ohm resistor, is shown in the blue trace on the scope image. Peak current was 26 milliamps, and the transistor continued to discharge the capacitor until conduction suddenly ceased at 6 milliamps (Many thanks to Luke in Australia for pointing out the correct current). After the transistor stopped conducting, the capacitor began charging again, thus starting a new cycle.
At no point in the circuit is there a need to follow the current flow that I can see. The circuit operation is pretty straight forward and Dick explains it well. Don't let this current flow direction thing start getting to you.

Ron

#### Tonyr1084

Joined Sep 24, 2015
7,251
I think the simplest way to describe how the circuit is working is that the capacitor charges up through the 1KΩ resistor. When sufficient voltage is reached the 2N2222 transistor (wired backwards with no control current on the base) breaks down and causes the LED to flash on, draining the capacitor sufficiently to fall below the breakdown voltage of the transistor.

Unless instructed to consider "Electron Flow" I'd just follow the conventional flow rule. Someone said there are times when a component is installed backwards by design. The example they used was a zener diode for instance. It has a break down voltage and when it is exceeded the excess voltage is dumped. They're used somewhat as a regulator or can be used as a voltage reference. But here again, we're following conventional flow. At no time should an engineer wish to have a circuit that must be understood in both conventional AND electron flow.

By the way, I've built that circuit. Messed around with different capacitors and resistors. As far as I can remember at no time did the LED go completely out. But it DID flash as advertised.

#### ThirtyWest

Joined Jul 15, 2017
150
Their behavior is characterized by the current through them and the voltage difference across them. The sum of the voltage drops across resistor and the LED is the same regardless of which comes "first", since addition is commutative. Hence the order doesn't matter.
I hear you, but how does the LED not burn out from (say) 12V hitting it before putting a resistor on it? It's hard to imagine

#### Tonyr1084

Joined Sep 24, 2015
7,251
Imagine a hose. Water is running. You kink it in the middle and the flow in the ENTIRE hose slows down because of the kink. Same is true of a resistor in a circuit. Whether the resistor is before or after the LED - doesn't matter. The amount of current that can pass through that circuit is restricted by the resistor.

Here's how you would figure the current in that leg of the circuit (and lets ignore the transistor for this exercise): You have 12 volts. Assuming the LED drops 3 volts that means that there is 9 volts effectively working on that circuit. The LED is supposed to be powered with not more than 10 mA (0.01 amps). 9 volts divided by 0.01 amps equals 900 Ω. So a 900 ohm resistor will limit the current to 10 milli-amps. ALL LED's drop some voltage. Depends on the type, the color and what not. Red LED's typically run around 2.2 volts (called a forward voltage drop) while White and Blue LED's typically run around 3.3 volts forward.

Like I said, if you kink the hose ALL the water slows down. Voltage is best considered as "Electric Pressure". Current is how much electricity is flowing (not voltage). Resistance does just that - RESIST the flow.

#### crutschow

Joined Mar 14, 2008
31,580
I hear you, but how does the LED not burn out from (say) 12V hitting it before putting a resistor on it? It's hard to imagine
The 12V is at one end of the LED but not at the other end, since it's reduced by the resistor after it.
It's the voltage across the LED that is of interest, not the voltage at one terminal.
If you understand that, then everything else will become clear.

#### WBahn

Joined Mar 31, 2012
28,192
I hear you, but how does the LED not burn out from (say) 12V hitting it before putting a resistor on it? It's hard to imagine
You continue to insist on the notion that devices see the voltage at one of their terminals and not voltage differences across their terminals.

The LED does NOT get hit with 12 V. Ever. There is a current limiting resistor. As current flows a largely fixed voltage drop appears across the LED terminals and the remaining voltage appears across the resistor (ignoring the voltage across the transistor for this).

Does this seem more comforting to you as far as keeping the LED away from large voltages since the resistor is between the LED at the negative voltage?

It shouldn't, because these are the EXACT same circuit! The ONLY difference is which node I happened to choose to call 0 V.

I could just as easily have made the bottom node 1000 V and the top node 1012 V and it would still have been the EXACT same circuit.