Conventional current vs electron current - elements in different order - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: October 16, 2019, 06:16:30 pm
content: Why is there no difference if we look at the circuit and say, there is no matter in which direction current flows? The elements of the circuit (resistors, capacitors, inductors) are in a different order. If we just look at the RC filter, there is a significant change, if we put first a capacitor and then a resistor, or first a resistor and then a capacitor.Thanks.

author: IanB
author_rank: Super Contributor
timestamp: October 16, 2019, 06:26:03 pm
content: Because the equations for the system are exactly the same in each case. There are some sign changes in numerical values (positive to negative) but these changes cancel out when you look at the overall signal response of the network.

author: IanB
author_rank: Super Contributor
timestamp: October 16, 2019, 06:29:00 pm
content: Also the elements are in thesameorder in each case. In typical filter network it is drawn so that the input signal comes in on the left and the output signal leaves on the right. The physical arrangement of the circuit doesn't change if you change the sign convention of current.

author: nForce
author_rank: Frequent Contributor
timestamp: October 17, 2019, 12:49:54 pm
content: Sorry, but I don't understand. If we see the schematics of the RC filter:Here if we start at V_in, first we get to the resistor then parallel connected capacitor and finally the V_out. If we change V_in and V_out, then first we get to the capacitor then resistor, and the output.

author: TimFox
author_rank: Super Contributor
timestamp: October 17, 2019, 01:09:24 pm
content: The input and output nodes are labelled intimeorder, i.e., the change at the input causes a change at the output. This has nothing to do with thepolarityof the voltages and currents.

author: schmitt trigger
author_rank: Super Contributor
timestamp: October 17, 2019, 01:37:54 pm
content: Quote from: nForce on October 16, 2019, 06:16:30 pmThe way you are describing it, the first instance is a high-pass filter, and the second is a low-pass filter.Their transfer functions do not change if you utilize conventional vs. electron current for the analysis. As a fact, that is the beauty of Laplace transforms. Once that you are in the s-plane the system you are describing could be electrical, mechanical, thermal.....even biological. The results are the same.I believe this is what you were asking.

author: ejeffrey
author_rank: Super Contributor
timestamp: October 17, 2019, 02:13:23 pm
content: Because electrons are a bit of a red herring. When we describe electric circuits we mostly care about the fields not the charges themselves. The charges establish the boundary conditions for the fields, but we don't really have to think about it. The charge does matter inside devices like semiconductors, vacuum tubes, and electrolytic cells but not within wires or at the circuit level.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 17, 2019, 02:30:19 pm
content: Quote from: nForce on October 17, 2019, 12:49:54 pmCurrent doesn't start out like someone driving from city to city.For current to flow into one terminal, it must simultaneously flow out the other one."Draw a battery across your input terminals with the positive to the top terminal & the negative to the bottom terminal.Imagine Conventional current flowing through the circuit.Draw in the polarities of the voltages across the components (to not get confused by the capacitor charging, imagine you are looking at the circuit during an infinitesmally small stretch of time.)The voltage across R will be more positive at the end connected to the positive batt terminal than the end connected to C.In the same way,the voltage across C will be more positive at the end connected to R, than at the end connected to the negative terminal.Now, imagine Electron flow.The voltage across R will be more positive at the end connected to the positive batt terminal than the end connected to C.In the same way,the voltage across C will be more positive at the end connected to R, than at the end connected to the negative terminal.The effect upon the circuit is exactly the same, whichever convention you use.In the early days, it was assumed that current flowed from positive to negative.A whole lot of theoretical & practical concepts were built on the basis of Conventional current flow, & these were too useful to discard, so were kept for most purposes, along with the assumed direction of current flow.In the heyday of vacuum tubes, when Electron flow was an important concept, Students would take about 20 minutes to understand the fact that there is no difference in circuit behaviour, whichever convention was used, & that it was sometimes easier to use one concept, & sometimes the other.Some people are so protective of Conventional current flow, that the very mention of Electron flow makes them react as if you insulted their mother!

author: Kirill V.
author_rank: Regular Contributor
timestamp: October 17, 2019, 03:25:49 pm
content: It is necessary to clearly distinguish the physical direction of the current and its theoretical direction.The theoretical direction that is used in the calculations does not always coincide with the physical.This should not confuse you if you are an engineer and not a physicist. The work of electrical circuits does not depend on your thoughts and feelings.By the way, engineering methods and models greatly simplify life in contrast to pure physical methods.

author: IanB
author_rank: Super Contributor
timestamp: October 17, 2019, 04:11:54 pm
content: Quote from: nForce on October 17, 2019, 12:49:54 pmBut why would you change V_in and V_out?If you build the circuit you may apply a signal generator at the V_in terminals and you may put an oscilloscope at the V_out terminals. You will not rearrange these physical connections according to how you imagine current flows. The real world doesn't care how you model things. It just continues to do what it does regardless. The input signal comes in on the left and it comes out on the right.

author: rstofer
author_rank: Super Contributor
timestamp: October 17, 2019, 04:25:57 pm
content: Quote from: vk6zgo on October 17, 2019, 02:30:19 pmI don't know if it is 'protective' or just trying to have a common frame of reference. It's nice to know about electron flow, it may even come up when thinking about MOSFETs, but the "Conventional Current Flow" allows us to follow the arrows on transistors and diodes and that seems handy.And, truly, it makes no difference but keeping everybody on the same page seems useful. Particularly in a Beginners forum.It isn't necessary but it is useful when dealing with Kirchhoff's Laws to have a uniform approach to writing the loop and node equations. But even there, nodal equations are often written as though all current are entering a node while knowing full well that there MUST be currents leaving the node. That works well because a negative numerical result will immediately indicate current leaving the node. That seems handy to me! Positive resultant currents are entering the node, negative resultant currents are leaving the node. Perfect!I guess we just need to be flexible in our thinking!

author: Bassman59
author_rank: Super Contributor
timestamp: October 17, 2019, 05:01:18 pm
content: Quote from: nForce on October 16, 2019, 06:16:30 pmYou do not consider the return current in your description.

author: pwlps
author_rank: Frequent Contributor
timestamp: October 18, 2019, 11:40:12 am
content: Quote from: nForce on October 17, 2019, 12:49:54 pmThis is a legitimate question but with all vague explanations given here I understand you are stuckYou forgot one thing here. To define the gain V_out/V_in of a circuit you need a condition on currents (otherwise there is no enough information, for example here we have 1 equation for voltage, one for current, two unknown voltages and two unknown currents, leaving two free variables, you need to fix one to get the voltage ratio). Usually this condition is I_out=0 i.e. unloaded output. If you reverse the circuit the gain will be different because it will now be defined with I_in=0. This is the ONLY reason for the difference when the circuit used in reverse direction. And as stated already in other answers it has nothing to do with the polarity of the signal itself.

author: Nerull
author_rank: Frequent Contributor
timestamp: October 18, 2019, 05:09:04 pm
content: Supply your RC circuit with AC - now current switches direction every half-cycle and there is no single polarity to even argue about, but the gain doesn't change. You don't have to switch all the components around in your model for the negative half of an AC signal - polarity has no effect on where your inputs and outputs are.

author: nForce
author_rank: Frequent Contributor
timestamp: October 19, 2019, 03:04:48 pm
content: Oh, I get it now. The problem I made is I changed the input and output of any kind of circuit. I should just switch the polarity of the battery at the INPUT. But why do we have then reverse voltage protective devices like a diode, if there is no matter in which direction the current flows?

author: IanB
author_rank: Super Contributor
timestamp: October 19, 2019, 03:34:48 pm
content: Quote from: nForce on October 19, 2019, 03:04:48 pmBecause it matters a lot which direction the current flows. Electronic devices are designed to work with a certain polarity of supply. If you change the polarity will not work and may be damaged.When you change the current model from conventional current to electron current you may not make any physical changes to the circuit. You may only change the equations you use to model the circuit in your mind and on paper.It should be clear to you that what you think in your mind makes no difference to how things actually work. It doesn't matter whether I think my car works on internal combustion or whether I think it works with magic pixies. My car continues to work the same either way.

author: Nerull
author_rank: Frequent Contributor
timestamp: October 19, 2019, 04:18:45 pm
content: Quote from: nForce on October 19, 2019, 03:04:48 pmAC was perhaps a bad example, since it involves an "actual" change in current direction and not just a mathematical one.Conventional current has positive charges moving from positive to negative, while electron current has negative charges moving from negative to positive. These are equivalent - both result in a net movement of charge from positive to negative. The direction of *charge* remains the same. When working at a high level, you only care about charge and not the actual charge carriers. It could be electrons, it could be positive ions, it could be a mix of both. It makes no difference. It is equivalent to consider a diode allowing positive charge to move from anode to cathode as it is to consider it allowing negative charge to move from cathode to anode. Unless you're working at the solid state physics level, you don't have to care what the actual charge carriers are doing or what they even are.Consider the water analogy and a one way valve. You could say a one way valve opens when the pressure on high side is more positive than the pressure on the low side. You could also say that a one way valve opens when the pressure on the low side is lower than the pressure on the high side. These are two ways of saying the same thing. Nothing about the valve or the actual flow of water has changed, just where you decide to stick your frame of reference to do the math.