Beginners idiot question - Page 1
Forum section: electronics / beginners

author: Mr D
author_rank: Regular Contributor
timestamp: May 25, 2019, 11:34:59 pm
content: Hi folks,Sorry to do this to you but here goes............I'm messing around with Every Circuit (online circuit simulator), and there's something really basic i'm not getting.Until now i've learnt that to have a valid circuit, you need a voltage source. Let's say a battery.If you connect the plus to the minus terminal, with a resistor in between you've got a valid circuit. It'll do something (heat up the resistor)But not according to Every Circuit (EC). EC demands that you place a "ground" icon in the circuit, and attach it somewhere.But i really don't understand why or what's the purpose.Attached you'll see a printscreen from Every Circuit. It contains 3 circuits. The bottom one is invalid according to EC but i don't understand why.The other 2 have the ground attached at different places in the circuit, and this alters the voltage potential at various places in the circuit. Again, i'm at a loss to understand why.I know this is a super noob-ish question but i'm really baffled by this.Thanks in advance!

author: magic
author_rank: Super Contributor
timestamp: May 25, 2019, 11:42:19 pm
content: The ground symbol defines which point is considered 0V.Every other point's voltage is reported with respect to ground.

author: Dabbot
author_rank: Regular Contributor
timestamp: May 25, 2019, 11:49:08 pm
content: All three are valid circuits. I think the requirement for a ground point, in this instance, is enforced by how Every Circuit performs its calculations.

author: jeremy
author_rank: Super Contributor
timestamp: May 26, 2019, 12:04:01 am
content: Yeah, it’s likely that the solver needs a fixed known voltage of 0V at a node to work.

author: AlfBaz
author_rank: Super Contributor
timestamp: May 26, 2019, 12:18:11 am
content: All spice simulators need all circuit paths to have a dc path to ground (0V).Some however, have default models such as capacitors with a high built in dc resistance.There are circumstances where a particular branch might not have a dc path to ground and generally spice simulators wont converge.In these instances you usually place a very high resistance to ground so that the circuit can be solved

author: Brumby
author_rank: Unknown
timestamp: May 26, 2019, 01:24:21 am
content: Quote from: Mr D on May 25, 2019, 11:34:59 pmThere is your answer. It is a software requirement - not an electrical one. All 3 of these circuits will function inexactlythe same manner in the real world.As for moving the "ground" connection around - this (generally) will make absolutely no difference to the operation of a circuit - as long as there is only ONE such point specified. Moving it about is like moving where you put the black probe of your multimeter ... and that's about it.Where you place a "ground" connection in more complex circuits is, however, sometimes much more important, particularly when it comes to shielding and/or interconnection with other circuits. This tends to NOT influence how a circuit functions - but to prevent/minimise unwanted artefacts from spoiling the show.PS. It was not an idiot question. It was a basic one - but if you don't get the basics nailed down (especially when you are starting out) then the rest of it will never make sense.We've all asked "idiot"/"stupid"/"dumb" questions ... sometimes after playing with electronics for 20 years. This can happen when we are faced with something we haven't encountered before or when we actually go into a circuit in detail.My philosophy is this: The only "dumb" question is the one youdidn't ask.

author: lordvader88
author_rank: Frequent Contributor
timestamp: May 26, 2019, 02:34:00 am
content: like they have said....and on the programing side tho, a program could look at a few components and a battery, and decide to make the negative terminal gnd, because thats how we do it most of the timeSo what u also asking or I'm thinking, is it's also a user interface topic.I sure wish LTSpice was prettier looking, and with more components, like I had another program i used all the time, and it had lots of things like a variable resistor, which takes years to learn in ltspice for me anyway, I'm being lazy tho

author: bitseeker
author_rank: Super Contributor
timestamp: May 26, 2019, 03:33:24 am
content: Quote from: Mr D on May 25, 2019, 11:34:59 pmYour understanding of a basic circuit's requirements is correct. If a ground was required for a real circuit, flashlights and handheld DMMs wouldn't work. Your voltage source with resistor example is the essence of a battery with a light bulb connected to it.A simulator is not the same as real life. In this case, it has an artificial requirement that ground be defined. It could have chosen an arbitrary point itself when one wasn't selected.Don't feel bad about asking such questions. There's nothing wrong with sanity checking.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 26, 2019, 05:46:28 am
content: It's just one of the idiot things simulators do!Another one is that as a default, LTspice shows an ac source as a DC one, then has a note alongside saying what it really is.I tried to like LTspice---- I really did!, but when it couldn't simulate a basic CR network, I gave up.It's a bit like trying to remove cylinder head bolts with a Stilson Wrench------it might work, but it will take 10 times as long, & a lot of scraped knuckles along the way!

author: Mr D
author_rank: Regular Contributor
timestamp: May 26, 2019, 07:56:38 am
content: Thanks folks,I'm still having a hard time wrapping my head around it.Ok, so am i right in thinking that in these circuits / discussions, the word "ground" could be replaced with "the negative terminal of the battery"?But if that's correct, then the circuit in the top right of my image doesn't make sense to me. Because before the current enters the 200ohm resistor, it has the option to go to ground (to the battery neg. pole), which is less resistance........ so why would any current flow through the 200ohmR? (which according to EC it does).Quote from: Brumby on May 26, 2019, 01:24:21 amI suspect the above quote is the key to the answer, but i'm not getting it yet!

author: timelessbeing
author_rank: Frequent Contributor
timestamp: May 26, 2019, 08:44:09 am
content: Quote from: Mr D on May 26, 2019, 07:56:38 amNope. You can attach the positive terminal to ground and it won't make a difference because it's arbitrary.

author: Cnoob
author_rank: Regular Contributor
timestamp: May 26, 2019, 08:53:59 am
content: The ground terminal is the simulator's reference point, and treats it at zero volts potential. Which is why in your top right hand circuityou have plus 833mV on the positive terminal and -166mV on the negative terminal.

author: magic
author_rank: Super Contributor
timestamp: May 26, 2019, 08:58:48 am
content: A battery is a source of voltage, not potential. Voltage is a difference between potentials. A battery creates a difference of 1V (or whatever) between its two terminals.Ground (the rocky stuff underneath you) is a source of absolute potential: it is huge and electrically neutral, any small object connected to ground looses its static charge and becomes neutral too.You can connect the positive terminal to ground and then the negative is at -1V. You can connect a resistor divider to ground and then each terminal is at ±0.5V.Not all circuits are connected to ground. For those that aren't, it's usual to choose some point which is called "ground" and measure all voltages in the circuit with respect to that "ground". The absolute potential of that "ground" with respect to the physical ground can be anything and it doesn't influence the way the circuit works.

author: mikerj
author_rank: Super Contributor
timestamp: May 26, 2019, 10:24:38 am
content: Quote from: vk6zgo on May 26, 2019, 05:46:28 amLTSpice shows a voltage source as a voltage source. You then define what kind of voltage waveform you want it to provide.Quote from: vk6zgo on May 26, 2019, 05:46:28 amThis is a PEBCAK problem, not LTSpice.Quote from: vk6zgo on May 26, 2019, 05:46:28 amIt's like trying to use a tool you haven't yet learned to use.

author: Brumby
author_rank: Unknown
timestamp: May 26, 2019, 03:52:41 pm
content: Quote from: Mr D on May 26, 2019, 07:56:38 amYes - you will see the answer in there, once you "get it".(When you read the following, take it slowly and be comfortable you know what is being said before moving on. If you have a question - ASK! Either I or other members will help you as best we can.)Here we must take a minute to wrap our heads around some basic terminology... We have:1. Ground2. Earth3. Chassis4. Common5. 0V6. Reference point (doubt you will see this)In many cases, these mean EXACTLY THE SAME THING - particularly as far as the circuit under study is concerned. What adds to the confusion is that the two terms "Ground" and "Earth" DO NOT ALWAYS refer to a connection to the soil on which your building is situated (usually through the Earth pin on your mains power outlet).Sometimesthey will, but not always. Quite often, these terms will be used when the term "chassis" would be more descriptive. But even then, you should check to see that there is an explicit connection to a metal framework and/or case before you takethatliterally. (And even if there is, it may not mean anything!)Fun, eh?The next point that you need to understand is that having ONE point connected to a Ground/Earth/Chassis will haveno fundamental effecton the circuit operation. Any such connections will need aclosed loopbefore you need to take them into consideration. A single connection cannot make a loop. As for the "Ground" or "Earth" allowing current to flow, even that requires a loop. In the case of a mains supply, that loop is made up of two parts. The first is where the pole transformer has wires carrying the normal active(s) and neutral running to your property and to the power points around the building. When your equipment is working properly, all the current flows through this system. The second part is made up of another wire from the pole transformer (connected to the neutral) going straight down to a rod buried in the soil below it - and - all the "Earth" pins of building's mains outlets connected back to a common point which then has a similar wire running down to another rod driven into the ground alongside the building.In normal operation, all the current coming "in" on the active will go "out" on the neutral (let's not get caught up on the nature of AC and let's leave 3 phase out of this) - but in a fault condition where the active comes into contact with an earthed part (that is a part that is actually connected to the building's earthing rod), In this case, the circuit loop goes from the pole transformer - through the active to the building - through the wiring to the power point - to the piece of equipment where it then takes a (fault) path to the earthed part. From there, the current goes back through the earth wire of the power cord - back through the earth wiring of the building - down to the earthing rod driven into the ground - andthrough the soilback to the earthing rod underneath the power pole, back up to the transformer neutral.NOTE: This only works when there is direct connection all through. When you have a circuit which has galvanic isolation - such as power from your typical mains transformer secondary - then there is no loop possible through the building earth system ... unless you deliberately connect to the mains side (which is not normally done).Where the terms: "Ground", "Earth", "Chassis", "Common", "0V" and "Reference point" are used, it is very often the case that they are talking about a point/track/conductor/metal part which provides a0V reference point(a place to put the black lead of your multimeter) - and that's all! No actual connections to the soil need be involved. Now, sometimes there will be a connection to a real "Earth" - but if your circuit is isolated on the secondary side of a transformer, a single connection to such an "Earth" cannot be part of a loop and thus will notand can nothave any influence on circuit operation.THUS - in the circuits originally posted, having ONE Earth point noted in two of them, there is no loop path for current to flow, so that Earth point means nothing (as far as circuit operation is involved). This is why the third circuit will operate in exactly the same way!SO - Why does this circuit simulator require an "Earth" point? That's easy - albeit a tad confusing.When we humans look at a circuit, we can pick out a likely spot to use for our 0V reference point - but for a simulator program, that's not necessarily a simple process. It could make a guess - but with a circuit having dozens of connection points, odds are the voltages it would show will be confusing. Certainly, it has enough information without such a point being defined to be able to described circuit operation. It could take a random point and use that - but when it wants to show you voltagesthat will mean something to you, it needs to know "where to place the black lead" (for want of a better phrase). The easy way to do this is to haveyouplace a 0V reference point on the circuit - and the often misappropriated "Earth" point is the natural choice.I hope that has helped ... and not helped you into further confusion.

author: Zero999
author_rank: Super Contributor
timestamp: May 26, 2019, 07:02:15 pm
content: Quote from: mikerj on May 26, 2019, 10:24:38 amYes, it appears to be the case of user error/misunderstanding.The original SPIC didn't have any symbols or graphics. The data was inputted and outputted in text only format, although many users of modern SPICE programs, myself included don't know that much about the internal commands, because it's no longer necessary for day to day usage. No doubt if one needs to use it to create their own models, they'll need a greater understanding of the internal workings.As far as the CR circuit is concerned: you need to tell LTSpice to start the simulation with the input voltage at 0V, otherwise it will calculate the steady state conditions, at start-up. This may seem clumsy but it's often handy to avoid waiting for capacitors to charge.

author: IDEngineer
author_rank: Super Contributor
timestamp: May 26, 2019, 07:22:12 pm
content: Here's another way to look at this question (and thank you for asking it, you're not the first person to scratch their head over this).Unless we're talking about an actualearth"ground" (as discussed by others above), "ground" is usually just another name for "what we all agree to be the reference point". That's it. Just an agreed-upon spot against which all other potentials are compared (measured). And it does not have to be "the negative terminal of the battery" or "the most negative point in the circuit" either... lots of telephone and other rack-mounted equipment runs on -48VDC, for example. What is "ground" then? If you put a meter on it, and measured referenced against the other supply rail, "ground" would be 48VDCpositive. There's also an older logic family, ECL, where the supply was -5.2VDC. Again, "ground" is the more positive terminal of the power supply.My point is that there is nothing special about "ground". It's just one of several agreed-upon names for the common reference point in the circuit. As noted by others, "common" itself is another one.As an analogy: Say you have a fire escape ladder going up the side of a multistory building. At first you're standing on the "ground", and then you start climbing the ladder. When your feet are even with the second story's floor level, you ask residents how high you are. Folks standing on the ground might answer "ten feet". But folks on the second floor might say "You're not high at all, you're at the same level  as me". And folks on the third floor might say "You're not high, you're BELOW my level . You're negative ten feet ". As Einstein might have said, "it's all relative" to your reference point. Since letting each viewer of the schematic pick a different reference point would lead to confusion, the industry has settled on some commonly accepted ways to designate where common/ground is for any particular circuit.As for your specific situation, as others noted it's a software thing. The authors of the software need something to use as their baseline reference point. The electrons in the circuit don't care, but it makes the software's job enormously more straightforward if it can start off with a known reference point. It also makes it easier for YOU to use the software... because otherwise how does the software report results to you? What if its author came from the days of ECL, and decided -5.2VDC was the baseline reference, and expressed all voltages relative to 5.2VDC? Everything would be mathematically correct but you'd be tearing your hair out doing the constant mental math to convert to whatever is YOUR favorite frame of reference.

author: Mr D
author_rank: Regular Contributor
timestamp: May 26, 2019, 09:13:57 pm
content: Many thanks to all so far!So i think, with reference to my original question, i understand why EveryCircuit is doing what it's doing.So could we now focus on ground in general in a circuit?Attached is a new example to highlight some difficulties i still have.In the image we have two circuits.The one on the left is a valid circuit. My understanding is that when you have two or more grounds in a circuit, you need to imagine a virtual or real wire connecting all the grounds. That's why the circuit lets current flow. From the pos. terminal to the neg. terminal, via the ground bus.The circuit on the right is also valid. But what i don't understand is why EC shows the current only flowing through the bottom horizontal wire.If the grounds are connected (as proved by the circuit on the left), then the current should flow just as happily through this ground bus, right?!So i'd expect, all other things being equal, that we'd have 50% of the current flowing through the bottom horizontal wire, and 50% through this real or virtual ground bus!?

author: AlfBaz
author_rank: Super Contributor
timestamp: May 26, 2019, 11:25:14 pm
content: Quote from: Mr D on May 26, 2019, 09:13:57 pmWell no.Both grounds are at the same potential so with no potential difference there is no current flowing.The key thing to take from Brumby's explanation is the "black lead" bit.If you go back to your original circuit and forget about the simulation, if you wanted to measure voltages you would place your multimeters black negative probe somewhere on the circuit and the red positive probe else where.In the simulation, where you place ground is where you want the simulator to place the negative black probe and your mouse pointer then becomes the red positive probe. Essentially all measurements will be with respect to where you placed your ground or more precisely where in your circuit you designated 0VI don't know about this EC simulator you are using but in ltspice you can place a "ground" symbol somewhere on your circuit, left click it and choose a new name for it. If you want it to be ground you simply call it '0' (as in 0V) and as described earlier it becomes the reference point. If you call it something else it becomes a "net" and by placing another net of the same name elsewhere in the circuit you create a connection between these points in the same way your 2 grounds connected together in the first example of your last post

author: Brumby
author_rank: Unknown
timestamp: May 27, 2019, 12:40:03 am
content: Quote from: Mr D on May 26, 2019, 09:13:57 pmAbsolutely correct!QuoteThe key to understanding is in this phrase: "EC shows". This simulator only shows current flowing through the explicit connection becausethat's all it needsto consider in order to describe circuit operation. You could remove either or both of the "Ground" connections and it would still work correctly. As far as circuit operation is concerned, ANY additional parallel conductors should not change the results. Remember, a circuit simulator will be considering any interconnecting wires as "ideal" - that is, zero resistance. Thus, what "EC shows" may not be the absolute truth when it comes to the real world - but it is quite good enough to produce valid results.QuoteYou aren't far off the mark here. In fact, with a small qualification, I will give you a thumbs up!Where you say "all other things being equal", you have hit the nail on the head - but I will go one step further and say what the things are that need to be equal: the resistance of the bottom horizontal wire and the resistance through the ground bus. If these two resistances are equal, then your 50/50 current sharing expectation is 100% correct. This is how it would work in the real world, but in the circuit simulation, these paths are "ideal".If you want to take into consideration the non-ideal realities in a circuit simulation, the simplest way is to include a low value series resistor (that represents the actual resistance of each conductor) in each path. If you wanted to do this properly, you should do some homework so you get the right values, but for an exercise, something like 0.001 ohms might be one example you could try. In some circuits adding such a resistance may actually be necessary for correct simulation - eg. including the ESR of a capacitor in a circuit where that is important.So, once again, your confusion has come about from the processes implemented within the circuit simulator. It usesidealcomponents.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 27, 2019, 01:06:15 am
content: The technically correct answer is:1. The simulation works by iterating matrix inversions.2. If no absolute reference is defined in the equation, the result is a singular matrix.3. "Singular" is a synonym for "non-invertible", i.e., there is no solution.Which is obviously a not-very-helpful answer to an idiot.Expanding on that some more:- Every component represents an equation, a relationship between variables in the system. Yes, this is a very mathy thing; simulators lie entirely in the mostly-abstract domain of applied mathematics.- Every node (set of common connections between components) represents a variable in the system.- To solve for the voltages and currents in the system, we need to know all of these relations. Then we put them all together into a matrix.- A matrix is just a system of equations. You can solve "ax + by = c; dx + ey = f" by algebra (rearranging and substituting), or you can group the variables in a common order and use an algorithm to solve it. If we let A = [a, b; d, e] (i.e., a matrix, in order of rows, then columns), and x = [x; y] (a 1-column matrix is also called a vector), and b = [c; f], we have the matrix equation Ax = b, with the solution x = A^-1 b (read: "x equals A inverse times b"). This is practically a trivial solution (it's essentially division by A), so you can see why it's so attractive.- The hard part is then: writing the matrix A in the first place, and, solving for its inverse.- A is generated from the schematic, itself not the easiest thing, but just a matter of programming. It can also be written out in plain text (a netlist), or, for that matter, by inspection (with some experience, of course!).- The matrix inverse is a PITA to calculate by hand, but it's always done the same way every time -- as long as we can tell a computer how to do this, we're set.So, "singular matrix" is very much like "division by zero". It's telling us that there are no unique solutions to this system -- in practice, this is quite understandable, as an isolated circuit can have any voltage at all, with respect to ground. The difference is, that circuit continues to work regardless of what voltage it's sitting on. This is partly a consequence of the analysis method we've chosen -- we could simply define a node arbitrarily, or rearrange the equations to remove that ambiguity. But it's easy enough to stick in a ground, in the designer's preferred location (normally a node with a lot of common connections), so that's what we usually do.Tim

author: Brumby
author_rank: Unknown
timestamp: May 27, 2019, 05:01:52 am
content: Just noticed this:Quote from: AlfBaz on May 26, 2019, 11:25:14 pmSorry, but that is an absolutely incorrect statement.Current CAN flow through a conductor without any potential difference across it, when it is part of a circuit that has current flowing because of the other circuit elements. In fact, this is the definition of a conductor (in the normal sense).

author: DimitriP
author_rank: Super Contributor
timestamp: May 27, 2019, 06:04:55 am
content: I'd like to swim in this spoonful of water too but my life preserver doesn't fit in it

author: Zero999
author_rank: Super Contributor
timestamp: May 27, 2019, 09:21:16 am
content: Quote from: Brumby on May 27, 2019, 05:01:52 amOnly if it's a superconductor.Simulators always assume the resistance between ground connections is lower than the wires, which also confusingly have zero resistance. As a thought experiment, imagine the connection you've drawn has a very low resistance, say 1pΩ which is 0.000000000001Ω and the resistance between the ground connections is really 0Ω. All of the current will flow between the ground connections.Strictly speaking, this isn't what actually happens. SPICE considers the ground nodes first, so if they're connected together with other wires, then the current in those wires is assumed to be zero.

author: Brumby
author_rank: Unknown
timestamp: May 27, 2019, 10:16:35 am
content: Quote from: Zero999 on May 27, 2019, 09:21:16 amI added that highlighted commentspecificallyto short circuit the superconductor criticism. Let's not make it any harder for the poor Op with pedantic distraction.

author: magic
author_rank: Super Contributor
timestamp: May 27, 2019, 01:19:02 pm
content: Quote from: Zero999 on May 27, 2019, 09:21:16 amOP is asking about simulation, not real cables.Quote from: Zero999 on May 27, 2019, 09:21:16 amMaybe that's what SPICE does but OP sees exactly the opposite behavior.It's irrelevant either way, it's just a simulation of "ideal" components which don't exist so it doesn't need to make sense.

author: ArthurDent
author_rank: Super Contributor
timestamp: May 27, 2019, 02:10:43 pm
content: Mr DQuoteThe replies to this topic are starting to make the question far too complicated. This is like when a young child asks: “where do babies come from?” and the parent goes into a very long and uneasy sex-ed explanation when all the child wants is: “the hospital”.There are lots of circuits that have absolutely no ground, or earth, or whatever you want to call it. An automobile is isolated from ground by the tires, an airplane in the air has no ground connection, and a space station has no connection to ground when in orbit. To check or troubleshoot the circuitry in each of these examples you have to choose a somewhat arbitrary point of reference if you are going to describe the voltages at various points in the circuit to others.If someone asks: “what is the voltage at B?” you have to ask: “with respect to what?” When you ask. above, whether to use “ground” or “the negative terminal of the battery”, both are not the answer, it is "the reference point specified". Check the link below for a better explanation.http://www.tpub.com/neets/book1/chapter3/1-18.htm

author: Macbeth
author_rank: Super Contributor
timestamp: May 27, 2019, 09:34:14 pm
content: OP should just breadboard (who even needs a breadboard for such a simple circuit?) and use the cheapest crappiest DT830 or other meter possible and figure it out for himself.

author: Brumby
author_rank: Unknown
timestamp: May 27, 2019, 11:30:42 pm
content: The Op has not expressed any difficulty with circuit operation - just the way it is being represented by the simulator he is using. His comment about current sharing through the common wire and the ground connections demonstrates he has a reasonable grasp of what to expect in a real circuit - it's just that the simulator isn't showing that:Quote from: Mr D on May 26, 2019, 09:13:57 pm

author: Brumby
author_rank: Unknown
timestamp: May 27, 2019, 11:50:02 pm
content: For the benefit of the Op, maybe it's time to roll this out again.....

author: vk6zgo
author_rank: Super Contributor
timestamp: May 28, 2019, 03:06:53 am
content: Quote from: mikerj on May 26, 2019, 10:24:38 amAn ac voltage source is not, in normal schematics shown with + & - signs----- a DC voltage source is.The very presence or non-presence of the polarity signs defines which it is.Having the default a DC source is misleading.Standby for multiple "Noobs" becoming confused & drawingrealschematics this (incorrect) way.QuoteWhen I learnt Electronics, the designers of LT spice were but a sparkle in their Father's eyes.We had a number of ways to determine the operation & performance of circuits.(1) Use our imagination & our knowledge of theory to visualise how it would work.(2) Work through the mathematics.(3) Steal someone else's design.(4) Build the circuit up & test it in the real world.(5) A combination of the above.If I have to devote a not inconsiderable part of my limited time on this Earth to learning how to use a particular underwhelming "tool", I'd rather use one I know how to use.And, by the way:- "Get off my lawn!"

author: Zero999
author_rank: Super Contributor
timestamp: May 28, 2019, 08:01:58 am
content: Quote from: magic on May 27, 2019, 01:19:02 pmYou're right.LTSpice doesn't do that. The simulator the original poster is using must do it the other way round: calculate the current in connections before the grounds.The simple way to avoid this confusion is to not connect more than one ground symbol to a wire.

author: Zero999
author_rank: Super Contributor
timestamp: May 28, 2019, 04:51:20 pm
content: Quote from: vk6zgo on May 28, 2019, 03:06:53 amAll right, I'll bite.You've missed the point. SPICE doesn't need schematics. They were added later and there is no default voltage source symbol. LTSpice has both AC and DC voltage source symbols. It's easy to choose the correct one and it's not the software developer's fault, if user makes a mistake and clicks on the wrong one. Even if the user has used the wrong symbol, there's a piece of text next to it clearly showing whether it's AC, DC or both (quite often you want a signal with both AC and DC components).QuoteThat's all well and good, until you work on a project containing hundreds of transistors, etched into a single piece of silicon. It becomes too costly to get an IC made, just to test a concept, hence the need for a virtual breadboard, allowing the designer to test it first. If it wasn't for SPICE, you wouldn't be able to type that message and post it, because the Internet wouldn't exist.QuoteThat's fine. No one is making you learn how to use it and by the way, get off the LTSpice threads!

author: Brumby
author_rank: Unknown
timestamp: May 29, 2019, 02:00:50 am
content: Could I please suggest that you not continue this tangent "discussion" on this thread.This is the beginners section and we don't need to bury the Op in stuff that isn't directly helping him.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 29, 2019, 04:10:39 am
content: Quote from: Brumby on May 29, 2019, 02:00:50 amWhy not! ---- That's what we always do!

author: AlfBaz
author_rank: Super Contributor
timestamp: May 30, 2019, 12:23:13 am
content: Just noticed this:Quote from: Brumby on May 27, 2019, 05:01:52 amGiven the context of the answer ie why something isn't something happening in a simulator and your stance on not confusing the OP, why then do you write a paradoxical statement that can be proven mathematically to be false and then allude to mysterious elements that are not part of thecurrent() problemLtspice shows zero current if you try to parallel "wires" and spice in general does not simulate connection properties. If you want a wire in your sim you'll have to model it

author: Brumby
author_rank: Unknown
timestamp: May 30, 2019, 01:24:37 am
content: Quote from: AlfBaz on May 30, 2019, 12:23:13 amApologies.My response was based on the real world situation, where I now see you were talking about "why something isn't something happeningin a simulator".Apologies to the Op as well.