Electronics > Beginners
Beginners idiot question
Zero999:
--- Quote from: mikerj on May 26, 2019, 10:24:38 am ---
--- Quote from: vk6zgo on May 26, 2019, 05:46:28 am ---It's just one of the idiot things simulators do! ;D
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.
--- End quote ---
LTSpice 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 am ---I tried to like LTspice---- I really did!, but when it I couldn't simulate a basic CR network, I gave up.
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This is a PEBCAK problem, not LTSpice.
--- Quote from: vk6zgo on May 26, 2019, 05:46:28 am ---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!
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It's like trying to use a tool you haven't yet learned to use.
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Yes, 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.
IDEngineer:
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 actual earth "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 48VDC positive. 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 [potential] as me". And folks on the third floor might say "You're not high, you're BELOW my level [potential]. You're negative ten feet [volts]". 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.
Mr D:
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!?
AlfBaz:
--- Quote from: Mr D on May 26, 2019, 09:13:57 pm ---then the current should flow just as happily through this ground bus, right?!
--- End quote ---
Well 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 0V
I 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
Brumby:
--- Quote from: Mr D on May 26, 2019, 09:13:57 pm ---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.
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Absolutely correct! :-+
--- Quote ---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.
--- End quote ---
The key to understanding is in this phrase: "EC shows". This simulator only shows current flowing through the explicit connection because that's all it needs to 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.
--- Quote ---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!?
--- End quote ---
You 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 uses ideal components.
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