### Author Topic: connecting to 0 volts.  (Read 1242 times)

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#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #25 on: March 16, 2020, 12:11:10 pm »
I do lose half the current down the left side,  because its 2 resistors versus one.  (which i wish wouldnt happen, but its true, and using potentiometres introduces error to it.)
If it were an exact ground. id have no current, because a-a=0.

But if the resistors were different you would get current there.   But as it sits in error at 5v,  that means it would be flowing towards the left, in my mind, because the potential difference is GREATER.

[EDIT]
Brumby,  they are not multimetre measurements, they are NOT current limited,  they are just raw voltages from batteries. at those positions.
« Last Edit: March 16, 2020, 12:13:20 pm by krayvonk »

#### Brumby

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##### Re: connecting to 0 volts.
« Reply #26 on: March 16, 2020, 12:13:38 pm »
Do us a favour .... show us the whole circuit - with all the components (including power sources) and all the connections.

#### Brumby

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##### Re: connecting to 0 volts.
« Reply #27 on: March 16, 2020, 12:17:56 pm »
Just for good measure - if all your power sources are batteries, you can stop using the word "ground".  It is irrelevant.  Completely, totally and wholeheartedly irrelevant.

#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #28 on: March 16, 2020, 12:18:35 pm »
It is the whole circuit, do you need me to show you the application as well?

just think of the resistors as potentiometres,  and the battery connections are solid and unchanging.

There is error in the result, because of load theft on the potentiometres,  when you increase the power of "operand b" itll actually draw more current down the side from "operand a", so it would give you a more contrastive result than a correct subtraction.

I wished the bare "ground pole" (which is the low PotentialDifference battery...) would short away because its 100% loadless,  but I forgot both operands have a resistor/potentiometre so that wouldnt happen.   So it doesnt work that well anyway.
« Last Edit: March 16, 2020, 12:22:16 pm by krayvonk »

#### Zero999

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##### Re: connecting to 0 volts.
« Reply #29 on: March 16, 2020, 12:23:24 pm »
It is the whole circuit.

just think of the resistors as potentiometres,  and the battery connections are solid and unchanging.

There is error in the result, because of load theft on the potentiometres,  when you increase the power of "operand b" itll actually draw more current down the side from "operand a", so it would give you a more contrastive result than a correct subtraction.

I wished the bare "ground pole" (which is the low PotentialDifference battery...) would short away because its 100% loadless,  but I forgot both operands have a resistor/potentiometre so that wouldnt happen.   So it doesnt work that well anyway.
The two resistors form a potential divider. The output resistance is equivalent to both their values in parallel, half of 1k, as they're both the same and the output voltage is 0V, half way between -12V and +12V. The equivalent circuit is simply 500 Ohms, connected to 0V, so the current taken by the 5V supply is 5/500 = 10mA. This is all assuming the other sides of the 12V power supplies are connected to 0V.

I agree, not drawing circuits clearly and doing things such as omitting the ground connection, can cause confusion.

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#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #30 on: March 16, 2020, 12:26:14 pm »
connected to 0v?   +12v connects to -12v.

#### Zero999

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##### Re: connecting to 0 volts.
« Reply #31 on: March 16, 2020, 12:32:46 pm »
connected to 0v?   +12v connects to -12v.

Yes, assuming the other sides of the 12V supplies are connected to 0V, the Norton's equivalent circuit for that potentiometer is just a 500 Ohm resistor connected to 0V. It does have 24V across it, because that's the potential difference between +12V and -12V.

All voltages are relative. What confused me when I initially replied, was I made the assumption that the other side of the 12V supplies are connected to +5V, rather than 0V.

And you haven't shown the complete circuit, because there's no 0V connection. If you entered that circuit into SPICE, it would refuse to simulate it.

#### Siwastaja

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##### Re: connecting to 0 volts.
« Reply #32 on: March 16, 2020, 12:35:52 pm »

I can see why this might be confusing to total beginners, though.

There are two different concepts mixed, without clear indication which is which.

There are numbers that directly represent voltage differences: the "9V" labels next to the batteries. They describe a voltage between two points in circuits (namely, the battery terminals). This is the only physical definition of voltage. A telltale sign in this picture is that the labels indeed are not next to any single point, they are in the middle of two points.

Then there are the numbers that represent a "potential" at a single point. These refer to each other; if one node says "+9V" and another says "-9V", the voltage between these two points is 18V (or -18V, depending on which way you measure). But you could call them +1009V and +991V, as well, and the result is the same. Voltages are differences, and the reference of such "single point potentials" is an arbitrary choice.

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#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #33 on: March 16, 2020, 12:37:31 pm »
Thanks for that Siwastaja,   my 12v,5v notation are battery terminals,  in that case.

I can see now that is a picture of 2 power sources in series.

It might not be a complete schematic (my retard drawing i mean) to go into a simulator,  but u can actually build it from that fine, its the full information...  if, you have a 12 v +- source and a 5v + source.

There is no 0v connection,  because the joke was I couldnt get 0v to work, because hunks of metal dont work as grounds, as weve been discussing, and have experimental proof for,  but i can get *5v* off a battery...   why would one work and not the other.
« Last Edit: March 16, 2020, 12:49:10 pm by krayvonk »

#### Zero999

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##### Re: connecting to 0 volts.
« Reply #34 on: March 16, 2020, 01:50:56 pm »
No, you can't really build it because it's ambiguous.

I entered it into LTSpice, guessing where the grounds and commons should go.

Firstly, consider the other side of each power supply is ground. It can be analysed as I described in my previous post. I added V4 and R3 to prove the potential divider is equivalent to a 500 Ohm resistor to 0V, as far as the 5V PSU is concerned. Note the currents are negative, because LTSpice shows them as the current is being taken from the PSU, i.e. the batteries are being discharged.

Secondly, assuming the other side of the 5V rail is ground and the 12V power supplies are floating, with the other sides connected to a different reference, hence the different symbol. No current flows out of the 5V rail, which lifts both 12V supplies up by 5V. This was what I assumed, when I initially replied to the thread.

Now let's change the 5V rail to 100V. The 12V rails shift up by 95V, giving 112V and 88V. I stuffed up on my arithmetic on my previous post.
« Last Edit: March 16, 2020, 01:54:37 pm by Zero999 »

#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #35 on: March 16, 2020, 01:57:43 pm »
Where do you guys learn all this stuff???   Why am I the odd one out.

#### Zero999

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##### Re: connecting to 0 volts.
« Reply #36 on: March 16, 2020, 02:13:59 pm »
Where do you guys learn all this stuff???   Why am I the odd one out.
School, college, university, books, work, the Internet etc.

Finally, if I draw the circuit, as you drew it, assuming the other sides of the power supplies are floating, the currents are all zero, because they're all open circuit. SPICE won't simulate it, because the voltages will all be undefined, as everything is floating, so I edited it in MS Paint.

#### krayvonk

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##### Re: connecting to 0 volts.
« Reply #37 on: March 16, 2020, 03:37:42 pm »
Thats a limitation in my book...   I need to write my own simulator.  (laugh if you will...)

#### Siwastaja

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##### Re: connecting to 0 volts.
« Reply #38 on: March 16, 2020, 04:35:46 pm »
There needs to be a closed loop. Without that, current can't flow. Without current, you can't do much electronics.

Closed loop is a line of wire and components that you can follow, make a full lap, and end up where you started.

#### Nerull

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##### Re: connecting to 0 volts.
« Reply #39 on: March 16, 2020, 05:49:58 pm »
Thats a limitation in my book...   I need to write my own simulator.  (laugh if you will...)

Perhaps learning basic electronics would be good start, first.

No simulator can fix your misunderstandings.
« Last Edit: March 16, 2020, 05:51:52 pm by Nerull »

#### Siwastaja

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##### Re: connecting to 0 volts.
« Reply #40 on: March 16, 2020, 06:09:44 pm »
Most simulators assume the basics are "right" and work improperly if they are not, not matching real-world results. SPICE simulators especially are pain-in-the-ass for total beginners.

If you want to learn by doing instead of reading (I recommend the combination of the both), play with the actual components instead: batteries, wire, lightbulbs, small DC motors, a multimeter. I started this way before I could read. Helps you get intuition so that wrong ideas won't stuck. The problem with reading is, many bookwriters suck in clear communication, and while technical details are plentiful and mostly right, stupid basic mistakes happen in the first chapters.
« Last Edit: March 16, 2020, 06:12:50 pm by Siwastaja »

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#### Brumby

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##### Re: connecting to 0 volts.
« Reply #41 on: March 16, 2020, 11:57:51 pm »
No, you can't really build it because it's ambiguous.

I entered it into LTSpice, guessing where the grounds and commons should go.

Firstly, consider the other side of each power supply is ground. It can be analysed as I described in my previous post. I added V4 and R3 to prove the potential divider is equivalent to a 500 Ohm resistor to 0V, as far as the 5V PSU is concerned. Note the currents are negative, because LTSpice shows them as the current is being taken from the PSU, i.e. the batteries are being discharged.

Secondly, assuming the other side of the 5V rail is ground and the 12V power supplies are floating, with the other sides connected to a different reference, hence the different symbol. No current flows out of the 5V rail, which lifts both 12V supplies up by 5V. This was what I assumed, when I initially replied to the thread.

Now let's change the 5V rail to 100V. The 12V rails shift up by 95V, giving 112V and 88V. I stuffed up on my arithmetic on my previous post.
My appreciation to Zero999 for taking the time to put this together (I just didn't have that yesterday) ... and to note that the word "assume" appeared in his efforts.  I was trying to avoid making assumptions.

There needs to be a closed loop. Without that, current can't flow. Without current, you can't do much electronics.

Closed loop is a line of wire and components that you can follow, make a full lap, and end up where you started.

This is the point I was making earlier - except this wording makes it extremely explicit.

#### james_s

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##### Re: connecting to 0 volts.
« Reply #42 on: March 17, 2020, 12:12:14 am »
It is the whole circuit, do you need me to show you the application as well?

just think of the resistors as potentiometres,  and the battery connections are solid and unchanging.

There is error in the result, because of load theft on the potentiometres,  when you increase the power of "operand b" itll actually draw more current down the side from "operand a", so it would give you a more contrastive result than a correct subtraction.

I wished the bare "ground pole" (which is the low PotentialDifference battery...) would short away because its 100% loadless,  but I forgot both operands have a resistor/potentiometre so that wouldnt happen.   So it doesnt work that well anyway.

That's not the whole circuit though, you don't have a circuit, you just have some resistors connected to arbitrary voltages with no return path. Normally "ground" is implied in a circuit, and if this is the case then you have 3 voltage sources relative to this point called "ground" and your meter is also referenced to this point. This is not clear from your schematic though because it isn't drawn.

I'm not sure how you plan to write a simulator when you are missing a fundamental understanding of some of the most basic and rudimentary concepts of electricity. Spend a few hours reading about Kirchoff's laws and that should get you started. You are missing the very, very basic low level concepts and without grasping those you will never understand the rest. It's like trying to understand calculating the trajectory of a rocket without grasping how gravity and Newton's laws of motion work.

Smf