# Negative Rail and 0 volts

Discussion in 'Homework Help' started by EpicFails, Sep 7, 2012.

1. ### EpicFails Thread Starter New Member

Sep 7, 2012
18
0
Hi all,
Im really struggling with the concept of 0 volts on a negative rail and the terms used to describe ground/earth etc. in a circuit. Im new to this forum and new to electronics as you can tell! I hope Ive posted this in a relevant area of the forum  apologies if not.
My main question is this: Does the voltage in a circuit coming from positive terminal of a battery have to all be used up before it gets to the negative rail, thus arriving at the negative terminal with 0volts?
For example: 9v leaves a battery and goes through a component and 6v is used up. It returns to the negative terminal still with a voltage of 3v.
Will this overload the system? Is it allowed or is this bad circuit design?
I should add that I understand that voltage is the difference between two points and that as such, 0 volts could be seen as a relative value and not actually absolute, but if different connections to a negative terminal all arrive with different levels of voltage, how can this be used as a point of reference
Also, in practical application, does it matter at what point connections come back to the negative rail of a circuit, i.e. does it matter which connections are closer to the negative terminal of the battery than others?
I appreciate this is pretty basic stuff, but I have not found an explanation I can understand for this yet. Im getting the impression that Ive misunderstood the basic concept of voltage altogether

Thanks!

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
I hope that this imagines will help you understand what negative voltage is

• ###### rys0.1.JPG
File size:
47.8 KB
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3. ### MrChips Moderator

Oct 2, 2009
12,625
3,451
Hmmm, think of electrical and electronic circuits as resistors in series across the power supply.

Voltage does not get "used up". Imagine a circuit consisting of two resistors R1 and R2 in series with a dual power supply that provides +10V and -10V at the supply rails.

The voltage drop across the resistors R1 and R2 do not get "used up" resulting in a voltage that is different from the -10V at the negative supply rail.

Without getting into details about the supply source impedances, the top rail will always be +10V and the bottom rail will always be -10V.

What we can determine is the voltage at the junction of R1 and R2. To do this we apply Ohm's Law.

The current is the same through both R1 and R2 and is (Vs1 + Vs2)/(R1 + R2)

The voltage across R2 is R2 x (Vs1 + Vs2)/(R1 + R2).

The voltage at the bottom arm of R2 is -Vs2 with respect to GND.

Hence we can determine the voltage V at the node with R1 and R2 (being careful to apply the signs correctly).

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4. ### ramancini8 Member

Jul 18, 2012
447
119
The battery voltage is allpied to the load. The full battery voltage is applied to the load. If you place a 6V load on a 12V battery the load usually burns up. You can connect two batteries back to back to get 2Vbattery or ground centertapped with plus and minus Vbattery.

5. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
There's a problem with the nomenclature. When you indicate a voltage polarity with an arrow from Node 'a' to Node 'b', the indicated voltage is nominally (as in, "by name")positive at Node 'b' relative to Node 'a' and is designated by Vba, which is a conventional shorthand for Vb relative to Va, or Vb-Va. The nice thing about the convention is that you don't need to indicate polarity at all since the convention makes that clear from the order of the subscripts.

Hence

Vba = (Vb - Va)

and

Vab = (Va - Vb).

Therefore

Vba = -Vab

In all three figures, Vab is unambiguously -1.5V and Vba is +1.5V, regardless of where the ground is, because both are relative measurements.

Now, Va and Vb are a different matter, because that is referring to the voltage on a single node, and voltage only has meaning in the context of a relative quantity, so to put a number to Va and Vb, we need an agreed upon reference node which has been assigned an arbitrary voltage. Usually, we pick a convenient node and declare that it is at 0V; we then refer to this node as "common", "ground", "ref", "earth" (though the last one is usually reserved for an actual earth ground, but not always).

So in Fig 1, Va and Vb are undefined. (But we still know that, however we end up defining them, that Vb is 1.5V greater than Va and, hence Vba=1.5V and Vab=-1.5V).

In Fig 2, we are declaring Va=0V, so Vb = Va + Vba = 1.5V.

In Fig 3, we are declaring Vb=0V, so Va = Vb + Vab = -1.5V

Notice how subscript math falls out directly from the conventional definition of subscripted voltages.

Vrw = Vr - Vw
Vrw = (Vr - Vs) + (Vs - Vw) = Vrs + Vsw

The inner subscripts go away and the outer ones survive if the inner ones are the same.

I don't know if this convention on the use of subscripted voltages has a name, but it is pretty universal. Of course, it only applies when you are using the subscripts to identify nodes, and not when you do something like Vdet, Vbatt.

As an aside, what is Vrr? Using the convention of subscripted voltages, it's zero. So a voltage with a repeated subscript is pretty useless. As a result, the convention says that a repeated subscript is not a relative voltage, but rather the designation for a supply or bias voltage. Hence Vcc is the supply voltage related to the collector of an NPN circuit and Vgg is the bias voltage associated with the gate in a MOSFET circuit. But this is a lot looser and not meant to be rigid; instead, repeated subscripts are a way of providing contextual information in a clean way that is usually not difficult for "practitioners familier with the art", so to say, to interpret correctly without a lot of information being made explicit.