Understanding simple voltage divider

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Hi

I understand the mathematical equation for voltage dividers

I just don't understand why the voltage drops ...

Can anybody explain or point me to something that will

John

Re: Understanding simple voltage divider



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If you tell us what you don't understand about it, or explain why the
voltage should not drop, we can point out the flaw in your argument.
Otherwise, we can only guess where the misunderstanding lies.



Re: Understanding simple voltage divider


On Mon, 14 Sep 2009 10:14:56 -0700 (PDT), John Rivers

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The answer to 'why' can be either a very deep or a very shallow one.
Which are you looking for?  Do you just want to see the equation
development from Ohm's law?  Or are you looking for something a little
deeper than that?

Jon

Re: Understanding simple voltage divider



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Hi John, welcome to sci.electronics.basics.

Ok, think what happens if you have a light bulb connected to a battery.
Now, you put a resistor in series with the light bulb.  Obviously, the light
bulb glows dimmer now with the resistor in series than it did when you put
full battery voltage to it.
The light bulb is like the "bottom" resistor in the voltage divider.  It
gets less voltage if you put an intervening resistor in the line.
The current has to go through both of them, and the each eat up some of the
voltage.  You only have so much to start with.  A zero sum game, as it were.
So each piece of load you put in series gets some fraction of the total
voltage.  If you add them up it comes to 100%.   You can refer to
Kirchhoff's laws for voltage annd current to get a better understanding of
it:
http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws



Re: Understanding simple voltage divider



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  The voltage you are applying to one side of the divider string
  flows through all of the resistors, at the end of the string,
the last resistor gets connected to the
other side of the voltage source, this is normally called the COMMON or
  0V (zero volts).

   In your case, your voltage source could be a 1.5 power cell or maybe
  a 9 Volt battery.

   Assume for the moment, the (+) terminal is connected to the top side
of your divider string, and the (-) terminal is connected to the low
side of the divider string. This creates a circuit in which current flows.

  Because your voltage divider is made up of resistors, you will not
completely short the voltage source, or, lets say you have an infinite
voltage source that will never lose voltage so that we won't need to
worry about that in this discussion.

   As soon as you get current flow, this means energy is getting used
and thus, you now can measure drops of voltage with in your strings
because you have inserts between the high and low points
of the voltage source.

   So, if you knew the current flowing through the string, you can use
      basic Ohms law..

   E=I*R, Just pick out what ever resistor you want in the string and do
a check with a volt meter across that R using both probes. If you add up
all the values, it should equal what your voltage source is.

   With out a current flow, you will not get a voltage divided effect
which means, if you were to take measurements along the string, using
a volt meter with infinite load resistance on the probes so not to have
any errors in measurements, with the (-) probe to the (-) source, you
would see the same value of your voltage source at any point.



  Keep in mind, the ideal volt meter does not exist, the closes you get
is a static meter, but that isn't applicable here, or some physic majors
here may have some other methods I've never heard of.

    Normal digital meter loading is 10Meg ohms. When ever you use your
meter to take measurements, you must assume the meter will act as a
  10M ohm resistor in the circuit. So conducting the test as above with
an open string, at the last string, you may see a slight drop, depending
on what your total resistance sums up to.

    This means if you were to use a 10 Meg R in series with a 9 Volt
battery to a DMM. You'll See 4.5 VOlts on your
  Display and not 9 Volts. Or, 1/2 of what ever the battery is.

   This is effectively a voltage divider too, because the DMM is acting
as the final resistor in the string going back to the Energy source.


    Hope that helped some.


Re: Understanding simple voltage divider



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Voltage is an abstraction, not a real thing; it makes handling
electric field problems easier to think of a quality of points
in space that makes a charge move from one point (high
potential) to another (low potential), just as a weight
moves from one point (high height) to another (low height).

In the case of a 'voltage divider', there are electric fields
inside the resistors, and any charge that comes into those
fields will be moved by them; the voltage at various points
in the circuit (the divider input, output, and common node)
reflects the expected motion of charges, with respect to
any attached circuitry (with its own set of voltages, one voltage
value per node).    So, if you connect a light bulb from common
to common, there's no voltage difference- the light doesn't
light up because no charge movement is expected.
Connect the same light bulb from divider output, it glows
(perhaps dimly); connect from divider input, it glows more
brightly.

The expression "the voltage drops" is ambiguous- it could indicate
time-dependence of a voltage value (it doesn't), or alteration due
to some changing of connections (again, not right); what is indicated
is rather that the work done on an electric charge by the internal
fields of the voltage divider components is in correspondence
with the charge moving in the direction of decreasing potential,
in this case from a (presumed) source of current at the voltage
divider input, to a (presumed) destination at the voltage divider
output.
Once at the output point, this charge is located at a lower
potential than at its source.

Re: Understanding simple voltage divider



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The simple answer is because current is flowing down hill and the
"potential" decreases. Same as a hill. The resistors are like two hills, one
stacked on another, you have created. The size of the hill is analogous to
the resistance. By choosing the hills you get to choose the graviational
potential point at "rest".


____
    \
     \ R1
      \
       \
        --*-
            \ R2
             \____

(use fixed with font to view this properly)


As you walk down the hill your potential energy decreases. At the point *
your potential energy depends on the ratio of heights. If R1 = R2 then you
have cut your potential energy in half. If R1 = 9*R2 then you have cut your
potential energy by a factor of 10. (you are at R2/(R1 + R2) of the way down
as R1 + R2 is the total height and you are at a height of R2.

In fact this analogy is almost exact because the gravitational potential and
electric potential are almost identical(not quite because the electric field
is not necessary "conservative" but in almost all practical problems it is).

Think of the electrons flowing down this hill. They have to flow from the
top to bottom(to complete the circuit which is necessary for them to be
flowing in the first place).

At * you are measuring the potential energy of the electrons. Because they
are only R2/(R1 + R2) of the way down(the resistance ratio is what matters)
you are at that potential(measured from ground.

R2/(R1 + R2) = 1/(1 + R1/R2) so all that matters is the ratio of
resistance(not entirely correct for practical matters though).

If we have

V---R1---*---R2---GND

And we know V and GND. Then R1 and R2 create "resistance" to the electrons.
If R1 is really small then it has very little resistance and hence * would
be close to V. Similarly for R2 but it would be close to GND.

If R1 = R2 then we would expect to be halfway between V and GND or V/2. Now
we might not expect it to be linearly related but it is. e.g., if R1 = 3/4
and R2=1/4 then the potential is (1/4)/(3/4 + 1/4) = 1/4

i.e., R1 dropped the potential by 3/4 and R2 by 1/4 and hence  we are at 1/4
of V(not because R2 = 1/4 though).

To make things more complicated. We could "normalize" the resistance values
if we assume there absolute values do no matter(but just the ratio). so that
we have

V---R1/(R1+R2)--*--R2/(R1+R2)---GND
       ||             ||
V------A--------*-----B---------GND

(A = the first and B is the second)

All I did here was scale there values by the total. In some sense, I am just
getting a "percent".

Now we have the potential at * is just B*V with respect to ground. (if it's
not ground then we just "shift" the two sides down so one is ground to)

Remember, we tend to take measurements with respect to something like
ground. Altitude is a measurement of height from "ground" and not the "sky"
or the moon. (This is why A*V is not right since it gives us a measure below
V which is generally useless because GND is the same throughout the circuit
while the V might change for different parts)






Re: Understanding simple voltage divider


Thankyou everybody

After reading all your posts I have a clearer understanding

and don't feel like I am following a formula like a parrot anymore

Re: Understanding simple voltage divider



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Note that electricity and magnatism follows very closely to gravitation. If
you are familar with basic physics then most analogies will be correct.

This is not completely true though since when dealing with motion of
charge(basically magnetic effects) but for most examples and special
examples(uniform motion or no relative motion) it is very similar.

You can see this because

F_gravity = G*M1*M2/r^2

and

F_electric = k*Q1*Q2/r^2

The only thing different is Q is charge which can be +- and M is mass which
cannot.  But much of the theory is the same because of the abstract
equivalence of the formulas. (again, this isn't completely true when adding
magnetics in the equation)

Most of the "circuit stuff" is large scale simplifications of the EM
model(maxwells equations) because generally most people don't know how to
solve a system of PDE's and most of the time general and simple rules can be
followed(kirchoff's laws, ohms law, etc..).

Note that a voltage divider is very important to understand because it is
used constantly. Note also to understand the effects of "loading" which is
another very important concept. Loading is when you attach circuits together
and they exhibt different characteristics than when separate. This is
generally bad if you don't want that and causes problems for most novices.

Take your voltage divider

10V
|
R1
|
*V
|
R2
|
0V

You know how to calculate the voltage at V. It's easy. But say we add a
resistance at *.


10V
|
R1
|
*--R3--0V
|
R2
|
0V


Now what is the voltage at V(or *)?? You can calculate this simply by
combinging the resistors and such. Notice that you will get a complex
formula(or maybe not but it's not super simple).

But if we know R3 is very large then R3 will not significantly effect the
voltage divider and we can possibly "ignore" it. (think if R3 = infinity
then no current would flow and it would be useless to even have it and if we
removed it then it would look exactly like the first)

You can see this from the formula too by analyzing the formula's
characteristics. You will see that as long as R3 >> R1 and R2 then the
formula reduces to the original voltage divider(approximately at least).

Hence we have not loaded the voltage divider. If R3 << R1 or R2 then we will
load the circuit(effectively making R2 unimportant) and the voltage at V
will be different.

This is an important concept because many times R3 will not be a resistor
but an amplifier and 10V will be a signal. If you overload the previous
stage it cannot drive it and it will affect all the results. If you don't
overload it you can connect the stages together without any problems.

This is why using a voltage divider is more than just chosing the ratio of
R1 and R2. You need to know if you want very large R1 and R2 or very small.
If it's too large for R1 and R2 then then if you connect anything at * you
might overload the voltage driver. If R1 and R2 are too small then you have
excessive power dissipation. Hence this is an optimization problem which you
can choose R1 and R2 so that the next "stage" can be "driven" yet you
minimize power dissipation.


Example


10V
|
R1
|
*--R3--0V
|
R2
|
0V


Suppose we want to "drive" R3(this isn't really good language as we
generally drive stages and R3 really isn't a stage but it is good enough)
without overloading(or just loading it) the voltage divider.

As long as R1 and R2 << R3 then V(at *) is just V = R2/(R1 + R2)*10

and since we know V we can calculate the current through R3, which is V/R3.
Note the current is proportional R3. and looks like we just attached a
battery to V which the voltage R2/(R1 + R2)*10. WE can change R3 without
screwing up V.

But if R3 is comparable to R1 and or R2 then the formula is

V = (R2 + R3)/(R1 + R2 + R3)*10

Now if we change R3 we end up changing V which makes it a complex problem.
In the first case R3 does not load the voltage divider and V is
approximately constant. In the second case changing R3 also changes V.

Making sure a circuit is non being loaded means that when you connect
stages, just not randomly but in a well designed circuit, that you can be
somewhat confident that the independent stages will work independently and
not cause "interference" with other circuits.

If you have an idea of mathematics then basically if one circuit is
represented by a function F and another by G, then the combined circuit is
represented by F + G rather than some complex function H(F,G).



Re: Understanding simple voltage divider


Thanks for that post Jon - that was my next train of thought ...

Re: Understanding simple voltage divider


On Mon, 14 Sep 2009 10:14:56 -0700 (PDT), John Rivers

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Consider two squares of aluminum foil glued to the floor, a few feet
apart. Connect a 10-volt battery between them. Connect the one on the
left to ground, too. Now one foil square is at 0 volts and the other
is at 10 volts.

Now paint a neat stripe of carbon-filled conductive paint between
them, not quite conductive enough to short out the battery or set the
floor on fire.

Stand barefoot on the grounded square; you're at zero volts. Start
walking towards the 10-volt square; every step, your voltage goes up.
Halfway between, you're at 5 volts. The voltage increases smoothly
from 0 to 10 along the path.

If you're 25% of the way across, you're at 2.5 volts. The resistance
looking backwards is one unit of paint, and the resistance to the
other end is three units. That makes a voltage divider


     +10 volts
      |
      |
      |
      |  3R
      |
      |
      |
     you    < 2.5 volts
      |
      |  R
      |
      |
     GROUND


  Vout = Vin * R / (R+3R)

which is simply the geometry of where you are along the path. R is how
far you are along the path, R+3R is the total length end-to-end. A
fourth of the way is a fourth of the voltage. You are a human
linear-taper potentiometer!


Some people will "teach" by presenting equations and assuming they're
done. Manipulating a page of equations doesn't mean you understand
what's going on; the understanding should come first, the equations
second, and the connection should be obvious.

John


Re: Understanding simple voltage divider


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Try thinking of voltage as water pressure in a pipe network.  Current is
equivalent to flow and resistance is equivalent to 1/pipe diameter (very
crudely).

R

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