Understanding simple voltage divider

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.

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

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:

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.

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.

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)

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

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

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

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

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