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Trouble understand Voltage Divider and purpose of two resistors

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rokuez

New Member
I have some questions about the voltage divider.

Voltage divider - Wikipedia, the free encyclopedia

Here is what I think is going on.

The voltage starts at the negative terminal i.e V in then goes to the first resistor Z1. & in my mind i'm thinking why isn't Vout determined by current * Z1?

I basically don't understand why a second resistor is used in a voltage divider.

In the equations at the bottom I understand V in = total current * total ressitance i.e z1 + z2


BUT I don't understand why V out = I * Z2 ??? Doesn't the voltage drop occur when the electrons encounter the first resistor i.e Z1 and their energy gets transfered to heat and dissipated in the air?

I understand the re-arranging of terms for I = V in / (z1 +z2)

I just don't get why V out = I * Z2..... in my mind I believe it should be I * Z1 because linearly speaking the current / electrons / energy first encounters Z1 and gets the voltage drop from that..? I don't even understand why a second resistor is needed to show this voltage drop/divider example because in my mind the first resistor does this.

Does this have something to do with electron flow vs conventional current flow? I've read about both terms but don't understand if it has any basis on why i'm not understanding this.

Please help me sort this out. I'm reading Electronics Self-Teaching Guide by Harry Kybett and Earl Boysen and i enjoy this cause it has lots of problems but not a lotta explanation.
 

ericgibbs

Well-Known Member
Most Helpful Member
hi,
My advice would be to stop thinking in terms of electrons.

Consider just two resistors of equal value, say 1KΩ each, connected in series as shown in the link diagram.

Now apply a 20Vdc voltage across the free ends of the two resistors.

A Current of 20V/2K = 0.01A [10mA] will flow thru both resistors.

If you now used a 'dc' voltmeter to measure across each resistor in turn, it would give a voltage of 10V.

This is from Ohms Law. V= I * R >> 0.01 * 1K = 10Volts.

The two resistors do not have to be equal in value, the values are chosen to give the required voltage at the junction of the two resistors.

Its important to note that the 'load' connected to the junction of the two resistors will have an effect upon the junction voltage.

The 'load' resistor is effectively in parallel with one of the resistors.

Do you follow this OK.?
 
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rokuez

New Member
I am still a little bit confused. after crunching some numbers from doing calculations with resistors of different values.


Say the two resistors are of different values.

V = 5 , I = 5/6 , R 1 = 1 , R 2 = 5


If I where to measure the Voltage drop across R1 I would get V 1 = 5/6

If I where to measure the Voltage drop across R2 I would get V 2 = 4 & 1/6




I guess my confusion comes from when I see the wikipedia examples I do not clearly see which voltage drop they are measuring i.e voltage drop from Z1 or Z2.

OR

are they not so much "measuring voltage drop" as they are using the voltage divider to create a new "voltage value/voltage source" to power something i.e is this example trying to show the most basic structure for how one gets a precise voltage "divided" from a large voltage source?




To me the voltage divider seems ambiguous because it is not clear which voltage-drop amount is being measured... the value of the drop across Z1 or Z2. Are they not really measuring voltage drop so much as determining how much voltage would be coming from V out?


Also another question. What if V out was a wire that came before Z1 instead of a wire coming out between Z1 and Z2. Would those equations still hold true i.e

Vout = V in * (Z2) / (Z1 + Z2)

or would it be

Vout = V in * (Z1) / (Z1 + Z2) sense the wire i.e V out is now before Z1, instead of inbetween Z1 & Z2
 

dknguyen

Well-Known Member
Most Helpful Member
To me the voltage divider seems ambiguous because it is not clear which voltage-drop amount is being measured... the value of the drop across Z1 or Z2. Are they not really measuring voltage drop so much as determining how much voltage would be coming from V out?
THe voltage drop being measured is the actual voltage across the resistor AND is the actual voltage at the point. THis means that the voltage that is being measured is the voltage across the resistor whose end is connected to ground.

See this picture:
http://en.wikipedia.org/wiki/File:Impedance_Voltage_divider.png
What is Vin referenced to? Vout or GND? GND obviously (it is obvious, right?). So what is Vout being referenced to? Also ground. Just like every other voltage in the schematic. Every labelled voltage in a schematic is being referenced to ground unless indicated otherwise. This indication comes in the form of an arrow or brackets highlighting the two points the voltage is being measured across. If it just has a name without anything else, it's being referenced to the circuit reference...which is ground.

are they not so much "measuring voltage drop" as they are using the voltage divider to create a new "voltage value/voltage source" to power something i.e is this example trying to show the most basic structure for how one gets a precise voltage "divided" from a large voltage source?
In reality though, you don't power something with a divider. THe divider equation given relies on the same amount of current flowing through both resistors. WHen you "power" something you are connecting a device to VOut that is drawing a lot of current. THis diverts current out of the bottom resistor into the device. THis changes the current flowing in both resistors to the point where the error is not negligible- it means output voltage changes with output current. If you use large resistors if the output current increase a little, VOut will droop a lot. If you use small resistors to eliminate this voltage drop you are wasting a lot of power doing nothing (a lot of current flows straight from +V to ground).

It does work to scale down voltage signals though (which are low current and the error introduced is tolerable.

Why don't you try drawing the resistive divider and all your various questions about wires coming out at different places as a loop diagram (rather than a top-to-bottom circuit). Then do circuit analysis. THis should answer all your question. FOrget about the resistive divider formula for now. Just do the circuit analysis. You'll come up with a better understanding about where the formula comes from.
 
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MrAl

Well-Known Member
Most Helpful Member
Hi,

Sometimes it helps to work in whole numbers as it allows a quick analysis and
thus a quick understanding of the circuit.

For your example, make Vin=5 and make R1=1 and make R2=4. This way
the voltage across each resistor is the same as the resistor value.
See how this works?

If you make Vin=6, make R1=1 and R2=5, and the same relationship
holds. VR1=1 and VR2=5.
See how the voltage distributes across the two resistors?
Part of it across R1 and part of it across R2. The voltage levels
are determined by the value of the resistors.

Another thing to note about electrical circuits:

1. Voltage appears *ACROSS* a circuit element.
2. Current appears *THROUGH* a circuit element.

This is a very very important relationship that should be
memorized because it helps understand other things too.
 
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ericgibbs

Well-Known Member
Most Helpful Member
hi,

My advice would be not to use the 'Z' parameter until you fully understand the basic Resistive divider using the 'R parameter.

Also choose a reference voltage point from which to make your calculations, in this sketch I have chosen the -V terminal of the battery as the 0V/Common reference.

Look at this image.
 

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rokuez

New Member
dknguyen; I see now that it is clear that these Voltage measurements i.e V out is referenced to ground and being measured across the resistor, Z2, that makes a lot of sense now. I wasn't aware that if Vout just has a name w/o anything else by default it is referenced to ground.



File:Impedance Voltage divider.png - Wikipedia, the free encyclopedia

When i first looked at this image I never figured by default that V out was voltage measured across Z2 . I just figured it was a the voltage at that point, between Z1 & Z2. But I suppose when measuring voltage it is always measured across a component?



One thing I do find confusing with this diagram

File:Impedance Voltage divider.png - Wikipedia, the free encyclopedia

is where would Vout be placed if you wanted to know the voltage across JUST Z1?

Would it be placed before Z1? Akin to how Vout is placed before Z2?


Also if one was measuring the voltage drop across Z1 and Z2 would that require a line enclosing Z1 & Z2?
 
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