question about voltage dividers

[drk]

for this last weeks i've been reading alot of theory to understand electronics, and one doubt that come across my mind, and that i didnt find any answer for it was about the voltage dividers.

when you use two resistors that have the same resistance, you get half of the voltage in the middle(after the 1st), but from looking at schematics, some have two 10k resistors, other have two 22k, etc. and it seems it doesnt make a difference in the voltage divider that the resistor value changes(as long as both are the same value). why this happens??
whats the difference beetwen two 10k or two 1M as a voltage divider for example? considering the same power source.
thanks

[R.G.]

The difference is how much the voltage at the middle of the divider "sags" when loaded.

To understand, you only need to know Ohm's Law, that the voltage across a resistor MUST be equal to the current through it times the resistance. In fact, resistance is DEFINED as the ratio of voltage to current, or R = V/I.

You also need to understand the other, equivalent ways of stating this V= I*R, I = V/R.

In the simplest divider, there are two resistors in series. So the current which goes through each one is exactly the same current as goes through the other. If the resistors are equal, then the voltage across each resistor must also be identical, and the only way that can happen is if each voltage is half of the total applied voltage on the divider. It cannot be any different because of the way we define resistance.

So far, so good. But a two-resistor divider which never has anything connected to it to use that voltage is pretty dull, no use at all. In all cases, there will be some external thing connected between the middle of the two resistors and somewhere else; this ends up being a leakage current either out of or into the middle point of the two resistors. We have to account for that lost (or added) current somehow.

Let's pretend that the lost/added current can always be modeled as a resistor to ground across the lower resistor. That makes the lower resistor look like a lower resistance, since for the same voltage across the both of them, more current will flow - they both get to "drink" from the available voltage.

That causes more current to flow through the upper resistor, so the voltage across it gets bigger, and so the voltage across the lower resistor gets smaller. The voltage at the middle point of the divider sags down some. How far it sags depends on how high or low the added resistance is. If the added resistor is a wire (that is, milliohms of resistance), then the divider voltage sags all the way to ground being "shorted" by the wire. If the added resistor is huge, like the resistance of half an inch of air, then there is no detectable sag.

What matters is how big the load resistor is compared to the two resistors making the voltage divider, and here we are at why some dividers are 10K/10K, some are 22K/22K, and some are 1M/1M. It all depends on how much loading you put on the middle of the divider.

See

http://geofex.com/circuits/Biasnet.htm

for more on exactly this same topic. It would help you a lot to read the rest of the articles at GEO.

[drk]

ok, i see why i didnt find anything when i read about it. in the examples of books, normally it only talks about a circuit with 2 resistors and powersource, nothing more.

so its like having an extra circuit in parallel with the lower resistor right? which reduces the resistance of the lower resistor(combined with the other part), increases the current through the upper, etc.
then, if we change one of the resistors value to match the extra "interference"(like one with higher value than the other), wouldnt it work too?

thanks for explaining, i now need to sleep a bit on the subject 

[R.G.]

ok, i see why i didnt find anything when i read about it. in the examples of books, normally it only talks about a circuit with 2 resistors and powersource, nothing more.

Yes. That is one perfect illustration of the difference between an EE degree and knowing how to design circuits. The two-resistors-only circuit shows the circuit in isolation. If you know how to design circuits, you know that there is always another load connected, in one form or another.

so its like having an extra circuit in parallel with the lower resistor right? which reduces the resistance of the lower resistor(combined with the other part), increases the current through the upper, etc.

That's correct.

then, if we change one of the resistors value to match the extra "interference"(like one with higher value than the other), wouldnt it work too?

It would indeed "balance" again. However...

That only works if you already know what the extra load is so you can balance it. What happens if you don't already know the value of the extra load, only that it's more or less than "X" ma?

The single most valuable lecture I got in my undergraduate EE program was from a professor who had worked in industry before retreating (as he himself put it) back into academia. He spent a lecture telling us about when we could ignore things.

It turns out that as a rule of thumb, us humans can usually be happy if we get between 9/10 and 11/10 of what we expect. That is, the result is within 10% of perfect. There are lots of cases where that's not true, but often it is. So if you make a voltage divider that you want to be exactly half of 9V, or 4.5000V, and it comes out between 4.05V and 4.95V, you'll likely be happy. At least the widget will probably work OK until you can tweak it in.

I'm forever tossing in math and making everyone's eyes glaze over, so I won't do that now, but if you do go through the math, you find that as long as the current in or out of the middle of the voltage divider is less than 1/10 of the current through the divider, then the voltage will be within 10%. So the job is reduced from knowing exactly what the error current is to knowing that it won't be bigger than "X". Once you know it will guaranteed be less than some boundary, you pick the two resistors to let more than 10X (or 50X or 100X) that current flow in the divider and all is well.

[drk]

im starting to noticing, that sometimes it's a bit dificult to use the knowledge presented in books with a real life circuit. the text examples are too simple(or the real life too complex  )

that 10% rule its a nice way to overcome the problem, you explain that in that article at geofex.com too, i sometimes do the same thing when i dont have that exact cap value for example, just use something close. gonna try keep this rule in mind
thanks for explaining