Why use a voltage divider and not just a variable resistor?

[BRetonDP]

Hello, everyone. I have a simple question I just can't get my head around.

Some time ago I was working with an LDR for a circuit that would be based on light.

Naturally, what I wanted was that the my LDR (connected directly to Vcc) gave me (on the other terminal) a signal that would be later converted into a High or Low to be used by the rest of the circuit. (This proved to more difficult than I initially thought so, but it is not the point of thread.)

So, as I researched some examples, I noticed the LDR was always used in a voltage divider. My question is: "why?" Why is the voltage divider used? Isn't the voltage drop at the LDR enough to be connected directly into whatever I want to use it?

I hope my question is understandable. If not, I will further elaborate. Thanks in advance for all comments

[David_AVD]

That voltage drop can only occur if something is drawing current.  That's the fixed resistor (going to V+ or 0V) in the case of the LDR example.

[Lightages]

Variable resistors, or potentiometers, have a metal wiper rubbing on a partially conducting material. This causes potential problems. The first one is that oxidation can occur at the contact pint which will change the resistance. Secondly the contact point can "burn" over time with the current that is passing through it and cause an unreliable contact. Another problem is that the contact can move with vibration or temperature.

A fixed, high precision and low temperature coefficient divider network is much more stable. Trim pots are needed sometimes but should be avoided if possible.

[BRetonDP]

I see. My question, however, kinda goes this way.
Attached are the circuits in question. At first, I connected something that looked like the second circuit (just the LDR), but all examples used a configuration similar to the one above. Is the circuit below even viable at all?

[Legion]

I can't tell if you're asking why use a voltage divider instead of a single voltage source and variable resistor or if you're trying to use the base of the transistor as a second impedance to divide your voltage? I'm not sure the transistor base can act as a second impedance to divide your 1.5V.

[BravoV]

Suggesting to read "thoroughly" the theory of voltage divider using "two" resistors, I believe once you understand it fully, this will become clear to you.

[digsys]

V=IR
The LDR is pure R (for this example), so it varies say 10K <> 1K. That's all it can do.
When you connect it directly in series from a battery to a transistor, you are just switching the transistor ON !
Even with 10K, in your case, you can't turn it off (Input "current" x beta > Load current).
By putting a 2nd resistor to 0V, you create a voltage divider and change how the transistor "sees" the LDR entirely.
It will now see, say 0.1V - 1.0V (using a 10K LDR, 10K 2nd resistor, and 2V supply.
You have 2 different issues there.

[BRetonDP]

Sorry, ignore the values of the components, I just sketched that up with the defaults. Also, the grounds should be common. What I'm trying to ask is, as you said "g "why use a voltage divider instead of a single voltage source and variable resistor"

[BRetonDP]

I see where you're going, digsys. I don't understand the

"It will now see, say 0.1V - 1.0V (using a 10K LDR, 10K 2nd resistor, and 2V supply.)"

at first glance; is that a subtraction going on there?

[Legion]

Sorry, ignore the values of the components, I just sketched that up with the defaults. Also, the grounds should be common. What I'm trying to ask is, as you said "g "why use a voltage divider instead of a single voltage source and variable resistor"

Because if you take a simple circuit with just a voltage source and a single resistor, it doesn't matter what the value of the resistor is, all the voltage will be dropped across it. If you measure the voltage after the resistor it will always read 0V whether it's a 100 ohm resistor or a 20M resistor. But if you have two resistors they will divide the voltage between them in proportion to their resistance values.

So with the LDR, no matter what it's current resistance is, by itself it will drop all the voltage. But if you pair it with a resistor they form a voltage divider. When the LDR's resistance is small compared to the paired resistor, the transistor base will see a high voltage. When the LDR's resistance is high compared to the paired resistor, the transistor base will see a low voltage.

[Frink42]

One of the reasons to use a voltage divider where one resistor is a fixed (known) resistor and the other is the LDR is when you're using a uC (an Arduino comes to mind because that's what I've used).

If you drive the LDR directly with the 5V and try to get an "analogread" (meaning, the analog pin would go to the "positive" side of the LDR) that pin will ALWAYS see 5V no matter how the resistor changes. It is effectively a voltage source in parallel with a resistor, hence, voltage in the resistor (LDR) will always be the same.

However, if you put another resistor in series with the LDR the LDR now changes the amount of current that flows through the circuit, affecting what the voltage is in that LDR (and in the other resistor).

tl;dr if you put a resistor in parallel with a voltage source it will always have the same voltage across its terminals, because that's what parallel means.

[BRetonDP]

Thank you all for taking the time to respond
I think I answered my own question running some simulations and doing some mesh anaylisis on the circuit.

[electronics man]

i dont understand why you have 2 lots of batteries in your circuit

[BRetonDP]

What do you mean? :O I guess I'm using the 5V battery to drive the transistor and enable another circuit with a 9V source to turn on the LED. Is anything wrong? 

[electronics man]

Why can't you run it all from 9v but if you can't then why don't you use a voltage regulator

[BRetonDP]

Oh, Ok, I understand what you mean now.
There was no specific reason, just sketched it quickly and found it more aesthetically pleasing like this; this diagrams are not implemented in reality

[David_AVD]

Oh, Ok, I understand what you mean now.
There was no specific reason, just sketched it quickly and found it more aesthetically pleasing like this; this diagrams are not implemented in reality

The schematic not reflecting the real intent / question can be the cause of much confusion on both sides.  Always better to draw exactly what you have / want.

[BRetonDP]

What I mean is that the circuit is not breadboarded or really supposed to do anything specific; my doubt circled solely on the resistor and transistor. This was just an example.

Anyhow, my doubt has now cleared up

[Frink42]

I actually believe what he meant was "why are you using two batteries to bias the transistor instead of using one+voltage divider.