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

Discussion in 'General Electronics Discussion' started by BretonDP, Apr 7, 2014.

1. ### BretonDP

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Apr 7, 2014
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 2. ### Harald KappModeratorModerator

11,446
2,628
Nov 17, 2011
Hello and Welcome to the forum.

A voltage drop across a resistor (an LDR is a kind of resistor) can develop only if a current flows according to Ohm's law: R=V/I.

If you connect one end of a resistor to a voltage source (e.g. battery) and measure from the other end of the resistor to the other pole of the battery, you will see the full battery voltage regardless of the resistance of the resistor, becaus no current flows through the resistor (o.k., this is a bit simplified because even a good voltmeter has a finite resistance and therefore a very small current will flow).
In a voltage divider setup there is a closed current path fromn the voltage source through the LDR, the second resistor (in series with the LDR) to the other pole of the voltage source. Therefore a current will flow and a voltage will drop across each of the resistors.

3. ### BretonDP

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Apr 7, 2014
I understand that, thank you.

What if, however, the other end of the LDR is going into, say, an IC (ignoring the middle state), or the base of a transistor? This is more the focus of my question In the original circuit, I connected the components so that one terminal of the LDR was in Vcc and the other end was going into a transistor base. Every other example that did the same thing used a voltage divider, and I wasn't understanding why the voltage drop from the LDR was enough to saturate the transistor. Is it because of the current?

Last edited: Apr 7, 2014
4. ### Harald KappModeratorModerator

11,446
2,628
Nov 17, 2011
The voltage drop across the LDR doesn't saturate the transistor. A bipolar transistor is controlled by the current flowing into its base. You can control the current by adjusting the resistor. This will, without other measures like feedback etc., give you only a rather crude control over the behaviour of the circuit as it strongly depends on tolerances aof the components, temperature, supply voltage etc.
If you want to have control over the circuit, you need to compensate these effects. One way is by using the LDr+resistor as a voltage divider and to evaluate the voltage from the divider using a stabel circuit like e.g. a comparator, preferably with a Schmitt-Triggger characteristic.

5. ### BretonDP

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Apr 7, 2014
Ok, I think I am understanding.

So, in my attachment, even though the circuit below is also viable, we only use the one with the voltage divider (the one above) because of the higher reliability and control? Or is the one below just not viable at all? (ignoring the values, I just sketched this up with defaults)

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6. ### Harald KappModeratorModerator

11,446
2,628
Nov 17, 2011
The top circuit will not work. It needs at least a conenction of the two batteries' "-" poles. Even with that connection in palce the base of the transistor will still draw considerable current from the voltage divider. It will also limit the voltage across R1 to the base-emitter voltage of the transistor (~0.6V). In essence, the circuit is not very different from teh lower circuit, only that some current is bypassed from the base of the transistor by R1.

Neither of both circuits gives you very good control over the behaviour with respect to parameter variations, temperature etc.

Here is a better example.. Note that this is just one example. Google will show you tons of circuits using LDRs as light detectors.

7. ### BretonDP

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Apr 7, 2014
Thank you for your insights, it has given me something to study Thank you for your replies.

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