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I am reading about the pull-down resister here.

Question 1.

The circuit schematic of a light sensor using a voltage divider circuit is shown as:
image http://www.doctronics.co.uk/images/vdiv3.gif
LDR has a resistance of 0.5k\$ \Omega \$ in bright light, and \$ 200k\Omega\$ in the shade.

In the shade, \$V_{out}\$ will be: \$\dfrac{200}{210} \times9 \times \dfrac{k\Omega}{k\Omega}\times V =8.57V\$
In the bright light, \$V_{out}\$ will be: \$\dfrac{0.5}{10.5} \times 9 \times \dfrac{k\Omega}{k\Omega}\times V=0.43V\$

So by applying the Voltage Divider Rule we come to know that the circuit gives a high output voltage in the shade and a low output in the bright light. So if we use this circuit with a bulb connected at the output then in the night the bulb should glow.

There is a problem. The bulb which is to be connected has its own resistance which might be lower than \$200k\Omega\$. Let's say the bulb to be connected has resistance \$100 \Omega\$. Since the bulb is in parallel with the LDR, so in the night the equivalent resistance of this parallel combination is approximately \$100\Omega\$.
Applying the voltage divider rule we will find that the bulb will not glow in both, dark and bright light. So the circuit is impractical and is of no use.

My question is:

  • Does the voltage divider circuit have no practical importance/use?

Question 2.

Here it is explained that we usually use a pull down resister of very high resistance of nearly \$10k\Omega\$. Rather using a high resistance resister we can left open the terminals at which pull-down resister is connected. By doing so we will obtain \$\infty\$ resistance for pull-down resister and whole of the \$V_{in}\$ will appear at \$V_{out}\$.

Why we use a resister of \$10k\Omega\$ not an open circuit for pulling whole of the \$V_{in}\$ at \$V_{out}\$.

Thank you.

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It looks like the issue you're facing is that you are trying to combine too many functions into a single circuit. As you have identified, if you put a small R_Load resistance in parallel, then you are effectively eliminating the contribution of the LDR to the circuit. You still have a voltage divider, but not dividing the way you want. Looking at the equation below, we can see that we will need a R_load on the order of magnitude of the dark resistance of the LDR to maintain the expected operation.

$$V_{out}=\dfrac{V1\times (LDR||R_{load})}{R_1+(LDR||R_{load})}$$.

schematic

simulate this circuit – Schematic created using CircuitLab

Taking your values, that means that we really want the effective R_load to be greater than 200 kΩ. To state this in another way, this gives us a high impedance signal only capable of driving a high impedance load. Our 100Ω light bulb is, by comparison, a very low impedance. Either we need to change the impedance of our voltage divider to a much lower value (which can be difficult or impractical), or we need to amplify the output of our voltage divider to be able to drive our lamp. In this case, amplifying things is very easy as it really just needs a single transistor, as shown below:

schematic

simulate this circuit

These circuits take the high-impedance signal of the voltage divider and turn it into a low-impedance drive capable of turning on your light bulb.

For button presses, you are usually driving a pin on a microcontroller that may have a very high impedance, probably greater than 1 MΩ. Also, instead of using a LDR that varies from 0.5 kΩ to 200 kΩ, we use a switch that varies between (approximately) 0 Ω and ∞Ω. If we are using the switch as the pull-down device, then we need a pull-up resistor to be able to drive the pin back to our supply voltage.

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  • \$\begingroup\$ Ok, means that whenever we have the voltage divider circuit at hand, we have to make the input impedance of the load very high. \$\endgroup\$ – user31782 Dec 6 '14 at 19:38
  • \$\begingroup\$ @user31782: I've read through this whole thing and I'm thoroughly confused as to the nature of your question and/or just exactly what it is you want to do. Ignoring everything that's been posted so far, could you explain, please, in a couple of sentences, what you need to find out in order to award the bounty? Thanks. \$\endgroup\$ – EM Fields Dec 6 '14 at 20:47
  • \$\begingroup\$ @EMFields I wanted to make sure that that website misses the point that the bulb to be connected at the output should have high enough resistance. I asked this question to some of my friends and teachers but they did not accept that the voltage divider requires a high input impedance circuit. That's it, I wanted to make sure that my understanding is correct. \$\endgroup\$ – user31782 Dec 6 '14 at 20:50
  • \$\begingroup\$ So as to what I need to find out in order to award the bounty-- Just an answer to my question. \$\endgroup\$ – user31782 Dec 6 '14 at 20:54
  • \$\begingroup\$ I'm still at a loss. Do you want a circuit where the lamp turns ON/OFF sharply as the LDR is more weakly/strongly illuminated? Do you want a circuit where the lamp brightens and dims more or less gradually as the LDR's illumination changes? Do you want a circuit and some math to explain the operation of a voltage divider? Something else? \$\endgroup\$ – EM Fields Dec 6 '14 at 21:00
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There are a number of conflicting things in your question, so it can't be answered directly.

First, are you sure the LDR resistance goes up with light? That is possible, but not how the common CdS sensors work. Those have a significant decrease in resistance with increasing light.

Second, the output voltages you get make no sense at all without even looking at the numbers closely. The bottom resistor is 10 kΩ, so is significantly larger than the LDR resistance (200-500 Ω) regardless of light. Therefore the output should always be just below 9 V. Clearly 420 mV can't be right without looking any further.

Third, trying to see where you got the output voltages from leads to more confusion. You have a bunch of dimensionless numbers without any indication where they came from on the left side, but the result is in Ohms on the right side. That can't possibly be right just from dimensional analisys alone.

The way this circuit works is to cause a voltage that is dependent on the resistance of the sensor. This is useful because most things you would put downstream of this circuit will want a voltage signal, not a resistance signal. The A/D input of a microcontroller is a good example of something that wants a voltage signal. You can think of this circuit as being a resistance-signal to voltage-signal converter.

A resistive voltage divider is a simple way to achieve this. The output of the divider is a function of the two resistances. With one resistance being a known constant, the output voltage is a function of the other resistance, which is a function of light level.

You say the dark resistance is 200 Ω. That means the voltage divider gain will be (10 kΩ) / (200 Ω + 10 kΩ) = .980. That times the 9 V input yields 8.82 V output. For the light case, do the same computation except that the LDR resistance is now 500 Ω:

(9 V)(10 kΩ) / (500 Ω + 10 kΩ) = 8.57 V

The reason that you're not getting much variation between light and dark is that the fixed resistor is mismatched to the variable resistance. Pick a fixed resistor near the middle of the variable resistor's range. The optimum case in terms of resolution is to pick one so that the min and max variable resistances result in voltages centered around midway. Depending on the situation, you may wish to sacrifice largest overall resolution to gain more resolution in a particular part of the range. However, in all optimal cases the fixed resistance is bounded by the min/max value of the variable resistance.

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