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Voltage divider circuit and the Pull-down resister.

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Anupam
#1
Mar12-14, 11:54 AM
P: 6
Am reading pull-down resister herehttp://www.doctronics.co.uk/voltage.htm.

PART-1

The circuit schematic of a light sensor using a voltage divider circuit is as shown:

LDR has a resistance of 500Ω, 0.5kΩ , in bright light, and 200kΩ in the shade.

in the shade, V_out will be: (10/210)9=0.42V
in the bright light, V_out will be: (10/10.5)9=8.57V by voltage divider rule

so by applying voltage divider rule we came to know that the circuit will give a high output voltage in the bright light and low output in the dark. So if we use this circuit with a bulb connected at the output then in the afternoon the bulb should glow.
But there is a problem the bulb which is to be connected has its own resistance which might be lower than 10kΩ. Let's say the bulb to be connected is of resistance 100Ω. Since the bulb is in parallel with the 10kΩ resister the equivalent resistance of this parallel combination is approximately 100Ω.
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 has no practical importance?

PART-2

Here(http://www.doctronics.co.uk/voltage.htm#switches) it is explained that we usually use a pull down resister of very high resistance of nearly 10kΩ. 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 ∞ resistance for pull-down resister and whole of the Vin will appear at Vout.

Why we use a resister of 10Ω not an open circuit for pulling whole of the Vin at Vout.
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phinds
#2
Mar12-14, 12:04 PM
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Quote Quote by Anupam View Post

Does the voltage divider circuit has no practical importance?
That's just silly. Of COURSE a divider circuit is useful. You just have to use it in a circuit that deals with it properly. In this case you need to follow the divider with an amplifier (could be just a single transistor)

PART-2

Why we use a resister of 10Ω not an open circuit for pulling whole of the Vin at Vout.
With no pulldown resistor, there won't be any current. Again, you need to put this construct in a circuit.

You are taking things on their own where you need to think about how they are used in a circuit
Anupam
#3
Mar13-14, 02:28 AM
P: 6
@phinds You say: "you need to follow the divider with an amplifier". As i understand you mean that since the current flow through the amplifier will be very small so its resistance will be >>10k\Omega. But then we can directly connect $V_{in}$ to the amplifier because its resistance is >>10k\Omega so whole of the input voltage V_{in} will appear as V_{out}. So we do not need voltage divider circuit. So there is no practical advantage of using a voltage divider having 10k resister.

Windadct
#4
Mar13-14, 08:56 AM
P: 553
Voltage divider circuit and the Pull-down resister.

You are using a SENSOR circuit to drive a LOAD - In some cases this can be done, in some cases not - then an AMP (as suggested above) or possibly a Relay needs to be used to provide the power to the load based on the condition of the sensor.
sophiecentaur
#5
Mar13-14, 10:53 AM
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Quote Quote by Anupam View Post
@phinds You say: "you need to follow the divider with an amplifier". As i understand you mean that since the current flow through the amplifier will be very small so its resistance will be >>10k\Omega. But then we can directly connect $V_{in}$ to the amplifier because its resistance is >>10k\Omega so whole of the input voltage V_{in} will appear as V_{out}. So we do not need voltage divider circuit. So there is no practical advantage of using a voltage divider having 10k resister.
Unless you actually know the resistance value of the load, how would you know what voltage corresponds to which temperature? Undefined or inadequately defined operating conditions are bad Engineering and will produce problems, you can be sure. The Load resistance and the LDR actually do constitute a voltage divider but what is the ratio?

Best practice is to follow with an amplifier with high input impedance (buffer) and use an explicit (calculated appropriately) value of pull-down, Then, if you happen to connect a different (in detail) amplifier, the sensor will still give you 'sense'.

Some cheap and nasty domestic equipment designs can use a single component to fulfil several functions, simultaneously. Try fault finding a circuit with a lot of those in it. A nightmare.


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