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New to electronics and need help

kyffin11

Oct 25, 2016
4
Joined
Oct 25, 2016
Messages
4
Hello i have an assignment in which to design a series circuit for a switch and LED and resistor which i have completed, but the next part is about redesigning it and using a LDR instead of a switch, i have just replaced it but i am wondering if this is correct.

more info: 6V psu, 400ohm resistor.
 

Harley

Oct 21, 2016
16
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Oct 21, 2016
Messages
16
Hi there, welcome to the site! Do you have pictures of what you are doing?
 

davenn

Moderator
Sep 5, 2009
14,254
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Sep 5, 2009
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14,254
hi ya
welcome to EP

Hello i have an assignment in which to design a series circuit for a switch and LED and resistor which i have completed, but the next part is about redesigning it and using a LDR instead of a switch, i have just replaced it but i am wondering if this is correct.

more info: 6V psu, 400ohm resistor.

LDR's don't do well passing too much current .... using one in a switching circuit will usually always involve a transistor

google light activated switch ( you could add LDR and transistor to the search) there are dozens of circuits available :)


Dave
 

analog_engie

Oct 26, 2016
1
Joined
Oct 26, 2016
Messages
1
Hello i have an assignment in which to design a series circuit for a switch and LED and resistor which i have completed, but the next part is about redesigning it and using a LDR instead of a switch, i have just replaced it but i am wondering if this is correct.

more info: 6V psu, 400ohm resistor.

Hi kyffin11,

There is so many ways you can accomplish this task. Like the others said, you'll have to use a supraconductor switch like a mosfet or a transistor to do it. Here are the thing I assumed for your task:

-The LED must have a current of 10mA
- The photoresistor has those caracteristics: 500KOhms when dark, 30Kohms when lit.
- You will use a transistor which has approximatly 200 of hff (current multiplicator)

what you want to do is drive the base of your transistor with the photoresistor.

The equation for the current through the led is quite simple:

hff * (V / R) Where R is equal to the resistance of the LDR, hff = 200 and V = 3.6

(*Notice I wrote V= 3.6V and not 6V. The reason for this is I took into account the voltage drop of the LED 1.8V and the drop between the base and the emitter of the transistor 0.6V. What is left from this is 3.6V)

if it's dark R = 500K

the collector current (Ic) of the transistor will be 1.44mA and the base current (Ib) 7.2uA
This is not enough to light the LED.

Now shine something bright onto the LDR: R=30K

Ic =24 mA and Ib = 120uA

Since your circuit doesn't allow more than 10mA through it, the transistor will saturate and you will get a nice brightness from your LED.

You can experiment by adding a 42K resistor in serie with the LDR to drive the gate. As a result, every current mentionned above will be divided by 2.4. Since your transistor saturation current is 10mA it won't affect the "lit LDR" LED brightness.

See for yourself!

Capture.PNG
 
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