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!