Circuit Analysis Help

[Rheostat.]

How do I solve for the resistance for R2?

 

[Unimportant]

7 hours ago, Rheostat. said:

How do I solve for the resistance for R2?

 

LDR1 and R2 form a resistive voltage divider, the formula for which is:

 

A transistor starts conducting at around 0.7Vbe.

With Vin = 5V and R1 = 220, you seem to be aiming towards 20mA of current trough the LED.

If we assume a wide ball-park gain of 100 for Q1, 20mA collector current will require 0.2mA base current. 0.2mA base current will drop 0.3V over R3.

So 0.7Vbe + 0.3Vr3 is about 1V.

 

Rldr1's value will drop with the more light it receives. Measure it's value with the amount of light you want your LED to come on.

Then use the above formula to calculate R2 with Vout = 1V, Vin = 5V and Rldr1 the value you measured.

 

The circuit is heavily dependent on Q1's gain, which can vary a lot between parts (even of the same type). You might want to use a potentiometer to make the setpoint easily adjustable.

 

[mariushm]

Well, that led is Red with typical forward voltage of 2v ... so 5v input, 2v on led ... 5v - 2v = I x R => I = 3/220 = 0.013 A so assume the circuit needs to let 15mA flow... or you can go with Unimportant's 20mA value.

2n3904  is a transistor with a wide range of hFe / gain ... again, what he says makes sense... I'd also assume around 100 gain.

 

Anyway... I always suggest this video to people who want to learn about transistors, maybe it provides some extra help: