Doing any calculation in this type of a circuit is not a easy job. Why ? Because we have too many unknowns.
But we could try assume some and this might help us.

First let us assume that the LED will light if Id >1mA and the transistor current gain is Hfe = 150 and Vbe_on >0.6V.
To get 1mA or larger at collector we need Ib > 1mA/150 = 6.6μA.
And If Vcc = 5V and R1 = 100kΩ the LED will start to light if Rldr is larger than
Rldr > 0.6V/((5V - 0.6V)/100k - 6.6μA) = 16kΩ.

So the BJT is turn ON and its work on active region (Ic = hfe*Ib).

But as Rldr resistance increase (the sun light is getting weaker), more current from R1 can now flow into the transistor base. And this will increase Ic current (Ic = Hfe*Ib), larger collector current will cause more voltage drop across Rc resistor. And this will drop the voltage at collector. And when this voltage reach 0.6V the transistor is on the edge between active region and saturation.
And this will happens if Rldr is larger than:
Ic = (Vcc - Vd - Vce)/Rc = (5V - 2V - 0.6V)/680Ω = 3.5mA

Ib = 3.5mA/150 ≈ 24μA

And new Vbe value is 26mV*ln(24/6.6) + 0.6V = 0.635V

Rldr > 0.635V/((5V - 0.635V)/100k - 24μA) = 32kΩ

And this means that if Rldr is >> 32kΩ the BJT will be in saturation (Full-ON) and the Ic current reach his maximum value.
Ic_max = (Vcc - Vd - Vce(sat))/Rc = (5V - 2V - 0.2V)/680Ω = 4.1mA
And the LED is also at the maximum brightness

As for BJT. The transistor will be in cut-off if Vbe < 0.5V for a ordinary small-signal BJT.
http://forum.allaboutcircuits.com/threads/npn-base-resistor.97834/page-2#post-731057

As for the saturation try read this
http://forum.allaboutcircuits.com/threads/bjt-saturation-question-s.81150/#post-577296
http://forum.allaboutcircuits.com/threads/which-method-npn.97963/#post-731172

 

Rldr > 0.6V/((5V - 0.6V)/100k - 6.6μA) = 16kΩ.


Didint quite catch this part. Maybe could u write how did u get that in seperate equations?

What is Rc resistor? Can u add a circuit?

 

The circuit looks like this :



Rdlr = Vbe/I2 from KCL we have I2 = I1 - Ib

http://forum.allaboutcircuits.com/threads/kirchoffs-law-question.69406/#post-481466 (KVL, KCL)

Using Ohms law we can find:

I1 = (Vcc - Vbe)/R1 = (5V - 0.6V)/100kΩ = 44μA

And from my assumption that the LED will start emits a light if LED current is larger than 1mA.

And because Iled = Ic we have

Ib = Ic/hfe = 1mA/150 = 6.6μA

We now can find Rldr value at which the LED starts to shine.

Rldr = Vbe/I2 = Vbe/(I1 - I2) = 0.6V/(44μA - 6.6μA) = 0.6V/37.4μA ≈ 16kΩ

And this diagram shows the situation


And as Rldr value increasing, Ib and Ic also will increase. But Vce will drop (Vce = Vcc - Vd - Ic*Rc).
And just before transistor enters the saturation region situation looks like this



But this kind of calculation are pointless in real life. We simply use a POT in series with R1 (reduced to 1kΩ) and "set" the circuit light sensitivity via a POT.

PS. Ie - the emitter current is always equal to the sum of a base and collector current.
Ie = Ib + Ic

 

I still cant understand how can a transistor change between modes if the polarity of the batter doesnt change.. If i connected positive to the base and collector, and negative to the emmiter. The one junction will be forward biasend and other will be reverse.. How it can become saturated ( both junctions forward biased ) . I hope you understand what im asking , in simple words: How can transistor change between modes if the polarity remains the same?

 

But do you understand that as Ic current increasing his value the Vce voltage drop because Vce = Vcc - Ic*Rc ??
Also did you read this post #7
http://forum.allaboutcircuits.com/threads/which-method-npn.97963/#post-731172

 

I still cant understand how can a transistor change between modes if the polarity of the batter doesnt change..

Forget about batteries. The power supply voltage has almost nothing to do with a transistor being in active or saturation mode.

If i connected positive to the base and collector, and negative to the emmiter. The one junction will be forward biasend and other will be reverse..

Your description would be easier to understand if you drew a schematic.

How it can become saturated ( both junctions forward biased ) . I hope you understand what im asking , in simple words: How can transistor change between modes if the polarity remains the same?

In Jony130's example where the transistor was biased in the active region. If you gradually (or not) change the circuit to increase the base drive until further increases in base current result in little or no increase in collector current, you will have driven the transistor into saturation.

 

Hmm. Okay.. When do i know exactly transistor is saturated? Which equation i have to use?

 

Transistor is in saturation if Vce << 0.6V. Notice that if Vce = 0.2V and Vb = 0.7V the base-collector junction is now in forward biased.
Vbc = Vb - Vc = 0.7V - 0.2V =0.5V. And in saturation Hfe*Ib don't hold anymore.
But if Vc = 1V and Vb = 0.7V ----> Vbc = 0.7V - 1V = -0.3V so the base-collector junction is in reverse biased, and the transistor is in active region.

 

Yeahh, i think im starting to understand howw that works, but can you show me in the circuit this place?
Vbc = Vb - Vc = 0.7V - 0.2V =0.5V. And in saturation Hfe*Ib don't hold anymore.
But if Vc = 1V and Vb = 0.7V ----> Vbc = 0.7V - 1V = -0.3V so the base-collector junction is in reverse biased, and the transistor is in active region.
Those letters Vb,Vc,Ve,Vbc just too many of them , so its very hard for me to understand where is that voltage
Tryed to write it down on the paper so it will be easier to understand , but still cant understand what exactly happened there.

 

Vc - voltage at collector , the voltage difference between collector and our reference point, ground.
Vb - voltage at base, the voltage difference between the transistor base and the ground
http://www.ittc.ku.edu/~jstiles/312/handouts/312_Introduction_package.pdf (start reading from page 3)

 

Hm. I think i understand it wrong, because if i calculate using the equations i get the negative voltage if Vc=0,2V and Vb=0,7V.

 

Hmm. Okay.. When do i know exactly transistor is saturated? Which equation i have to use?

Use the circuit below to do an experiment. Attach your voltmeter across the collector (+ lead) and base (- lead) terminals. Adjust the pot to ground and gradually increase while monitoring the CB voltage. Report what you find.

 

Use the circuit below to do an experiment. Attach your voltmeter across the collector (+ lead) and base (- lead) terminals. Adjust the pot to ground and gradually increase while monitoring the CB voltage. Report what you find.
View attachment 96463

Having problems with multimeter atm, i will try this circuit on the paper and i will see if i can calculate everything by hand...
By the way can u check my last message, where i calculated the Vcb.. I was supposed to get positive voltage but its negative for me :/

 

By the way can u check my last message, where i calculated the Vcb.. I was supposed to get positive voltage but its negative for me :/

Your calculations are correct. Because what you have calculated is Vcb --> The voltage difference between collector and base. And you treat the base as a reference point (you connected your virtual voltmeter like this: Red probe to collector and Black probe to base).
Do you see the difference between Vcb and Vbc ?

 

Your calculations are correct. Because what you have calculated is Vcb --> The voltage difference between collector and base. And you treat the base as a reference point (you connected your virtual voltmeter like this: Red probe to collector and Black probe to base).
Do you see the difference between Vcb and Vbc ?

Yeah, and how do i know which do i measure ? V

 

Yeahh, and how do i know which side do i measure? Vcb or Vbc, to know if this junction is forward or reverse biased? But if the Vc is 0,2V, and its parallel to Vcb+Vb, and Vc=Vcb+Vb(because they are parralel) so if we sum up Vb+Vcb we should get 0,2V. If Vcb would be 0,5V, Vcb+Vb=0,7+0,5=1,2V. Thats not equal to 0,2?

 

Yeahh, and how do i know which side do i measure? Vcb or Vbc, to know if this junction is forward or reverse biased?

Draw a diode between base and collector. And from there you should see how to measure it.

But if the Vc is 0,2V, and its parallel to Vcb+Vb, and Vc=Vcb+Vb(because they are parralel) so if we sum up Vb+Vcb we should get 0,2V. If Vcb would be 0,5V, Vcb+Vb=0,7+0,5=1,2V. Thats not equal to 0,2?

And this is why Vcb is negative Vb + Vcb = 0.7V + (-0.5V) = 0.2V

 

Draw a diode between base and collector. And from there you should see how to measure it.


And this is why Vcb is negative Vb + Vcb = 0.7V + (-0.5V) = 0.2V

Yeah, imaging this as a diode helped! B is like a negative and C like positive, so if we measure the voltage reverse, we get the negative voltage

 

But I hope that you have your "diode" anode at base and cathode at collector.

And the best explanation of saturation you can find here
http://forum.allaboutcircuits.com/threads/bjt-saturation-question-s.81150/#post-577296

 

Yeah, Its also strange how can voltage be measured from both negative points? Lets say Vc=0,2V, but both measuring points : collector and ground is negative?