This graph shows that the Ic/Ib ratio is constant at 10:
View attachment 170574
And shows the rise in Vce when base current is insufficient to saturate the transistor.

How does that graph show that the Ic/Ib ratio is constant at 10?

Each curve is for a specific value of Ic. Since the horizontal axis is Ib, if the Ic/Ib ratio were constant, each curve would be a vertical line.

The only point where Ic/Ib is 10 is at the right-most end of each curve -- and that's just because that's where they chose to stop plotting the line. They could have continued plotting it and the curves would have continued to go downward.

 

How does that graph show that the Ic/Ib ratio is constant at 10?

It's inferred because they stop showing data when Ic/Ib is 10.

If you want to be nitpicky, here you go:

 

That still doesn't mean that Ic/Ib is somehow a constant. What would that even mean? That the device isn't in saturation if Ic/Ib is 11 or 9?

The Ic/Ib ratio is a continuous function several different parameters, most notably Vce, Vbe, and temperature. That set of curves merely shows that they CHOSE to characterize the device by ADJUSTING the operating point until they achieved an Ic/Ib ratio of 10 before taking their data.

There's nothing magical about an Ic/Ib ratio of 10 -- it is quite arbitrary. By convention, it is usually chosen as the nominal operating point when talking about performance in saturation and so device manufacturers tend to use it as long as it is at all reasonable for that device. This is simply for consistency so as to make it easier to quickly compare different devices.

High power transistors often use a different ratio precisely because a value of 10 is often NOT a reasonable choice.

 

All this talk about saturation is probably moot, considering the single stage circuit being discussed here, I think if the TS wants the output Q to be in saturation they will need another stage.

 

That still doesn't mean that Ic/Ib is somehow a constant.

Where did I say that?

 

Where did I say that?

Uh... how about here:

This graph shows that the Ic/Ib ratio is constant at 10:
View attachment 170574

 

Hello, I am new to the site and to transistors. So I have read about the transistors few articles already and watched few tutorials(about the formulas of determining Ibmax and showing some example circuits)and decided that I have some small knowledge to build a simple circuit. I use 6v lead acid battery, transistor 2n3904, two resistors(300 Ohms, one for base of the transistor and one for the led that I try to lit), LDR at max resistance 50kOhms and one led with Imax=25mA. I have chosen resistor with resistance 300 Ohms for the base, because I calcula ted that with 6 volts I will have current at the base around 20mA and this will be enough to saturate the transistor making it possible for current to flow from the collector to the emitter.
I have looked at the datasheet of the transistor too, in order to check for Icmax, which is 200mA. I have calculated Ibmax at beta 20 and got 10 mAmps and at 6 volts I get that I need 600 ohms resistor.
So thought using something like 300 Ohms(the closest to 600 I had) will be okay, also I have 2 kOhms but I am afraid that this might let too less current in the base
The problem though is that the led is not being lit/glowing.
Am I wrong in the calculations ?
Datasheet that I have used: http://html.alldatasheet.com/html-pdf/11470/ONSEMI/2N3904/365/2/2N3904.html
Can you please tell me if I have made some of the readings wrongly(Vbe = 0.65V, Icmax= 200mA, hfe= 20, should I always take the lowest value of the hfe to determine resistor on the base) ?
The circuit that I have realized already in physical model on breadboard:

View attachment 170306

P.S. I have tried the same with 2 kOhms resistor(just from desperation, thinking that I give too much current with 300 ohms and this might fry the transistor) The result is the same(the led or the load in the case is not being lit)
Also if you know good books about transistors, their working and formulas explained simply, I am open for suggestions(the articles are not enough in-debt).

Hi pancake95,

I see you are having a few problems with your LDR (Light Dependent Resistor)/LED circuit.

Afraid to say that the circuit you have posted will not work, but not in the way that you are experiencing.. Instead, the blue LED should be on all the time regardless of the light falling on the LDR (assuming the LDR has a dark resistance of 50k and an illuminated resistance of 1k). But not to worry, there is a simple circuit that will do your job and the calculations are simple to follow..

So my advice would be to discover what is wrong with your circuit.

The first thing to try is to remove all components from the base of the transistor and then connect a 2K2 resistor between the base of the transistor and the 6V line. The LED should illuminate. If the LED does illuminate that indicates that the circuitry is functioning correctly up to the base of the transistor.

If the above test is successful move on to testing the LDR. With an Ohmmeter (DMM switched to Ohms) measure the resistance of the LDR in the dark: should be around 50K. Shine a bright light on the LDR: the resistance should drop to around 1k.

Let us know how you get on.

 

Uh... how about here:

I don't think a reasonable person wouldn't have interpreted that statement the way you did.

I meant that a beta of 10 is used regardless of collector current in the transistor. And we all know that that's a conservative number, but it's what the manufacturer suggests for operating conditions. And of course we wouldn't operate the transistor outside of the manufacturer recommended conditions.

 

I have been playing with the transistors I bought and one thing that I realized that as stated earlier in this post is that the LED will illuminate when there is light over the LDR and will turn off when there is no light over the LDR.
So I googled some solutions and saw that people are using circuit like this to make the LED turn on when it is dark and turn off when it is light(the picture given from me).

My questions are these:

How the LED turns on when there is base resistor of 100kOhms? There is not enough current to saturate(it is 0,00006 Amps I will get 0,0006 Amps as collector emitter current).
The LED is full on, I read somewhere that when it is light the circuit get shunted and does not goes to base but instead through the LDR(which in the moment is with less resistance than the base resistor) as the electricity is "lazy".
But then again it comes the question how is possible for the LED to be fully on with 100k base resistor when we calculated that around 4k is enough(and it was enough, but the dark sensor circuit did not worked with the 4k before the LDR, and I get it that there is no way to be shunted when r1 is less than the LDR all the time).

Also is this circuitry enough for being used for home lightning with the lead acid battery that I have ?

And my last question is how to make the circuit be on or off and not make the LED "fade" with the slightest change of lightning over the LDR?

P.S. I tried the circuit with removed LDR and only 100k resistor connected to the base(removing the voltage divider and leaving only this resistor) and it was fully illuminated again, I do not understand how is it possible when the base current gets so small that there is no way the collector-emitter current to be enough for the LED to be fully on.

 

How the LED turns on when there is base resistor of 100kOhms? There is not enough current to saturate(it is 0,00006 Amps I will get 0,0006 Amps as collector emitter current).

hi p95,
You are assuming that the transistor is in the saturation mode by using a Gain of 10.
If you have a DVM, check the Collector voltage and you should see that it is in the Active region, so allow a Gain figure of 100.
So thats 9Vbty - 0.7Vbe = 8.3V ... 8.3v/100k = 83uA.

So, 100 * 83uA = 8.3mA.

E

Added a LTS sim, to show the Collector current and voltage.

 

And my last question is how to make the circuit be on or off and not make the LED "fade" with the slightest change of lightning over the LDR?

Use a comparator instead of a transistor to drive the LED.

 

Your schematic did not say "2.85V blue LED" so I calculated with a 2.0V red LED.

With the 100k base resistor then the transistor is not saturated since the 100 ohm resistor value in series with the LED is much too low requiring a high current for saturation. If the base current is high enough to saturate the transistor then your blue LED will almost burn out with a current of 6V - 0.4V - 2.85V)/100 ohms= 27.5mA.

For an LED current of 5mA then the resistor in series with the 2.85V blue LED is (6V - 0.2V - 2.85V)/10mA= 295 ohms, use 300 ohms.
then for a base current of 1mA you need a base resistor that is (6V - 0.7V)/1mA= 5.3k. Use 5.1k.

But an LDR does not have a low enough resistance when it is in light then the LED will never turn off.

 

The one thing not mentioned is that most LED devices will illuminate some at a current FAR SMALLER than the rated current for maximum output at published lifetimes. So while the brightness is impressive itis probably far less than the rated amount. And a transistor does not need to be near saturation to conduct some current, which will provide SOME illumination, as you have seen.

To reverse the action so that the LED switches ON when darkness happens, the circuit is changed so that a low resistance of the LDR removes the bias from the transistor. If the LDR is connected from the base to the emitter and the 100K resistor is connected from the positive source to the base this will happen, although the resistor value may need to be changed.