Quenstion about resistor around transistor(HELP)

HI. I want to know how to choose suitable resistors(value) around transistor.
this contents is from Electronics all in one for Dummies.
Please refer the attached file, and answer the question.

Contents in Book
- R1: 330ohm resistor limits the current through the LED to prevent the LED burning out. You can use Ohm's law to calculate the amount of current that the resistor will allow to flow. Because the supply vol is 6v, and the LED drops 1.8v, the voltage across R1 will be 4.2v (6-1.8). Dividing the voltage by the resistance gives you the current in amperes, 0.127A

- R2 : You can use Ohm's law to calculate the current at the base. Because the base-emitter junction drops about 0.7v, the voltage across R2 is 5.3v

Question 1 : I don't know how to use Ohm's law when there is transister in schmatic. In the Contents, "Because the supply vol is 6v, and the LED drops 1.8v, the voltage across R1 will be 4.2v (6-1.8)." this means it will ignore transistor's inner resistors when using Ohm's law? or it uses Ohm's law from voltage source(+6v) to point "a"??

Question 2 : "Because the base-emitter junction drops about 0.7v, the voltage across R2 is 5.3v" -> how could i know base-emitter junction drops is about 0.7v?? i experiment with various resistors, and this junction drops seems to change every time. Or could i know this in transistor's data sheet?(if it is, could you let me know where can i find it?? i attach the datasheet. please refer sentence(or paragraph) indicating this.)

Question 3 The calculating(Ohm's law) will be different if i put R3 under the point "c"?? i mean R3 between point "c" and ground. How it will change?? please let me know ...

Question 4 : How can i choose certain resistor's value like 330 Ohm, 1K in this picture?? is it just intuition??? or simulate with Circuit Simulate Software before?? I don't know which resistor i have to use when using transistors.

I stuck in depreesion because of this. I can not go further cuz i can't understand how to use and apply transistor T-T... Please Help me !

Before addressing your Qs lets make some obvious statements: in this application the transistor works as a switch ie. it replaces mechanical On-Off device. So it works in completely non-linear situation: voltage of 0.7V applied to B(ase) – switch-ON, voltage below 0.7V – switch-OFF.
In mechanical devices you assume that the voltage across a switch (ON) is 0V – ideal switch has 0-resistance.
The same assumption applies here, however in practice voltage across saturated transistor will be 0.1V-0.5V and for simplifications you just assume it is 0V

How could you know the B-E junction drops about 0.7V?
If you haven’t study electronics you don’t know. However, brief Google search on n-p junctions (and in a simplified model an NPN transistor is N-P-N series of junctions) will give you an idea of 0.7V voltage drop – so, now you know!

The above should address Q1 and Q2..
(*)In Q3 you ask whether you can place another resistor between ( c ) and ground.
Lets start with the LED’s current. At this moment it will be (6V-1.8V)/R1.
If you add another resistor, R3 (c—ground) the LED’s current: (6V-1.8V)/(R1+R3).
Once you calculated the LED’s current you have to add I*R3 + 0.7V = U(base) and re-calculate the value of R1.
To be more accurate, you have to add I(base) current to I(Led) and re-calculate the voltage drop across R3 and repeat the calculations from (*).

In Q4 you asked how to choose resistor values. Well, you start with the estimation of the LED current from its data sheet.
Then, most of transistors can be assumed as current amplifiers with the amplification Betta of roughly 100: I(collector) = I(base) * 100
So if the LED’s current is 10mA, the Base current can be estimated to be 10mA/100, roughly 0.1mA

And with all the above assumptions OHM applies ..

:wink:
IanP

The calculation of 4.2V/330 ohms= 127mA IS WRONG. The correct answer is only 12.7mA.

The beta of a saturated 2N2222 transistor is 10 (not 100) on its datasheet. Then the saturated transistor here needs a base current of 1.3mA.

I also answered this post at Electronics Lab a few minutes ago.