please help transistor amplifier

michael1978

o sorry joney i make you tired, i understand now, so if load current is 50μ we chose voltage divider current 10 larger then the load current, and we have

R2 = 1.7V/(10*Iload) = 1.7V/((this is 10 time larger)10*50μ(base current)) = 1.5V/500μA = 3.4K this i understand

but this i dont understand
I_R1 = 10*Iload + Iload = 11*Iload = 550μA

10*Iload(50μA)+Iload(50μA)=11(how we get 11 now)*Iload(50μA)=550μA

i dont understand that 11 how we get? 10*Iload(50μA)+Iload(50μA)=0.55 not 11 sorry i dont know where i make mistake, where come this 11, i dont know how to calculate, i know 11*Iload is 550μA
thnx for answer

Jony130

Solve this pure math problem
B = 10A + A

Also from KCL we see that R1 must provide current for R2 resistor and also the current for the load.

So if Iload = 50μA and we pick voltage divider 10 times load current we have this:

I_R1 = Iload + 10 times load current = 11 times load current = 11*50μA = 550μA

michael1978

i am with calculator in my hand, i make mistake with my calculator but i dont understand i do it 0.050+10*0.050=0.55 how to do it ? CAN YOU TELL ME PLEASE HOW TO CALCULATE WITH CALCULATOR,

michael1978

Can you tell me how to calculate with calculator joney

Jony130

LoL
Use Google
https://www.google.pl/search?q=0.050...-a&channel=rcs

michael1978

haha i rely dont find i find one like min 0.050+10*0.050=0.55, joney were i make now mistake, with my calculator pfff i am shame

JONEY can you tell me please how to calculate in my calculator please

michael1978

Joney i know you know how to do it with calculator, can you tell me because i have in my hand calculator, i can find nothing in google the same result please....

Jony130

Simply use your brain instead of a calculator.
And if you still don't know how you should quit EE and start learn math.

michael1978

JONEY i use me brain, why you dont tell me to calculate with my calculator, you just help me, i thank you for help, just tell me a little bit, and i use a calculator, you help so much, help also this one

Jony130

But I don't understand what is your problem?
I_R1 = 50uA + 10*50uA = 50 + 500 = 550uA = 0.55mA = 0.00055A

michael1978

joney i know this from begin to calculate this is easy, but my problem was other you forget,
i ask you for this

-------------------------------
o sorry joney i make you tired, i understand now, so if load current is 50μ we chose voltage divider current 10 larger then the load current, and we have

R2 = 1.7V/(10*Iload) = 1.7V/((this is 10 time larger)10*50μ(base current)) = 1.5V/500μA = 3.4K this i understand

but this i dont understand
I_R1 = 10*Iload + Iload = 11*Iload = 550μA

10*Iload(50μA)+Iload(50μA)=11(how we get 11 now)*Iload(50μA)=550μA

i dont understand that 11 how we get? 10*Iload(50μA)+Iload(50μA)=0.55 not 11 sorry i dont know where i make mistake, where come this 11, i dont know how to calculate, i know 11*Iload is 550μA
thnx for answer
----------------------------------
joney this 11 is my problem

Jony130

OMG
Look here
We have this equation
B = 10*A + A = 11*A
The first part tell as that we have a ten apples and we need add one more apple.
So we end-up with eleven apples .

michael1978

ah you mean so
10*0.0.5+0.0.5=
10+0.5+0.5=11

you mean like this

michael1978

ooh joney boy, i make again mistake of is right, ooo thank you try a lot to explain me

Jony130

In our cases we have 11 apples
B = 10*A + A

And apples are equal to 50uA of load current.
So we substitute for A = 50uA
B = 10*50uA + 50uA = 550uA
And this is equal to 11*A = 11*50 = 550uA
the 11 tell as that the R1 current is 11 times large the load current.

michael1978

thank you joney for losing time for me and help i go to sleep tomorow work have a nice sleep

michael1978

thank you ver much joney now i start to understand