This is the problem...

Finding for the Voltage Input. I get 3.23v, but i know it's not the correct answer. pls guide me. thanks

 

Did you include the voltage drop across the LED in your calculations?

 

i think the voltage drop of LED is neglected, as it only just need 20mA on Ic , The problem states that What voltage (Vin) does it need to produce that 20mA on Ic

 

i think the voltage drop of LED is neglected, as it only just need 20mA on Ic , The problem states that What voltage (Vin) does it need to produce that 20mA on Ic

The question doesn't say to neglect the LED voltage which can vary from 1.5 to 4 volts depending on the colour of the LED. Thus you have to include it since Vcc is just 10V and 2V across the LED can be quite significant.

Do you have the correct answer?

 

Hmm,
Ib_min = 200uA so if safety margin is 2*Ib_min = 400uA
Vin = Ib*Rb +Vbe = 2V+Vbe
Or
Vin = 200uA*5K + Vbe = 1.7V----> 2*1.6V = 3.4v

 

I got Viin=2.32V.

But I used IB (min), for the calc.

 

Allowing around 3V for the LED Vf the Rc=300 Ohm and +Vcc of 10V will allow about 22mA with the transistor in saturation.

To allow exactly 20mA with 100Beta, if you double the volts as instructed Jony has it almost right.

Double only the volts across the resistor.

Hmm,
Ib_min = 200uA so if safety margin is 2*Ib_min = 400uA
Vin = Ib*Rb +Vbe = 2V+Vbe
Or
Vin = 200uA*5K + Vbe = 1.7V----> 2*1.7V = 3.4v

1 Volt across the resistor *2 +Vbe = 2.7 Volts

 

The problem is set up badly, but you know what?

You will get a lot more poorly thought out questions and proposals when you get out into real work. Know what you need to know and decipher the intended question, even when they don't know what they are trying to ask. It is a like a puzzle game sometimes, and if you learn to have fun with it and have a sense of humor about it you will find things hilariously fun instead of horribly annoying.

 

Thanks for all your replies, and your time to answer my problem. Our instructor solved it with an answer Vin = 3.93v

 

How did the instructor handle the 1.5 - 2V drop across the diode?