# transistor amplifier help

Discussion in 'Homework Help' started by manish1991, May 7, 2014.

1. ### manish1991 Thread Starter New Member

Jun 17, 2013
11
0
hello, I am trying to design a transistor amplifier with the following characteristics:

Vcc=9V
C1 = 100nF
C2 = 100nF
hfe = 350
ic (quesent) = 18mA

I have to find the resistor values of the circuit, however cannot seem to find the correct value for RL. Its probably easy but i can't seem to remember for the life of me.

I have managed to calculate R1 and R2 values along with RE.

My calculations:

IB = (5.14X10^-5)A
IC = 18.03mA

Firstly:
R2=(VRE+ VBE)/(10 X IB )
R2=(1+ 0.6)/(10 X (5.14X10^-5)

R2=3.11 KΩ

Secondly:

R1=(VCC-(VRE+VBE))/(11 X IB )
R1=(9-(1+0.6))/(11 X (5.14X10^-5)

R1=13.09 KΩ

RE=VRE/IE
RE=1/(18.08X10^-3)

RE=55.31Ω

Can anybody show me how to calculate RL as I am stumped??

I have uploaded a circuit diagram

Thank you

http://i695.photobucket.com/albums/vv313/patel5557600/ScreenShot2014-05-07at132542_zpsf1242ee7.png

Oct 2, 2009
12,624
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3. ### manish1991 Thread Starter New Member

Jun 17, 2013
11
0
Not quite sure how to use the load line

I used the following...

RL = VCC/IC
RL = 9V/18X10^-3
RL = 500ohms

Would I be correct?

4. ### #12 Expert

Nov 30, 2010
16,665
7,310
No. You've calculated, "What amount of resistance will use up the entire 9 volts?".

5. ### PRS Well-Known Member

Aug 24, 2008
989
35
You have given us an incomplete specification. You need to include the required gain. Otherwise there is a large number of values RL could be. At the collector we have (Vcc - Vc)/Ic = RL. Notice that without the gain specification RL is a variable.