am my steps about my question solution about bjt correct?

[circuit975]

Hello there
I am faced with such a question


1) Calculate and determine the value of the resistor RE that will ensure that VC=6 V.

2) Redraw the transistor model of the equivalent circuit for small amplitude AC signals. Determine the voltage gain Av of the circuit.

I solved the question as follows ;

VB = [24K / 24K+82K] x 16V = 3,62 Volt

RB = 24Kx82K / 24k+82k = 18,56K

6 = 16 - ICx5,6k
Ic = 1,78mA

IC = ßxIb = 150xIb
Ib = 11,86 microamper

IE = 11,86x10^-6 x 151 = 1,79mA
re = 26mV / 1,79mA = 14,5
VB - VBE - IBxRB - IExRE = 0
3,62 - 0,7 - [ (11,86 x 10^6) x 18,56K ] - 1,79mA x RE = 0
RE = 402K

Av = -ß x Rc / ßre + (ß+1)RE
Av = -150 x 5600 / 2,175 + 0 (The reason why it is 0 is because the Re resistor and capacitor are in parallel, I know that it is an ineffective element in AC analysis.)

This is my solution, are my solution steps correct? or if there is a point where I made a mistake when converting units, can you warn me, I need to correct my shortcomings and learn the right one. thanks a lot in advance

 

[Diver300]

RE will be less than 5.6 kOhms, as the voltage is less than the collector resistance and the current is slightly more, so you've got something wrong.

As a first estimate, ignore the base current. The emitter voltage will be 3.62 - 0.7 = 2.92 V. The emitter current will be 1.78 mA, the same as the collector current, so the emitter resistor needs to be 2.92 / .00178 = 1.6 kOhm

If you want to be more accurate, you can add in the base current to the emitter current, making it 1.792 mA. Also the base current will reduce the base voltage by 18560 * 0.00001186 = 0.22 V, so the emitter voltage becomes 2.70 V and the emitter resistor is 1.5 kOhm

However, the base-emitter voltage of 0.7 V will depend on temperature, so adjustments of 0.2 V or so probably don't matter on a real circuit.