Process Control - Challenge 3.6 - Error in solution

rme_2001

Hello everyone,

I have a question about challenge 6 from chapter 3 from the Process Control Text·where the following question is asked:

If R_E is decreased to 100 Ohm, with a V_in 5V and R_B 47K Ohm and h_FE 100, will I_B still be able to drive the transistor into saturation?

Based on a transistor in a·common-collector circuit, my·calculations:

I_B = (V_in - 0.7) / (h_FE R_E + R_B)
I_B = (5 - 0.7) / (100 * 100)·+ 47.000
I_B = 4.3 / 57.000
I_B = 0.0754 mA

I_C = I_B·h_FE = 7.54mA

I_SAT = (Vdd - 0.3)/R_E
I_SAT = 4.7/100
I_SAT = 47mA

I_C < I_SAT Saturation current can't be reached

However, the Solution sais the current can be reached, with (i think) an error in calculating the I_SAT
But since i'm very new to the whole concept, i just wanted to double check here.
They use the following calculation:

I_SAT = (Vdd - 0.3)/R_E
I_SAT = 4.7/100
I_SAT = 4.7mA

While typing this i've pretty much drawn the conclusion they divided by 1000 and therefore based the conclusion on the wrong I_SAT .
But ah well, i allready typed it, so maybe someone else can double check.

Thanks in advance [noparse]:)[/noparse]



_·