Topic: BJT (PNP) transistor
Given : Beta (current gain) is very high
I_base = I_emitter / beta = 0 mA
Question : How did they arrive to the voltage V_10 = 0. 73 V Using the voltage divider rule .

My calculations arrive me to V_10 = 3.37 V .
My work is incorrect , the correct solution is 0.73 V

Question

Topic: BJT (PNP) transistor 
Given : Beta (current gain) is very high 
             I_base = I_emitter / beta  = 0 mA
Question : How did they arrive to the voltage V_10 = 0. 73 V Using the voltage divider rule .

My calculations arrive me to V_10 = 3.37 V .
My work is incorrect , the correct solution is 0.73 V

mrhoola · Answer

attachment

kenljw · Answer

The Bsse equivalent circuit is VBB = 3[150/(150 + 91)] - 3 [91/(150 + 91)] and RBB = 150||91 K
VBB = 3/(150 + 91) [150 - 91] = 177/241 = .73444
RBB = 56.64 K
Beta = hFE
IE = IB (hFE +1)
IC = IB hFE

mrhoola · Answer

Very well explained .

This is the first time I have ever seen the voltage divider method being applied in this manner . 

Is there anything that you might suggest that would make this seem more intuitive ?

kenljw · Answer

I'm sure your familiar by now with loop/mess and nodal analysis, but generally with just two nodes complex or not voltage/current rules are best aplied. In this case the loop is from one source to another threw the bias network and one should use the principle of superposition to solve for Thevenin/Norton  equivalent voltage/current source and the Thevenin resistance which is equal to Norton resistance.  The only need to use loop/mesh or nodal analysis is when dealing with circuits with tree complex nodes or more that can't be reduced to two.