Calculating Emitter Current - Voltage-Divider Bias

[?n0t_A_nUmb3R?]

I am aware that there are two methods of working out I_E for a Voltage-Divider circuit.

The first is:
https://www.google.co.za/search?q=V...redoc.org%2Fdocs%2Findex-114597.html;2048;730
And then work out V_E = V_B - V_BE

And the second is:
https://www.google.co.za/search?q=v...rcuits.com%2Fvol_3%2Fchpt_4%2F10.html;480;257
(V_TH and R_TH)

I then tried to calculate a specific problem using both methods. The results were not consistent and varied.
My question is can we use both methods (first and second above) interchangeably?
Are there any differences between the two methods?

 

[mfb]

I then tried to calculate a specific problem using both methods. The results were not consistent and varied.

Then please show what you did and which setup you mean. Otherwise it is hard to tell what went wrong.

 

[?n0t_A_nUmb3R?]

There is a difference of 0.31mA. Is this still acceptable?
Can I also use these methods interchangeably? (Voltage Divider Method and V_TH/R_TH Method)

 

[mfb]

I don't understand your equation for I_E. A difference of 6% is a bit large for rounding errors.

 

[?n0t_A_nUmb3R?]

As I know it to be, it is basically simplifying the circuit using Thevenins Theorem.
Problem is I don't know where to use it.
From the examples I've worked it seems as if I can use this method OR the voltage divider method.
But in some instances, there is a considerate difference in my calculations.
My lecturer is out of office at the moment. So unfortunately I cannot email him.

 

[?n0t_A_nUmb3R?]

I do realize that the Thevenins method is for loading effects and the voltage-divider method is for unloaded effects. But I have come across questions where they do not mention whether the circuit is loaded or not.

 

[NascentOxygen]

You must inevitably arrive at the same answer, unless you make different or wrong approximations along the way, or through oversight or blunder.