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Calculating Emitter Current  VoltageDivider Bias 
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#1
Jun2014, 02:34 PM

P: 12

I am aware that there are two methods of working out I_{E} for a VoltageDivider circuit.
The first is: https://www.google.co.za/search?q=VB...l%3B2048%3B730 And then work out V_{E} = V_{B}  V_{BE} And the second is: https://www.google.co.za/search?q=vt...ml%3B480%3B257 (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? 


#2
Jun2014, 03:26 PM

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#3
Jun2314, 04:34 PM

P: 12

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) 


#4
Jun2314, 05:11 PM

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P: 12,113

Calculating Emitter Current  VoltageDivider Bias
I don't understand your equation for I_{E}. A difference of 6% is a bit large for rounding errors.



#5
Jun2314, 06:16 PM

P: 12

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. 


#6
Jun2314, 06:21 PM

P: 12

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



#7
Jun2314, 11:35 PM

HW Helper
Thanks
P: 5,496

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



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