Help calcultating Transistor Bias Point

[Microemission]

bias point analysis

Hi, i need some help calculation the transistor bias point by hand as i do not know if i can replace the 2 resistors R1 and R2 by the thevenin equivalent, Vth=((5v*2)*1.6k)/(1.6k+1.2))=5.7v and equivalent resistance R1//R2=685Ω, like we do when the supply is normally to GND.
The doubt is if is still valid when the supply is symetrical.
If this is valid then this result in the equation:
-5.7+685*Ib+0.7+100Ie=0<->-5.7+685*Ib+0.7+100(β+1)Ib=0, then solve for Ib, which gives 94uA for β=520, correct?

Tkx very much

 

[happy2005]

If V3 signal source internal series resitance is zero (the schematic shows that it is zero), the base voltage must be zero volt. Because, all independent ac sources is zero in DC analysis.So,
IR3 =-0.7/100 , IR4=-0.7/100
Ie=IR3+IR4=14mA
Ib=Ie/520=26.9 uA

 

[zhi_yi]

should we use the superposition method to find the emitter current through the R3? in DC circuit analysis for finding the IR3, the loop would become like this, and we can ignore the R4, is it true?

 

[tyassin]

If V3 signal source internal series resitance is zero (the schematic shows that it is zero), the base voltage must be zero volt. Because, all independent ac sources is zero in DC analysis.So,
IR3 =-0.7/100 , IR4=-0.7/100
Ie=IR3+IR4=14mA
Ib=Ie/520=26.9 uA

Hey

The current IR3 is not -Vbe/R3. Because you have 0V at the base, then you have -0.7V at the emitter, but I think you are missing the -5V from the source. The current in R3 is:

IR3=IR4=(Vbe-Vdc))/R3=(-0.7-(-5))/100=43mA

Ie=86mA
Ib is found easily from Ie

Regards

 

[happy2005]

Thanks tyassin. your explanation is true. But your solution is still wrong. Because lower-end of R4 is connected to ground, not -5V. So the current in the R4 is in reverse direction. The correct answer is:
IR3 =4.3/100=43 mA, IR4=-0.7/100= -7 mA
Ie=IR3+IR4=43mA+(-7mA)=36 mA
Ib=Ie/520=69.2 uA
Regards

 

[zhi_yi]

hehehe... thank you for the calculation.. the answer from happy2005 is correct i've tried to calculated it and simulated it in multisim and there is an error in my previous dc analysis circuit i'm sorry

 

[lichengcao]

at Bias Point:Vb=0.7v,Ve=-2.5v,then Vbe=3.2v>>0.7v so Ve ->0v;so I4=0,Ie=I3=[0-(-5)]/100=50mA.=>Ib=Ie/520=96uA

 

[zhi_yi]

at Bias Point:Vb=0.7v,Ve=-2.5v,then Vbe=3.2v>>0.7v so Ve ->0v;so I4=0,Ie=I3=[0-(-5)]/100=50mA.=>Ib=Ie/520=96uA <-- quote

why the Vb = 0.7V? i think that the Vb must be 0 V since the base terminal is shorted to the ground in DC analysis, why Vb = 0.7 V?