BJT Biasing - Simple easy 1 minute answer

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ally34
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Homework Statement


Find IB, IC, IE, VCE, VBC, and VEB for the NPN transistor in the circuit below for β = 50, 150, 500. Let VCC = 10 V, R1 = 100 KΩ, R2 = 400 KΩ, RC=2 KΩ, and RE = 0 Ω. Assume VBE = 0.7V.

Homework Equations



V = iR
IE = Ib + Ic

The Attempt at a Solution



My question is that if RE = 0 ohms, then is Ie automatically 0 amps?

But the how does Ie = Ib + Ic apply? Because I got 92.25uA for Ib and 4.56mA for Ic?

Help! :]
 

Discussion

Forgot to put the figure :]
 
Attachments
  • Circuit schematic of an NPN BJT voltage-divider bias network: R2 from VCC to the base, R1 from the base to ground, RC from VCC to the collector, and RE from the emitter to ground.
    Capture.JPG
    3.7 KB · Views: 565
ally34 said:
My question is that if RE = 0 ohms, then is Ie automatically 0 amps?
No. A piece of connecting wire from emitter to ground = 0 Ω :smile:
 
NascentOxygen said:
No. A piece of connecting wire from emitter to ground = 0 Ω :smile:


Awwww!

:[ So Ie = Ib + Ic is correct?

I would use this to calculate Ie?
 
ally34 said:
Awwww!

:[ So Ie = Ib + Ic is correct?
Yes.

I would use this to calculate Ie?
It is involved in the calculations.
 

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