- #1

davidbenari

- 466

- 18

## Homework Statement

https://drive.google.com/file/d/0Byf8Tfw7vme8N1ZZaUFzOWQtb0k/view?usp=sharing

Determine IEQ ICQ IBQ

My problem is that this circuit is overdetermined and cannot be solved. (See below)

## Homework Equations

##\beta = 75##

##I_E=76 I_B ##

## The Attempt at a Solution

The current IB is calculated as ##I=\frac{1}{500}=.002mA##

The current ##I_E## is ##I_E = 76*.002 = .152 mA##

Similarly ##I_C= 75 * .002 = 0.15 mA##

The voltage ##V_{CE}## is calculated by knowing the current in the emitter and starting upwards from the -3V node, in other words

##V_{CE}=3-(-3+4.8*.152)=5.27V##Knowing this we see that there is a contradiction. The voltage drop across the BE is not .7 .

Is this a sound way of proving that the BE voltage doesn't HAVE to be 0.7?