EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: advark on September 10, 2020, 09:30:06 pm
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Hi everyone,
I'm having some problem solve this circuit and I'm wondering what am I missing. Assuming Vbe is 0.7V and Hfe is 100. I want to calculate Ib and Ic.
First off, I get VR3
VR3 = Vcc *( R3 / ( R3 +R2 ))
= 12 *( 1K / ( 1K +10K ))
= 1.09V
VR2 = Vcc - VR3
= 12 - 1.091
= 10.91 V
Since Vbe is 0.7V and VR3 is 1.091V then VR4 must/should be 0.391V, because VR4 +Vbe should equals VR3.
So Ib (which is the same as IR4) must be 0.391 mA (VR4/R4).
The same goes for IR3 = VR3/R3 = 1.01 mA
Then again, IR2 = VR2/R2 = 1.091 mA
According to Mr. Kirchkoff, IR2 = IR3+ IR4 but 1.091 = 1.01 +0.391 is false... :wtf:
What am I missing here??? My guest is that because R3 is in parallel with R4 my initial formula is wrong. But how can I account for the Vbe drop in that formula (or is there another formula)? :-//
Thanks for any hints.
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VR3 = Vcc *( R3 / ( R3 +R2 ))
= 12 *( 1K / ( 1K +10K ))
That's just an approximation. It doesn't account for the current through the base resistor.
Here's a simulation of the transistor biasing with the assumption that Vbe = 0.75V -- you can redo the calculation for a more conventional value of 0.6 or 0.7:
[attach= 1]
Just create variables for the voltage and current for each node, write out the relationships between them and then solve the system of equations.
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Yeah some textbooks just assert that the voltage at the base node will be the same as a voltage divider rule, but that is only approximately close when the current through the resistors is orders of magnitude greater than the base current, therefore introducing only a small error.