Single Stage BJT Common Emitter Amplifier

Thread Starter

NiCeBoY

Joined Aug 20, 2008
59
r1 & R2 in parallel i get 5k.
The base voltage VB i use VB=(VCCxR2)/(R1+R2)

So am i rite that IB should be IB= VB/5K
 

Wendy

Joined Mar 24, 2008
23,415
Yes and No, I calculated using the values showed in your posted drawing.

Rt = 1 / ((1/5.1KΩ) + (1/47KΩ)) = 4.6KΩ

But this doesn't count into the base current. It is the Vb / (β X Re) that is the base current.

To figure the exact numbers then add the Base to Ground resistance (β X 1KΩ) into the R1 and R2 voltage divider.

R2 in parallel the Base to Ground looks like

R2bg = 1 / ((1/Base to Ground)+(1/R2)) = 1/((1/110KΩ)+(1/5.1KΩ) = 4.87KΩ

Using this value into the R1//R2 voltage divider

Vb = (Vcc X R2bg)/(R1+R2bg) = (15V X 4.87KΩ)/(47KΩ + 4.87KΩ) = 1.41V

Using these values to calculate Ibe is

Ibe = Vb / Base to Ground resistance = 1.41V/110KΩ = 12.8ua

Again, since β is such a variable these values can vary a bit. And this doesn't even count other variables, like tolerances of the resistors. The exact Base Emitter drop is also a major variable, but I was taught in school to round off to .7V. The reality is it is between .6 and .7V, depending on transistor family type.

These equations work, but they only get you in the ball park. Sometimes there is no substitute to building a prototype to pin it down, and if you're parts have a wide tolerance like the 10% resistors, there can still be problems. The Vbe is pretty tight within a family of transistors, which helps.
 
Last edited:

hobbyist

Joined Aug 10, 2008
892
Nice work Bill.

You took the time and patience, to work with this person all the way through, this thread.

This is a good example of what this forum is for.

Keep up the good work...
 
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