Why is Vbc = Vbe - Vce ?

(Vb - Ve) - (Vc - Ve) = Vb - Ve - Vc + Ve = (Vb - Vc)

The base region is between the collector and emitter regions. It is a stack. one part of that stack is the base-emitter junction (Vbe). the other part of that stack is the base-collector junction (Vbc). Add them together and you have the total stack voltage (Vce), or Vce=Vbe+Vbc. Rearrange.

Vce = Vcb + Vbe

Vcb = Vce - Vbe

Vbc = Vbe - Vce

Polarity is important.

Maybe this will help. View the ASCII diagram with mono-spaced font such as the WinXP default Lucida Console or Courier:

__ __ | | | | | | | | | Vbc | | | / | | |/ | |___| Vce | |\\ | | \\ | | | | Vbe | | | | | |__ | __|

Good answer, but misses the subtlety picked up by Andrew. Rearranging

Vce = Vbe + Vbc

gives

Vbc = Vce - Vbe

not

Vbc = Vbe - Vce

The equation in question is a reference to setting up a four-resistor biasing network for a BJT common emitter circuit. See:

slides 20-22, where the equation is derived from a mesh analysis.

network for a BJT common emitter

Please note that Vbc has the opposite sign from Vcb.

Vce = Vcb + Vbe Note: Vcb = - Vbc

Vce = -Vbc + Vbe

Vbc = Vbe - Vce

Vbc = Vbe + Vec

from B to C is the same as from B to E and E to C

Kirchoffs node law.

Bye. Jasen

network for a BJT common emitter

As indeed I did note when I referred to Andrew Holme's post - "...the subtlety picked up by Andrew." The reason for the polarity reversal is deeper than a typical thread in this forum would address. For those interested, here's another web reference to the Ebers-Moll large signal model analysis:

It gives me a headache. Just read Andrew's post.