Why is Vbc = Vbe - Vce ?

  • Thread starter jalbers@bsu.edu
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jalbers@bsu.edu

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I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
 
On Wed, 8 Oct 2008 11:51:46 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:

I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
Click to expand...
(Vb - Ve) - (Vc - Ve) = Vb - Ve - Vc + Ve = (Vb - Vc)

--
Rich Webb Norfolk, VA
 
jalbers@bsu.edu wrote:
I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
Click to expand...
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.

--
Regards,

John Popelish
 
<jalbers@bsu.edu> wrote in message
news:d0d4a3fb-fdcf-41f7-8aeb-983a0eff1e6d@z6g2000pre.googlegroups.com...
I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
Click to expand...

Vce = Vcb + Vbe

Vcb = Vce - Vbe

Vbc = Vbe - Vce

Polarity is important.
 
<jalbers@bsu.edu> wrote in message
news:d0d4a3fb-fdcf-41f7-8aeb-983a0eff1e6d@z6g2000pre.googlegroups.com...
I have seen in various places the equation Vbc = Vbe - Vce for
NPN and
PNP transistors. It may seem obvious to others but I just
don't see
how this is true. Could someone please throw me a bone?
Thanks
Click to expand...
Maybe this will help. View the ASCII diagram with mono-spaced
font such as the WinXP default Lucida Console or Courier:

__ __
| | |
| | |
| | |
Vbc | |
| / |
| |/ |
|___| Vce
| |\ |
| \ |
| | |
Vbe | |
| | |
|__ | __|
 
On Wed, 08 Oct 2008 15:07:53 -0400, John Popelish <jpopelish@rica.net> wrote:

jalbers@bsu.edu wrote:
I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks

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.
Click to expand...
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:

http://www.csus.edu/indiv/o/oldenburgj/EEE102/Chapter5/BJTs.ppt

slides 20-22, where the equation is derived from a mesh analysis.
 
Charlie Siegrist wrote:
On Wed, 08 Oct 2008 15:07:53 -0400, John Popelish <jpopelish@rica.net> wrote:

jalbers@bsu.edu wrote:
I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
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.

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:

http://www.csus.edu/indiv/o/oldenburgj/EEE102/Chapter5/BJTs.ppt

slides 20-22, where the equation is derived from a mesh analysis.
Click to expand...
Please note that Vbc has the opposite sign from Vcb.

Vce = Vcb + Vbe Note: Vcb = - Vbc

Vce = -Vbc + Vbe

Vbc = Vbe - Vce
 
On 2008-10-08, jalbers@bsu.edu <jalbers@bsu.edu> wrote:

I have seen in various places the equation Vbc = Vbe - Vce for NPN and
PNP transistors. It may seem obvious to others but I just don't see
how this is true. Could someone please throw me a bone? Thanks
Click to expand...
Vbc = Vbe + Vec

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

Kirchoffs node law.

Bye.
Jasen
 
On Wed, 08 Oct 2008 19:24:47 -0700, Dan Coby <adcoby@earthlink.net> wrote:

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:

http://www.csus.edu/indiv/o/oldenburgj/EEE102/Chapter5/BJTs.ppt

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

Please note that Vbc has the opposite sign from Vcb.
Click to expand...
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:

http://www.seas.upenn.edu/~ese319/Lecture_Notes/Lec_2_BJTLgSig_07.pdf

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