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.
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.
Reply by Jasen Betts●October 9, 20082008-10-09
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
Vbc = Vbe + Vec
from B to C is the same as from B to E and E to C
Kirchoffs node law.
Bye.
Jasen
Reply by Dan Coby●October 8, 20082008-10-08
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.
Please note that Vbc has the opposite sign from Vcb.
Vce = Vcb + Vbe Note: Vcb = - Vbc
Vce = -Vbc + Vbe
Vbc = Vbe - Vce
Reply by Charlie Siegrist●October 8, 20082008-10-08
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.
Reply by pimpom●October 8, 20082008-10-08
<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
Maybe this will help. View the ASCII diagram with mono-spaced
font such as the WinXP default Lucida Console or Courier:
__ __
| | |
| | |
| | |
Vbc | |
| / |
| |/ |
|___| Vce
| |\ |
| \ |
| | |
Vbe | |
| | |
|__ | __|
Reply by Andrew Holme●October 8, 20082008-10-08
<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
> 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.
--
Regards,
John Popelish
Reply by Rich Webb●October 8, 20082008-10-08
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
(Vb - Ve) - (Vc - Ve) = Vb - Ve - Vc + Ve = (Vb - Vc)
--
Rich Webb Norfolk, VA
Reply by jalb...@bsu.edu●October 8, 20082008-10-08
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