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jalbers@bsu.edu
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 Posted: Wed Oct 08, 2008 6:51 pm Post subject: Why is Vbc = Vbe - Vce ? |
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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 |
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Rich Webb
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| Posted: Thu Oct 09, 2008 12:05 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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On Wed, 8 Oct 2008 11:51:46 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:
[quote]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
[/quote]
(Vb - Ve) - (Vc - Ve) = Vb - Ve - Vc + Ve = (Vb - Vc)
--
Rich Webb Norfolk, VA |
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John Popelish
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| Posted: Thu Oct 09, 2008 12:07 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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jalbers@bsu.edu wrote:
[quote]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
[/quote]
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 |
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Andrew Holme
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| Posted: Thu Oct 09, 2008 12:09 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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<jalbers@bsu.edu> wrote in message
news:d0d4a3fb-fdcf-41f7-8aeb-983a0eff1e6d@z6g2000pre.googlegroups.com...
[quote]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
[/quote]
Vce = Vcb + Vbe
Vcb = Vce - Vbe
Vbc = Vbe - Vce
Polarity is important. |
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pimpom
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| Posted: Thu Oct 09, 2008 1:48 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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<jalbers@bsu.edu> wrote in message
news:d0d4a3fb-fdcf-41f7-8aeb-983a0eff1e6d@z6g2000pre.googlegroups.com...
[quote]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
[/quote]
Maybe this will help. View the ASCII diagram with mono-spaced
font such as the WinXP default Lucida Console or Courier:
__ __
| | |
| | |
| | |
Vbc | |
| / |
| |/ |
|___| Vce
| |\ |
| \ |
| | |
Vbe | |
| | |
|__ | __| |
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Charlie Siegrist
Guest
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| Posted: Thu Oct 09, 2008 7:08 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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On Wed, 08 Oct 2008 15:07:53 -0400, John Popelish <jpopelish@rica.net> wrote:
[quote]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.
[/quote]
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. |
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Dan Coby
Guest
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| Posted: Thu Oct 09, 2008 7:24 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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Charlie Siegrist wrote:
[quote]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.
[/quote]
Please note that Vbc has the opposite sign from Vcb.
Vce = Vcb + Vbe Note: Vcb = - Vbc
Vce = -Vbc + Vbe
Vbc = Vbe - Vce |
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Jasen Betts
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| Posted: Thu Oct 09, 2008 7:53 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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On 2008-10-08, jalbers@bsu.edu <jalbers@bsu.edu> wrote:
[quote]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
[/quote]
Vbc = Vbe + Vec
from B to C is the same as from B to E and E to C
Kirchoffs node law.
Bye.
Jasen |
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Charlie Siegrist
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| Posted: Fri Oct 10, 2008 7:47 am Post subject: Re: Why is Vbc = Vbe - Vce ? |
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On Wed, 08 Oct 2008 19:24:47 -0700, Dan Coby <adcoby@earthlink.net> wrote:
[quote]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.
[/quote]
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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