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

# Why is Vbc = Vbe - Vce ?

Started by ●October 8, 2008

Reply by ●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 ●October 8, 20082008-10-08

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? ThanksThe 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 ●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? ThanksVce = Vcb + Vbe Vcb = Vce - Vbe Vbc = Vbe - Vce Polarity is important.

Reply by ●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? > ThanksMaybe 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 ●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 ●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 ●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? ThanksVbc = 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 ●October 10, 20082008-10-10

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.