# Why is Vbc = Vbe - Vce ?

Started by October 8, 2008
```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

(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

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
>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

Vce = Vcb + Vbe

Vcb = Vce - Vbe

Vbc = Vbe - Vce

Polarity is important.

```
```<jalbers@bsu.edu> wrote in message
>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     |    |
|      |    |
|__    |  __|

```
```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.

```
```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
```
```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
```
```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