Numbers are OK.
Only for PNP Vec is 0.6V greater then Veb
For NPN Vec is smaller then Veb.
If you don't believe my measurements do the measurement yourself.

I showed you why your numbers are mixed up. It is the same for PNP and NPN transistors. This is for an NPN transistor:
1) The reverse-biased emitter-base breakdown voltage is 6.0V.
2) The forward biased collector-base junction voltage is 0.65V.
3) The total of the two voltages is 6.65V because they are in series.

Some of your collector-base junction voltages are wrong because they are 1.5V instead of 0.65V.

 

Maybe Vec is smaller then Veb because negative-resistance region in emitter-collector avalanche breakdown, and PNP don't have this negative region.

 

I measured a 2N3904 NPN transistor and you are correct, Jony.
The Vec has negative resistance which caused the Vec to be less than Veb.
I used a fairly high current of 26mA to make the negative resistance strong.

 

Jony130,
I studied your second explanation and that has cleared up the operation of this circuit for me! Thank you for the extra detail and step by step explanation, thats what I needed!

My only other question is about the sudden change or lack of change in the voltage of charged capacitor... Can you explain this event to me a little more or point me to a link that explains this characteristic. I understand that it does happen, but I am trying to understand why and how...

BTW thanks to all have particpated.. nice to have people to ask about this stuff who are willing to respond.

 

My only other question is about the sudden change or lack of change in the voltage of charged capacitor

This is a general principle how the capacitor work
The current in the capacitor is proportional to the rate of change in voltage.
I=dU/dt

In this circuit when we first connect empty capacitor (Vc=0V)



The voltage on capacitor cannot change suddenly form 0V to 1.5V.
We need time to voltage on capacitor to grow (t=5*R*C).
We have similar situation when we discharge the capacitor in A position of S1.
Can you tell me what is the voltage at LED D1 after switching S1 in A position.
http://www.allaboutcircuits.com/vol_1/chpt_13/2.html

 

Hey Jony130,

3volts?

 

Hehe, yes 3V

 

Jony130,
I checked out the link too. I understand what happens as we increase the voltage on the capacitor and I can visualize the flow of electrons... but when they discuss decreasing the voltage do electrons actually race back towards the negative battery terminal?

Lets say this is a 9 volt battery, and that the potentiometer is moved to the highest volts so the cap is charged to 9 volts... the negative side of the cap will have obviously more electrons that the positive side... As we decrease the pot and reduce the voltage on the cap is the cap pulling electrons from the positive terminal and pushing electrons towards the negative terminal? If so is this happening because as we lower the voltage the cap instantaneoulsy it is still more positive on the positive side and more negative on the negative side than the battery until is has adjusted the potential between the plates (discharged)?

 

As you can see I'm not a big fan of electrons flow convention.
I always analyse circuit with conventional current flow is from + to -.
But yes, you're right. When we decrease the pot and reduce the voltage the more electrons will flow to the positive plate of a capacitor and this repels the same amount of electrons from negative plate towards negative terminal of a battery.
Look at Lecture from the best prof of physics
http://www.youtube.com/watch?v=7NUbsQt-G9U&feature=player_embedded (10.11)
http://www.youtube.com/watch?v=E185G_JBd7U&feature=player_embedded
(11.30)

 

Sorry for the delay, my day job is keeping me busy...

Cool thanks for the links I will check them... you have been very helpful!

Wish I could afford MIT