Hi,

How do you get 0.6v from emitter to base PN junction? I've managed to understand most things up to this point? The text is showing me that theres a PN junction drop of 0.6v, but how?

CALCULATING Ee - Since the emitter to base PN junction drops 0.6 V, Ee equals:

 

This is just a value that is most commonly used when dealing with silicon diodes (usually between 0.6 and 0.7). The junction has an exponential I-V relationship and when used in the ideal case, a good approximation is that this voltage will be roughly 0.6 - 0.7.

It is a property of the transistor, and not something you are to solve for using standard KCL/KVL.

Hope that helped,
JP

 

Lots of physics... I think what you are looking for is in this link:
http://www.electronics-tutorials.ws/diode/diode_2.html Look for "depletion layer"

"The external voltage required to overcome this potential barrier that now exists is very much dependent upon the type of semiconductor material used and its actual temperature. Typically at room temperature the voltage across the depletion layer for silicon is about 0.6 - 0.7 volts and for germanium is about 0.3 - 0.35 volts."

 

The PN junction drop(barrier potential) actually cannot be measured by potentiometer(without connecting it to a power source). It is the voltage that needs to be overcome to make the junction forward biased.
Only in forward bias there is current flow and you can measure the voltage which would remain constant at 0.7-0.8 V.
http://sub.allaboutcircuits.com/images/03410.png
Since this barrier potential opposes the flow of carriers(current) there is no way to measure this barrier potential.
You can measure potential across two points only when there is a current flow.


So you may get a question how did they decide the value if it cannot be measured. Its quite simple.
When connecting to an external voltage source(in forward bias) there is no current flow until 0.6-0.7 V. Once this value is exceeded there is a visible increase in current. So we know that till 0.6-0.7 V the barrier is present and is overcome once the voltage is increased beyond that. Hence we can conclude that the barrier potential is 0.6-0.7 V. This is the first experiment in lab for an undergraduate programme in electronics.

 

The basic I/V diode equation can be found here:

http://pveducation.org/pvcdrom/pn-junction/diode-equation

Also here:

http://en.wikipedia.org/wiki/Diode_modelling

This is what determines the PN voltage drop. Note that is very dependent on temperature.

 

Thank you all for your prompt responses! I've looked all the hyperlinks you've provided me and read your responses, thank you so much for clarifying this minor road bump for me !

With Regards,

Alex