Properties of transitors used as diodes
If you follow BJT modelling, then you can conclude these facts


Emitter is doped highest but has considerable bigger area than base and lesser than collector
Base is doped considerable more than collector but less than emitter and is having least area
Collector is doped least and has maximum area among all three.


Now this will lead to Band diagram as shown below.

 

Thus for considerable amount of drift &amp; diffusion currents as per study from  Energy levels , Base and collector must be shorted to make a P-type node of Diode and Emitter is heavily doped already hence it will make N-type node of Diode

NOTE : I might be not able to explain well , but I hope this facts come useful to understand



added later on 

Why to short Base-Collector , Instead of Base-Emitter

Lets say following abbrevations for sake of explanation
BE: Base Emitter Junction
CB: Base Collector Junction 

Consider the BE as single entity,Now since emitter is doped more heavy compared to base BE junction will act as N-Type element, since u are left with collector which is already N-Type element , Thus it will result in bad diode.
Now on other hand for CB as single entity, base is more heavy doped compared to collector, Hence CB will act as P-Type element, you are left with emitter which is already comparative doped as N-Type element. Thus this can result in well functional Diode.



More ....

$$
I_E = I_B +\ I_C \\
I_C = \beta  I_B \\
I_C = \alpha I_E \\
I_E = (\beta +\ 1)I_B \\
I_C=I_s e^{\left(\frac{V_{BE}}{V_T}\right)} \\
\text{Normal values of}~\beta = 99 ~ \&amp; ~ \alpha = 0.01\\
I_E \approx I_C \approx \text{few hundreds}~mA \\
I_B \approx \text{few}~\mu A
$$
On study of above equations, we can have few conclusions


emitter &amp; collector contribute major amount of currents. compared to base
base has to be in picture as its connecting link for collector &amp; emitter


PS: this lab document has some good insights