Consider the NPN BJT. Normally, if you want to use it as a diode, you short base to collector and treat the emitter as the cathode. But in principle, you could also use the base emitter junction by itself (collector floating), or even the base collector junction (emitter floating). I guess you could even tie the emitter and base together and treat the collector as the cathode. What would the properties be of these diodes? Anything that might be useful in special circumstances?
Electronic – Properties of transitors used as diodes
analogdiodestransistors
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At the moment of discharge, there is current flowing from the charged object, through the transistor, and finally to ground or other lower potential.
Ohm's law is in effect here, and the higher the resistance, the higher will be the voltage observed across the two terminals of the "device under destruction".
FET's have a very high resistance on their gate lead. For MOS FET's, that resistance is extremely high. A BJT has a much lower resistance, although, when you are dealing with static discharge, there's no assurance you won't blow that up as well.
As a side effect, BJT circuits usually have relatively small resistors across their terminals to bias the device, and that will limit the IR drop even further.
So as you mentioned it says that the transistor is essentially two diodes.
You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:
As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.
What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!
Related Topic
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Best Answer
If you follow BJT modelling, then you can conclude these facts
Now this will lead to Band diagram as shown below.
Thus for considerable amount of drift & 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 ....
$$ I_E = I_B +\ I_C \\ I_C = \beta I_B \\ I_C = \alpha I_E \\ I_E = (\beta +\ 1)I_B \\ I_C=I_s e^{\left(\frac{V_{BE}}{V_T}\right)} \\ \text{Normal values of}~\beta = 99 ~ \& ~ \alpha = 0.01\\ I_E \approx I_C \approx \text{few hundreds}~mA \\ I_B \approx \text{few}~\mu A $$ On study of above equations, we can have few conclusions
PS: this lab document has some good insights