I have little confusion regarding barrier potential (or built-in voltage) and cut-in voltage of diode. Are these are same or what is the difference between built-in voltage and cut-in voltage in diode?
Electronic – Semiconductor diode barrier potential
diodessemiconductors
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You won't be able to measure it with a voltmeter, but there will indeed be a potential difference. In a P-N junction with no external voltage applied negative charges from the N-doped region will migrate to the P-doped region, and vice versa. The blue region will be negative charged, the red one positive. The difference in charge causes a 0.7V across the barrier. (Will depend on the donor and receptor concentration.)
For the first answer, this barrier is caused due to the different materials used to construct the transistor. Think of each connection in the transistor as a diode.
Now, the source may be of n type material(i.e. have extra free electrons) and the substrate might be p type material (i.e. have extra holes - positive charges). Now if you don't apply any external voltage, there will be an equilibrium state where the electrons at the border of the n-type material will cross over to the p-type substrate. This would then result in the formation of a charge neural depletion region, and any new electron which now needs to cross-over to the p-type substrate would require an additional external voltage.
This voltage is your barrier potential.
When you apply an external voltage, this barrier potential reduces as electrons are able to flow more freely, thus turning the diode on.
Why do I refer you to diodes, it's because a transistor can be considered as device with two back-to-back connected diodes. Once you understand how a diode works, transistors become easy!
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Best Answer
Built in voltage and cut in voltage are not the same thing at all.
Built in voltage is a physical parameter that depends on how the diode is built, for a p-n junction it holds: $$ V_{bi}=V_T\ln{\left(\frac{N_AN_D}{n_i^2}\right)} $$ where \$V_T=\frac{kT}{q}\$ is the "thermal voltage", I don't know how it's called in English, \$N_A\$ and \$N_D\$ are the carriers concentrations and \$n_i^2=p_0n_0\$ is the intrinsic concentration.
Cut in voltage is not a physical parameter at all. It can not be computed in any way, it does not (strictly) depend on the diode\$^1\$. It's just a voltage over with a great bunch of engineers would say "ok now that diode is conducting". Cut in voltage, symbol \$V_\gamma\$, is assumed to be equal to 0.7V in most cases, for schottky diodes it can be much lower, something around 0.3V or whatever.
Just keep in mind the huge difference:
\$^1\$Obviously it does depend on how the diode is built but there's no such a relationship as there is for \$V_{bi}\$.