Measuring the barrier potential of a p-n junction Is it possible to measure the potential difference across the depletion region (barrier potential) of a p-n junction using a voltmeter? 
 A: An ordinary voltmeter has finite input impedance which simply means that, to measure a voltage across, there must be some (tiny) current through the voltmeter.
Thus, to measure the built-in potential of a diode with a voltmeter would require that the built-in potential 'drive' a (tiny) current through the voltmeter.
But that would require that the diode, a passive device, is an energy source!
Yes, the power would be tiny but one could, in principle, leave the diode connected to the voltmeter for an arbitrarily long time and thus, extract arbitrarily large energy from the diode.
Further reading:  Why isn't there a potential difference across a disconnected diode?
A: The barrier potential is balanced by the metal to semiconductor contact potentials in the circuit.To show the reading a small amount of current must flow through the circuit.It causes Joule heating,but since there is no external source of energy,it must cause simultaneous cooling of the p-n junction.But due to 2nd Law of Thermodynamics it is not possible to derive work by cooling a body below its equillibrium temperature.Thus our assumption that current can flow is wrong,so though there is a barrier potential,no current flows in a open-circuit unbiased p-n junction,hence no potential can be detected by a voltmeter no matter how sensitive.
A: The current-voltage characteristic of an ideal semiconductor diode is given by:
$$ 
I = I_0\ (e^{\frac{qV}{kT}} - 1),
$$
where $kT/q \approx 25$ mV at room temperature. It does not explicitly depend on the built-in voltage. The ideal curve looks the same for all such diodes.
What the diode test on a multimeter does is to send a small current through the diode, often about $0.2$ mA. This gives a typical forward voltage of the diode in electronic circuits. But that value of the current is a bit arbitrary. One could also have chosen $0.2$ nA which gives a lower forward voltage.
It is not the built-in voltage. That is in the value of $I_0$.
A: 
(source: pveducation.org)
This is called the build in voltage (sometimes $V_{bi}$) and is used to find the different across the junction under 0 bias.  $k$ is the Boltzmann constant, $T$ is temp in Kelvin, $q$ is charge of an electron in coulombs i believe, $N_A$ is number of acceptor ions, $N_D$ is donor ions, and $n_i$ is intrinsic carrier concentration for Silicon usually.
If you need the difference under some bias, that's a totally different story and can get more complicated.
Hope this helps!
A: NO... A simple voltmeter cannot measure the built in potential of a pn junction. because, when we connect the voltmeter to the two p and n sides it will show the sum of all the contact potentials at the all possible junctions in the system and which will be equal to zero. But, a digital multimeter can be used to measure that. digital multimeter predicts the value by using the C-V characteristic of the depletion capacitance formed at the junction.
A: It is not possible.
In order to show the reading of voltmeter, a small amount of current should be passed. Since there is no external energy sources , there will be heating effect in the circuit.
