Voltage across PN-junction - Page 1
Forum section: electronics / projects

author: Arznei
author_rank: Contributor
timestamp: March 26, 2019, 06:53:37 pm
content: Hey there,I'm currently studying pn-junctions and think I somewhat understand whats going on on a basic level. However there is one question I just can't wrap my head around. So considering an ideal PN-junction where a depletion region is formed between the p-type and n-type substrate. An electric field forms inside the depletion region to counter the diffusion current based on the concentration difference between n- and p-type.Now in my textbooks it says that outside the depletion region the E-field is negligible if the dotation is high enough. If I now were to calculate the voltage across the diode (so, integral over the E-field across the whole diode) I must get *some* voltage right? Because in the depletion region there is a purely negative (or purely positive, depending on which way you look at it) E-field and outside of it the E-field is basically zero.However if I connect a multimeter to any diode I of course will *not* measure a voltage. So, what gives?My current theory is that there is nothing to sustain an ongoing current, so the internal resistance of my voltmeter is causing a reading of 0V. Is that correct or am I missing something?Were I to connect an ideal voltmeter to an ideal pn-junction, would I actually measure its diffusion voltage?

author: Kleinstein
author_rank: Super Contributor
timestamp: March 26, 2019, 09:05:05 pm
content: There is no simple way to measure the diffusion voltage with a "normal" voltmeter. The diffusion voltage is more like the difference in the electro-chemical potential between the two sides of the diode (P and N doped). Even an ideal infinite impedance voltmeter would read zero voltage if there is no light, AC or similar excitation to the diode - there is just no source of energy.The PN junction can indeed be quite confusing if it comes to the details. The concept of the diffusion current can also be a little misleading, as there are not actually currents flowing that cancel out, but effective forces that are balancing.It gets even more confusing in the non equilibrium state, e.g. with an external voltage or light.

author: DaJMasta
author_rank: Super Contributor
timestamp: March 26, 2019, 10:04:30 pm
content: And if there were currents flowing, even tiny ones, when the device had no external excitation, all PN junctions would have a shelf life, since they would gradually return to neutral charge when adjacent to each other, otherwise with a current flowing and fixed doping, they'd be a perpetual energy machine.

author: moffy
author_rank: Super Contributor
timestamp: March 26, 2019, 11:44:24 pm
content: I get your point. Like a thermocouple, two dissimilar metals producing a voltage based on temperature. My guess is that the E field is abrupt at the junction but slowly decays to zero through the p and n regions. I could be completely wrong though!

author: rfeecs
author_rank: Frequent Contributor
timestamp: March 27, 2019, 12:25:58 am
content: Yes, the contact potentials cancel the built-in potential. It's confusing because the pn junction band diagrams never show the whole diode, including the contacts. Have you studied ohmic contacts? They also have a built in potential. You have to tack them on each end of the band diagram, then when you integrate across the full diode, you get zero voltage.

author: OwO
author_rank: Super Contributor
timestamp: March 27, 2019, 08:48:40 am
content: ^ yes, the contact points from metal to silicon also form a junction with a built in potential, cancelling the potential of the pn junction. My understanding is that this metal-semiconductor junction is ALSO a diode (schottky barrier), but by controlling the amount of doping you can make its depletion region very thin and thus it gets punched through with any modest voltage. (This is what people call an ohmic contact)