I think potential difference across disconnected diode should be zero but how?   the last graph tells that there is a potential difference between the two ends of a diode. then how can be it zero?
77221
This image from wikipedia, explains that there occurs a potential drop across a , and an electric field confined to the depletion region.  I already know the reason for the existence of this drop and the calculation of the difference but I have two questions regarding this drop.    1) If the  regions are externally connected using a perfectly conducting wire, why will not any current flow? Connecting the two regions should equalize their potentials (since wire is a resistance less) conductor, and therefore the gradient at the junction is destroyed resulting in a  which is obviously against conservation of energy as the semiconductor has non-zero resistivity. Where will extra potential drops will be created so that Kirchoff's voltage rule holds without any current and the built-in potential difference $V_o$ of the juction persists?    2) (probably naive) If an external Bias is applied, of say $|V|&lt;|V_o|$, then the pn potential difference across the junction will just reduce by that amount ($|V|$). Assuming that the external voltage source is ideal without any resistance, what will be the potential drops which would sum to zero in this case? Will drop due to the resistance of semiconductor play a par in it? If yes, then there cannot be any ideal resistance-less semiconductor junction, can there?
39717
Why does voltage remains same over parallel circuit. If a resistor is connected in the circuit some of the charge should be transformed into heat and make a lack of charge after the resistor (in my sense). So, what's the reason in it?
97469
Say we have 2 charges set up like above. Why is $V_3=0$ ?   I can understand mathematically that the scalars fields cancel out, but I don't understand it physically. If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy. However because the Electric Potential is 0, thus the electrical potential energy is 0.  How can $V_3 =0V$ if a charge placed in between the two charges will gain kinetic energy?
33108
Hi I have a Diode Bridge as outlined in the image below. As you can see I am getting a negative voltage on one pair of diodes instead of this voltage being flipped to positive.   Can you see anything wrong with my design or does it look like the simulator program getting it wrong?    Using Every Circuit Simulator on Android.
77294
This is a question regarding the physics behind the observation.  I have guessed the answer to the question, but I may be wrong, so I want to wait for the responses before posting it.  Some major considerations:   A battery's positive terminal does have a positive potential. ie, a test positive charge will repel it and a test negative charge will attract it. Vice versa for negative terminal. From the paper below (Section 1.2.1), it seems abundantly clear that the battery will have positive and negative potential on respective terminals. Given 'point 1', above, connecting the positive terminal of battery A to negative terminal of battery B will lead to current flow in the conductor. If the potential difference only comes into effect when the circuit is closed using the terminals of the same battery (due to cell chemistry), then does that mean that 'point 1' will not be valid after this experiment (connecting positive terminal of A to negative terminal of B)? If it is only a tiny quantity of current that flows, then will it ever stop? And what determines this quantity? (Equation would be nice)   Related paper:
36218
&ndash; Schematic created using   Does any current flow from point A to point B?   My guess is no. Because both of those points are connected to a ground symbol by blank wires, they both have potential of 0V.
97109
Can anyone help me identify this component? It just says 0 or D on it (I think it's a zero). I tested it for conductivity with a multimeter - doesn't seem to have any (tried both polarities in case it's a diode). Tested for resistance: nothing (i.e. infinite resistance), although I'm not 100% sure I was touching the contact points correctly, so don't take this as an ultimate measurement. Thanks in advance!
70481
While studying photoelectric in my school, my teacher drew a graph of current versus the potential difference across the two electrodes:    I am not able to understand why do we get saturation current. I know that at saturation current all the electrons emitted from the metal surface are able to reach the other electrode. My question is suppose we increase the potential difference between the two electrodes further. I think that one of the following two things should happen:   current may increase as now the electrons in the wire will be under greater potential difference thereby reducing the time in which the electrons complete one round to reach the same point again and hence increasing current. ($V=IR$) current may not be continuous (steady current might not be achieved) as the time taken by the photoelectrons to reach the positive electrode may be more than the time taken by the electrons inside the wire to go from end of the wire nearer to the positive electrode to the  other end.   So in no case should we get a graph as shown in the link.  It may be that I have not understood the photoelectric effect completely. If that is the case please tell me where am i wrong?
110405
I have always had the idea that the forward voltage drop in a semiconductor diode was related in a simple way to the bandgap energies in the semiconductor.  However this is apparently not the case:    germanium has a bandgap of 0.66 eV, but germanium diodes have a typical forward drop of 0.2 V silicon has a bandgap of 1.12 eV, but silicon diodes have a typical drop of 0.6 V   I'm aware of the Shockley equation describing the current in a diode as a function of diode voltage drop $V_D$ and temperature $T$, $$ I(V_D) =I_0 \left(  e^{eV_D/kT}  -1 \right)  \approx I_0 e^{eV_D/kT} $$ where the scale current goes something like $$ I_0 = A \,T^3 e^{ -{E_\text{gap}}/{kT} } $$ and the constant $A$ depends on the geometry of the diode, the degree of doping, the charge mobility, and probably some other stuff, too.    I recognize that there's some arbitrariness to the approximation of a "turn-on voltage": the exponential grows so fast that if your choice for the current threshold differs from mine by a factor of a thousand, we'll only disagree about the turn-on voltage by about a couple tenths of a volt. However, I've had for years the impression that there's something fundamental about silicon that gives silicon diodes a forward drop of roughly 0.6V. Is that the case?  Or is there some constellation of design choices that conspire to give the same drop both across most p-n diodes and across the p-n junctions of bipolar transistors?  I was motivated to ask this question by a  about forward voltage drops in LEDs. I was expecting to answer that question with some data from a student using LED turn-on voltage and the wavelength of light to measure the Planck constant. However those data are a lot more complicated than I expected: in fact, most of my LEDs apparently emit multiple wavelength components with comparable strength, and there doesn't seem to be much correlation the turn-on voltage between and the most prominent color in the LED spectrum.   I don't seem to be able to say much more than "LEDs have turn-on voltages between two and three volts."  I have read a little bit about  on Wikipedia, which suggest the potential barrier $\phi_B$ across an interface is different from the band gap, but I can't figure out why.
77312
Using conventional notation in the diagram below, I understand that the circuit is not complete because the diode is in reverse bias. But knowing that the electrons actually flow from the negative to the positive terminal(direction of the arrows), wouldn't the electrons take a path from the negative terminal of the battery, go through the lamp and stop at the diode(the red path in the picture). If they would, then why doesn't the lamp light up with the first electrons that are able to flow through it before the voltage in the wire is built up and the current stops.
82819
I am having a hard time answering one particular question about our experiment.. In our experiment, R1 and R2 were set to 1Meg each and later to 10k... I understand the need for R1 and R2 a bit. Without R1 and R2, the voltage sharing wouldn't exactly be 50-50 for both D1 and D2 because no two diodes are completely identical. D1 and D2 will both have the same leakage currents (without R1 and R2) since they are just in series. However, they probably will have non-identical IV curves, so this particular leakage current will result to V@D1 /= V@D2.  The question I am having a hard time is that, why is V@R1 + V@R2 /= 10v when R1 = R2 = 1Meg?... One the other hand, those two voltages add up (to 10v) when R1 = R2 = 10k... I included the 60 ohm source resistance in my diagram for completeness. However, as I can see, both D1 and D2 are reversed biased and thus, they offer a very large (reverse resistance) which should be much greater than the 60 ohms. Even with the parallel combination of 1Meg and D1 reverse resistance, it should still be much greater than the 60 ohms. I tried thinking of an answer in terms of the RD1reverse//R1 = Req1 and RD2reverse//R2 = Req2. Req1 + Req2 (series) should still be much more than 60ohms and I thought that the 10v should still show up at the node of D1 cathode. Yet in our experiment, V@R1 + V@R1 &lt; 10v.  Can anyone point me out if I am thinking this in a wrong way? Some tips/first step hint would really be appreciated  Edit: question answered thanks to @CL. Assuming D1 and D2 are open during reverse bias for simplicity and noting that Rmultimeter = 10Meg, V@R2 (shown on multimeter) = 10v * (1Meg//10Meg)/((1Meg//10Meg)+1Meg+60) = 4.76v measured.
10384
Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path? Some clarification:  I understand that some current will flow through the resistance. A depiction of this phenomenon through Ohm's laws does not answer the 'why' part of the question as it does not illuminate the process behind this phenomenon. There seems to be disagreement among the answers on whether the charge repulsions are enough to cause this effect. Cited research would be appreciated to resolve this issue. To refine what I mean by this question, I think I should give an example of why I am curious as to the causal background of this phenomenon. If charge repulsions do in fact repel 'current' in the example of the short circuit, then why does the same phenomenon not happen in other parallel circuits. (Again, I know that not all circuits behave ideally) I believe the question is expressed in the most straightforward way possible. This should, however, not affect the scope of the answers. I do not wish to see or think on the level of an electron as circuit behavior always includes a cumulative effect due to the nature of electrostatic interactions.
4669
If voltage is a potential difference, describing how much energy will be lost per unit charge as charge carriers pass through the resistor, why would the rate at which charge flows (i.e. current, $=dq/dt$) affect the amount of energy lost per unit charge?  I suppose I am asking why a quasi-"extensive" property of the system (current) would affect a quasi-"intensive" property of the system (voltage drop across the resistor).  Apologies if the answer is obvious; I was having trouble understanding why the rate of charge flow would factor into voltage drop whilst trying to rationalize Ohm's law for myself.
30815
If I have a plot of current ($y$ axis) against 1/Resistance ($x$ axis).  The circuit it is measured from is a simply 2 resistors connected in parallel to battery, where the potential across the resistors is $1V$. One resistor remains fixed, and the second varies, the resistance plotted on the graph is only the resistance of the second resistor.   Does this mean that the gradient will be voltage, but will it be $1V$ or less, as the resistance on the graph is only one from one of the resistors?  Also what will the $y$ intercept signify? $I = V/R +c$, what does the $c$ represent?
100149
In the following image:   Why can't current flow across the following wire? It's a simple question, but I've kind off always wondered.  Thanks!
55888
According to , if there is a potential difference, $V$, across a resistor then there is a current, $I$, flowing through it.   Since we assume that points along the connecting wire are at the same potential, how can current, $I$ flow between points at the same potential, $V$?
75650
For a localized charge distribution the potential is set to zero far away from the charge distribution (at infinity)  Now, when grounding a conductor, i.e. connecting it to Earth, it is said that we are setting its potential to zero.   Why Earth's potential is zero? And if it is zero, is it zero even if the potential is still zero at infinity?