In a circuit if the ideal diode has both the cathode and anode voltage same, will it conduct or not? Similarly, in a constant drop model if the voltage drop is exactly equal to 0.7 will it conduct or not?
Electronic – what happens when both anode and cathode voltages are same for a diode
diodes
Related Solutions
Electronic – what happens with diode voltage drop when connected directly to constant voltage supply
What would happen is that the diode would self-destruct in a (possibly) spectacular fashion.
Every real-world battery has some internal resistance, as do real-world diodes. This resistance, together with the diode-drop, would determine the current flow.
Since there would be a very large current flow for most common batteries, the diode would be unable to dissipate the energy, and will overheat and fail.
Taking a common example: A 1N4001 diode connected directly across a alkaline "AA" battery:
- The battery's internal resistance is ~200 mΩ.
- The contribution of the diode's internal ohmic resistance is negligible.
Therefore, the current flow can be solved for fairly trivially, particularly if you ignore the fact that the diode drop varies depending on current.
The simple solution is \$\frac{1.5V - 0.7V}{0.2Ω} = 4A\$ (fig 1), so approximately 4 amps of current would flow.
With 4A current flow, and the 0.7V diode drop, the diode would be dissipating \$4*0.7 = 2.8W\$ (fig 2).
We can then look at the diode's thermal resistance (\$R_{θJA}\$), which is specified as 100 K/W. This means that for every watt dissipated, the diode's temperature will increase by 100 K (kelvin).
Therefore, with a 20°C ambient temperature, the diode's temperature will be \$20° + 100°*2.8W = 300°C\$ (fig 3). 1140° is well past the point where the diode would be incandescent, and it will promptly self destruct.
Edit:
Basically, the critical thing here is there are no perfect voltage sources. If you connect a diode across a perfect voltage source, you will get the voltage of that perfect voltage source across the diode, for the infinitesimal period of time before the diode self-destructs due to self-heating.
However, all real world voltage sources (such as a battery) have a internal resistance. It's that internal resistance, together with the resistance of the wires leading to the diode, the internal ohmic resistance of the components within the diode, and the actual diode-drop itself that must be considered when trying to determine the instantaneous voltage across the diode the instant it's connected (well, that's ignoring cable and battery inductance, but that's another matter).
Further Edit:
I'm using \$0.7V_{F}\$ as a simplification of the real diode forward voltage, for the sake of ease of calculation (and because I'm too lazy to work out all the math). In a real situation, the diode drop will depend on the current, so the actual forward current of the diode will be somewhat lower.
If you want to know the exact forward voltage to a higher degree of precision (and if so, why?), you can instead replace the perfect 0.7V forward voltage drop with the ideal diode equation, and calculate the voltage drop of that in series with the internal resistance of the battery.
The datasheet has a graph of forward voltage versus \$I_{F}\$, which you have already found.
If the same circuit in ideal model had both \$D_1\$, \$D_2\$ conducting, does that mean in constant voltage model \$D_1\$ and \$D_2\$ will conduct?
Need not be.
I see a few mistakes in the solution you have given here:
In the KVL, you missed one DC source. There are two voltage sources (+5V and -5V) in the loop.
The value of \$I_1\$ will be in milli-amps (0 ~ 1mA).
Voltage at node B is not \$5k \times I_1\$. It is \$5 - 5k \times I_1\$
Best Answer
This is actually a good question, and should not have been downvoted. The answer is that it could be any (forward) current (including zero)- you cannot tell what the current is from the voltage. Same as with the ideal wires we draw on a schematic. That is a consequence of the 'ideal' assumption- it's not a physical reality.
If you are trying to analyze a circuit with ideal diodes you have to treat them sort of like wires that are there if the voltage drop across them would be in the correct direction if they were removed.
The only time this comes up in reality is with a superconducting wire- if there is zero volts (exactly zero volts, not too low to measure) then there could be +/- any current through the wire (up to some limit in reality).