# When both sides of a diode are saturated can the diode conduct backwards?

Every explanation of diodes I can find say current cannot meaningfully flow from the cathode to the anode. However an NPN transistor is fundamentally two diodes and the only way it can work is if one of the diodes allows current to flow cathode to anode.

My question is this. If I have a single diode and both sides are at the saturation voltage can current flow from the cathode to the anode? Here is an example I can think of:

If the diode is a true one way valve then the voltage between P2 and P1 will be a constant 5V no matter what. However if the motor was to suddenly draw more power than the power supply can supply, if the capacitor is charged both sides of the diode will be at +5V. Would that allows the capacitor to discharge to provide the motor the additional power? The only reason I can image you would do this is if you had additional circutry you didn't want the capacitor slowing down.

• yes, for a short time..Look up diode reverse recovery.. tldr: the reverse conduction will -over time - deplete the saturation and make the junction insulating again Commented Oct 19, 2022 at 5:17

Every explanation of diodes I can find say current cannot meaningfully flow from the cathode to the anode.

There are at least two reverse current flows, but most diode users are able to ignore them (until the application demands that they can't)

However an NPN transistor is fundamentally two diodes and the only way it can work is if one of the diodes allows current to flow cathode to anode.

A BJT (bipolar junction transistor) is not two diodes. It contains two closely spaced junctions, which are similar to the single junctions you find in diodes. If you bias these junctions one at a time, then they behave like diodes. The junctions are so closely spaced however that the 3-terminal behaviour is much more complicated than the '2 diodes connected together' picture that's drawn as a 'lie to children'. If you want to push the model, then you could argue that the operation of the BE diode modifies the operation of the CB diode, because of the way charge in one can move into the other, but that's a dead-end really as it poses more questions than it answers. A BJT is a three terminal device, understand it through all three.

When a diode is reverse biassed, a reverse current can flow, usually written as IS. This is usually very small, pA or fA levels are typical for small signal diodes. That's not a meaningful current for many people or applications, but if you're doing doing a design with those current levels, you need to design for it. This current is not a defect, it appears in the Shockley diode equation which describes the ideal diode current flow at all bias voltages.

When a junction diode is forward biassed, the junction fills with carriers. When the diode is then suddenly reverse biassed, these carriers have to be swept out of the junction, and this movement of carriers is a current, which can be very large. This is the reverse recovery current, and will usually last us to ns before the charge is depleted. The current will then suddenly cease, or more accurately, drop to the IS level, often causing a very sharp voltage transient with signal components into the MHz or GHz. The 1N400x rectifier diode series are notorius for having a very large charge storage, and could often cause noise when used as the rectifier for audio amplifiers.

In your particular diagram, both of these diode conduction mechanisms would appear to be irrelevant, with the motor implying 'large' currents, and 'long' time constants.

There are multiple questions and uncertainties here:

• How transistors work has been explained in other questions. In short, they work because they are two diodes in a special arrangement, essentially, very close together. There are other reasons for a diode to conduct in reverse, including plain old leakage current (nothing is ideal).
• V(P2, P1) may be defined as constant anyway. If this diagram appeared in an electronics textbook, likely V1 would be an ideal voltage source, and so whatever D1 and C1 do is irrelevant to M1, which has a fixed 5V across it no matter the load.
• Conversely, if you supply a different definition for V1, then M1, D1 and C1 may matter.
• The state of the diode and capacitor is not given. In steady state, the capacitor (with nothing else connected to it) will charge up to a bit less than 5V. If the motor draws enough current that the supply is momentarily reduced, the diode in the off-state will do nothing.
• If an additional load on the capacitor keeps the diode conducting, or the 5V source happened to be rising a significant amount so as to momentarily charge the capacitor, then the diode could turn on momentarily, and then a sudden load on M1 could cause reverse recovery. Even for the worst diodes available, this will do nothing significant: some ~µC of charge will be withdrawn from the capacitor, a drop in the bucket.
• It's not clear what your intent was, if the capacitor could supply the load momentarily despite the diode being basically backwards (a question of how ideal, or not, diodes really are? or how they behave in general?), or whether any particular thing could really happen given the lack of definition in the diagram.