# How do protection diodes work?

There is something I don't understand with protection diodes. I am talking about a series diode between the + and the load of the circuit to prevent damaging components when there is a wrong polarity on the supply, like this :

Normally the current through the diode grows exponentially with its voltage accross, with a forward voltage of 0.7 V (for example).

I understand that if there was no load, then there would be 12 V accross the diode, resulting in a very large amount of current which would quickly destroy it.

But when there is a load, and that the circuit is working properly :

• Why is the voltage across the diode always 0.7 V ? I understand that all diodes have a forward voltage due to their depletion area, but I don't understand why if I supply the diode with a load behind, the voltage across the diode will always be 0.7 V and across the load 11.3 V. What sets this value ? Why not something else ?
• If the load needs a current of 10 mA or a current of 1 A, then the current through the diode will be 10 mA or 1 A, and given its characteristic curve the voltage across the diode will be very different. It seems like a contradiction to the above point.

There is clearly something I am missing there, hope you could enlighten me. :)

The voltage through an a diode is not always 0.7V. It's an approximation that we use that is "good enough" for most applications. The voltage across a diode is proportional to the the current through the diode.

If you look at a typical diode I-V curve such as the image below, as the current If increases, Vd doesn't change a whole lot. The knee of the curve, the point at which the diode begins to conduct is ~ 0.7.

In you're circuit, you are right- The load voltage would be ~ 11.3V. It's an approximation because Vd is dependent on the current through it.

If your circuit draws 1A, then the diode voltage maybe 0.9V (it depends on the diode) and the voltage at the load would be 11.1V instead of 11.3V.

• Thanks a lot I get it now :) – AladdinSane Feb 29 at 20:13

I am talking about a series diode between the + and the load of the circuit to prevent damaging components when there is a wrong polarity on the supply, like this :

Your circuit is very like a half-wave rectifier.

Normally the current through the diode grows exponentially with its voltage accross, with a forward voltage of 0.7 V (for example).

Correct.

I understand that if there was no load, then there would be 12 V accross the diode, resulting in a very large amount of current which would quickly destroy it.

You've tripped up here. If there was no load then there would be 12 V between LOAD+ and LOAD- on your diagram. The voltage across the diode would then be zero.

If you short-circuited the load then there would be 12 V across the diode, a very large current would flow and, if not limited by the power supply, the diode would burn out.

But when there is a load, and that the circuit is working properly : Why is the voltage across the diode always 0.7 V ? I understand that all diodes have a forward voltage due to their depletion area, but I don't understand why if I supply the diode with a load behind, the voltage across the diode will always be 0.7 V and across the load 11.3 V. What sets this value ? Why not something else ?

It's an approximation. The forward voltage does vary with current.

Figure 1. Relationship of voltage and current in a forward-biased diode. Source: Instrumentation Tools.

The linked article explains that, "As you can see in Figure (a), the forward current increases very little until the forward voltage across the pn junction reaches approximately 0.7 V at the knee of the curve.After this point, the forward voltage remains nearly constant at approximately 0.7 V, but IF increases rapidly. As previously mentioned, there is a slight increase in VF above 0.7 V as the current increases due mainly to the voltage drop across the dynamic resistance. The IF scale is typically in mA, as indicated."

If the load needs a current of 10 mA or a current of 1 A, then the current through the diode will be 10 mA or 1 A, and given its characteristic curve the voltage across the diode will be very different. It seems like a contradiction to the above point. There is clearly something I am missing there, hope you could enlighten me. :)

Hopefully we've gone some way to enlightenment.