# How does diode clamp protection work?

I am not able to understand how diode clamp protection works.

Considering the diode clamp protection as shown in Figure 1: When the input voltage (Vin) becomes more than power supply voltage (Vss = 5V), say Vin = 12V, then the equivalent circuit becomes as shown in Figure 2. But what I did not understand is that the potential difference between the terminal Vout and the ground is still Vin=12V, isn't it? Could someone please explain this to me?

Same question is asked here How does a diode clamping circuit protect against overvoltage and ESD? but I could not understand it and I do not have enough points to ask for clarification comment there.

Thanks. • What is missing on your schematic is series resistance, add a resistor in the left branch of the circuit, so in series with Vin. Then in figure 2 the 5 V will drop across that resistor. Without any series resistor the diode D1 would need to carry an infinite amount of current and simply be destroyed. Also ESD protection in ICs relies on some series resistance to be present. Jan 23 '19 at 14:08
• Thanks @Bimpelrekkie, I have made the modification as you suggested.
– Ajay
Jan 23 '19 at 14:19
• Now think what the voltage at Vout is going to be in figure 2. Jan 23 '19 at 14:26
• @Ajay - Welcome to the site :-) FYI, you don't mean to label the positive supply rail as Vss, as used in your text and diagram. I recommend that you read this previous question (and do some more research yourself) to learn about the meaning of terms like Vss (and Vdd, Vcc etc.). Good luck and, again, welcome :-) Jan 23 '19 at 18:53
• Thanks @SamGibson for letting me know, I will take care of this in future.
– Ajay
Jan 25 '19 at 13:24

$$\V_{out}\$$ is connected to the 5V supply in Figure 2. This means that the voltage at $$\V_{out}\$$ must be 5V.
If you are wondering why there is no conflict between $$\V_{in}\$$ and the 5V supply then the answer centres around $$\R_2\$$. If you look at the voltage across $$\R_2\$$ you know that $$\V_{R_2}=7V\$$ as there is 12V on one side and 5V on the other side. Because of Ohm's law this means that there is a current flowing through $$\R_2\$$.