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.
Best Answer
\$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\$.
A current flowing through a resistor means a voltage drop, the bigger the drop is - the larger the current flowing through the resistor is (and the more energy the resistor will turn into heat as a result!).