Dear all, please could you explain me function of ESD protective diode? This diode is connected to ground and after some higher input signal, which is not allowed, this overcurrent is going through the diode to ground (this overcurrent is divided to two sides - ground and normal way, so it is lower as if the protective diode is not there).
Is it truth?
Thank very much
Clamp diodes on CMOS are connected to all inputs and to Vdd & Vss. The Schottky diode drop protects the CMOS inputs from see a voltage outside the rails by this diode drop.
The only risk is that ESD has been known to be in the order of pico seconds and if the ESD rise time is faster than that of the diode, the diode will not protect.
There are standard equivalent circuits for ESD to represent a charged body conducted via finger and charged cart for test purposes.
If your input or output is connected to a long cable.. slowing down the rise time of ingress via ferrite chip or CM ferrite for a wire pair, is highly recommended. Of course this also reduces egress and rise time of signal as well as reduces EMI in both directions as well as slowing down the signal so the protection diode can function at device speed.
Clamp diodes are also connected to Vdd for open drain or open collector switches for the same reason to protect the switch output. If a high voltage is desired on pulse then a zener or TVS is used.
It is also known that reflections on flip flops can toggle the output so buffers are recommended:-D for line driving instead of FF.. an ESD event can also back drive FF's used as drivers, so again buffers are used.
Sorry, I think diodes, which protect circuits against overvoltage and these diodes are connected to ground. Could you please explain me, how it works?
Thanks very much