ltspiceschottky
I am trying to simulate a voltage clamp using two Schottky diodes.
I am trying to recreate the case when a resistor (R1) shorts (0 ohms)… I would expect to have around 3.3v at out port. Without adding the series resistance PS, I got 12V. Adding I got around 7.9v.
So I am suspecting I have to set some parameters to have a better simulation. any idea?
Those diodes protect the pins from overvoltage, undervoltage, and electrostatic discharge. Basically, they prevent the voltage on the pin from crossing outside of the supply voltages, in this case above 3.3 or below 0.
If you want to feed in a 5v signal, put in a series resistor of maybe 100 to 1k ohms or so. The diode will clamp the voltage and the resistor will prevent too much current from being drawn from whatever is sourcing the signal. This only works for low speed signals. If you need something fast, then getting a dedicated level shifter chip is a better idea.
Here is the real behavior of almost the same circuit, installed in a real device.
The differences are that the driving source impedance is unknown (V1 is ideal in the simulation), and the load impedance corresponding to R3 in the diagram is unknown.
The behavior is nowhere near the rosy picture painted by the simulation. (On the other hand it is good enough that the everything works in the DUT).
The horizontal divisions on the graticule represent 500 nS intervals. The upper trace is the transistor's collector: one division is 5V. The bottom trace is from the base: one division is 2V. As can be seen, the rise time is quite slow, taking over 500 nS to rise to more than 90%. What might be considered a solid high logic level is probably not established for some 250 to 300 nS.
Best Answer
What's happening is that when you have R1 set to zero, and 50 ohm impedance on the supplies, the 3.3V and 12V supplies are forming a voltage divider (they're modeled with a series resistance, as you set.)
So you're getting the voltage midway between them plus the diode drop Vf, so:
To visualize this better, set the supply resistances to zero, and add the 50 ohm series resistances explicitly. Then you will see that voltage divider at work.
That all said, a dead-short to +12V is not really a realistic test case. In the real world the part would actually get powered up to 12V and would be destroyed.
Instead, the protection diodes are meant to absorb transients, not protect against DC shorts. For that scenario, look into ESD tests that define a specific human body model to characterize the performance of protection circuits.
More about that here: https://en.wikipedia.org/wiki/Human-body_model
And some detail on modeling ESD in Spice: https://www.youspice.com/simple-spice-esd-generator-circuit-based-on-iec61000-4-2-standard/