I2C level shifter isolation

Question

I am designing an FPGA-based SOM. There are I2C slave and master devices on the board. The I2C bus on the board includes an MCU that acts as a system manager and a bridge from UART to I2C, a PMIC device configured by I2C, a component containing an RTC and temperature sensor, the FPGA itself, the clock generator, and the output to the external environment from the mezzanine connector outside the board.

The FPGA has a power-up sequence. FPGA IOs are not supposed to rise before the VCCIO voltage, so I have to go to isolation on the bus. For this reason, I decided to make a simple MOSFET-based level shifter so that it would take up little space and be cheap.

I made a SPICE simulation for a component containing two N-channel MOSFETs in a single package. The line was designed to meet the fast-mode I2C standard. Already the RTC/temperature sensor component and FPGA are using 1.8 V; other devices are using 3.3 V.

In theory, I should be able to provide both isolation and level shifting, because if 1.8 V VCCIO is not available, the MOSFETs will be cut off and there will be no leakage from the 3.3 V I2C line to the FPGA pin (a leak that may cause problems).

However, in the simulation I made with the SPICE model offered by the manufacturer of the MOSFET, while there is no problem during normal operation when VCCIO is removed and there is no switching on the 3.3 V side, 3.1-odd V leakage appears on the FPGA side. Additionally, if there is switching on the 3.3 V side when there is no VCCIO, I still read 1.8 V on the FPGA side.

What exactly do you think is the reason for this? Am I missing something? What are your possible solution suggestions?

Note: All parasitics on the bus were made taking into account the FPGA's IBIS model, component datasheets, and possible path widths and lengths in the PCB stack.

Normal operation is the same for SCL and SDA:

Normal operation, VCCIO side waveform:

Normal operation, 3.3 V side waveform:

With no VCCIO (floating) and no switching:

VCCIO side waveform with no VCCIO (floating) and no switching (3.3 V side already showing flat 3.3 V):

Transistor drain current waveform with no VCCIO (floating) and no switching on the 3.3 V side:

When there is no VCCIO (floating) and there is switching on the 3.3 V side:

VCCIO side waveform with no VCCIO (floating) and 3.3 V side switching (3.3 V side same as normal operation):

Answer 1

I am not sure why your last waveform looks like that. I am able to simulate correctly. Here is my simpler test circuit.

The DMN2991UDR4 NMOS Spice model was downloaded from here and was renamed to DMN2991UDR4.mod and included in the circuit using the .inc directive. And here is the waveform.

S1	S2	Outputs
ON	OFF	Both 0V
OFF	OFF	Both raise to 3.3V and 1.8V, respectively
OFF	ON	Both 0V
ON	ON	Both 0V

Note: V(sw1) (red) is set to 2V instead of 1V like V(sw2) (cyan) just to make the waveforms clearer when plotted together. V(sw1) = 2V switches S1 OFF and V(sw2) = 1V switches S2 OFF.

Now, if the gate and the 1.8V side are pulled down to 0V like below

The waveforms become

The 1.8V side remains low (blue waveform) regardless of whatever signal that appears on the 3.3V side, which means the 1.8V side is isolated from 3.3V side like what you have expected.

Edited

If you need to wait till the FPGA to complete its power-up sequence and you are able to spare an I/O pin, then here is a solution using one PNP, one capacitor, and 2 resistors

And here is the waveforms

The FPGA turns ON (FPGA_Ctrl=1.8V) the voltage shifter when it is ready and then it is powered down (FPGA_Ctrl=HiZ) as shown in the purple waveform. It takes a while for the C3 to discharge, but eventually the 1.8V side is isolated. The signal bumps you see when it is completely isolated are less than 90mV in the simulation.