SCEA144 December 2024 LSF0101 , LSF0102 , LSF0108 , LSF0204

6 What if my I/Os Operate Lower Than V_REFA?

To maintain proper functionality of the internal FET of the LSF, the voltage biased at V_REFA node needs to be the lowest voltage seen at the I/Os on A and B side. However, there can be cases where one of the I/Os operate lower than V_CCA and that voltage rail is also unavailable. One example of this is when the input voltage at the I/O channels have a V_OH,min that is substantially lower than V_CCA for worst case conditions. To maintain that the FET operates in the proper state, V_REFA needs to be biased to similar voltages as V_OH,min to achieve signal integrity. One method of resolving this is to create a voltage divider network with the V_CCA rail to maintain proper voltage at the V_REFA node. For single supply operation with only V_REFB rail, please refer to device data sheet.

Figure 6-1 Voltage Divider Network on V_REFA

Use 1MΩ for R₁ to reduce leakage current.

Assume the below variables:

V_x= Desired voltage for V_REFA rail
V_CCA= Voltage at A-side power supply
V_TH= 0.8V (typ. threshold value)
R_BIAS = 200kΩ (external)
R₁= External pullup on A-side power supply
R₂= External pulldown on A-side power supply
V_REFB = B-side power supply

Then the below equation can be used to solve for R₂ given known V_x:

Equation 1.

R_{2} = \frac{V_{x} \times R_{1} \times R_{B I A S}}{R_{B I A S} (V_{A} - V_{x}) + R_{1} (V_{R E F B} - (V_{x} + V_{T H})}

Take an example where V_REFA desired voltage is 1.4V but only a 1.8V power supply is available. In this case, we use R₁= 1MΩ to reduce leakage and V_x to be 1.4V since it is the voltage required. Then the pulldown, R₂ can be calculated through:

Equation 2.

R_{2} = \frac{1.4 V \times 1 M Ω \times 200 k Ω}{200 k Ω (1.8 V - 1.4 V) + 1 M Ω (3.3 V - (1.4 V + 0.7 V)}

Equation 3.

R_{2} = 237 k Ω

6 What if my I/Os Operate Lower Than VREFA?

6 What if my I/Os Operate Lower Than V_REFA?