TCA6424A: What is the rise and fall time of the Port Pins when output changes?

Hey Johnny,

The port outputs rise/fall times are dictated by the capacitance on the bus.

Lets look at figure 19 to get a better understanding of how the rise and fall times can be estimated. We can see two FETs which control the output portion of the device (Q1 and Q2). If the Q1 FET turns it, it pulls the signal high. If Q2 turns on, it pulls the signal low. Now we will have some amount of capacitance on the Pxx port pin so what we will see is an RC. C is the parasitic capacitance while R is Ron of the FET.

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Below is taken from 6.4 Electrical Characteristics. We can use the VoH/VoL and the IoH/IoL to estimate the Ron when the FET conducts.

Example1: Lets say VccP is 3V and we see the device sources 8mA. Then Ron=(Vccp-Voh)/I or (3-2.6)/8m = 50 ohms

Example2: Lets say VccP is 3V and we see the device sinks 8mA. Then Ron=(VoL-GND)/I or (0.25)/8m = 31.25 ohms

We now know the Ron for pull up and pull down.

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Assume parasitic capacitance is 10pF (8.5 max due to our device and lets say 1.5pF from board).

I will make 2 assumptions:

1) first the device's port pin is connected to a high impedance pin/device

2) if we calculate the rising time then the signal was previous low (GND) and if we calculate the fall time then the signal was previously high (VccP).

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Rising Calculation:

The equation for a charging cap is Vcap=Vsupply(1-e^(-t/RC))

Lets say rise time is defined from 0V to 90% of Vcc.

This means we want to see Vcap= Vsupply x0.90

Now we can solve for time.

0.90Vcc=Vcc(1-e^(-t/RC))

0.90=1-e^(-t/RC)

e^(-t/RC)=0.10

-t/RC=ln(0.1)

t=-RCln(0.1)

Filling in R for 50 ohms from previously and 10pF for C

time = 1.15 nano seconds

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Falling Calculation:

The equation for a discharging cap is Vcap=Vsupply x e^(-t/RC)

Lets say rise time is defined from %100 to 10% of Vcc.

This means we want to see Vcap= Vsupply x 0.10

Now we can solve for time.

0.10Vcc=Vcc x e^(-t/RC)

0.10= e^(-t/RC)

ln(0.1)=-t/RC

t=-RCln(0.1)

Filling in R for 31.25 ohms from previously and 10pF for C

time = 0.720 nano seconds

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I hope this answers your question,

-Bobby