Second-order time constant

[mastevano]

Hello Everyone! How do you calculate a second-order time constant. I have the following circuit with two caps and two resistors. Initially, everything is connected (the switch is closed). Then the switch (which is actually a FET) is opened. I would like to determine how long it takes for the uP pin to read a "low".

A schematic is attached (in visio format).

Added after 5 minutes:

For some reason it won't allow me to add an attachment. Basically the circuit has four legs in parallel.

1) A resistor to ground
2) A capacitor to ground
3) A resistor in series with a cap to ground.
4) Inductive Load

The node in between the cap and the resistor in leg 3 goes to the uP node.

A FET is turned OFF and I want to determine how long it will take for the uP to see that it is off.

 

[zorro]

Mastevano,

There should be something that limits the current when the switch is closed (a resistor in series with the battery or with the inductor).
Regards

Z

 

[mastevano]

Hi!

Thanks for the response. I am actually using a smart highside power switch which can handle the high current. I also have a diode connection from the output to ground to protect the device from negative transients.

 

[Kral]

mastevano,
You need to supply more detail on the inductive load. With a "perfect" inductor with zero DC resistance, the initial voltage across the inductor/cap combination will be zero. When you open the switch, the bulk of the initial inductor current will flow through the suppression diode, which will limit the voltage to -.7 volts. The uP input will be at zero volts at the instant you open the switch.
~
Without the suppression diode, the intial uP input would still be zero, but the L/C circuit would exhibit a damped oscillation, starting out in the negative direction from zero.
Regards,
Kral

 

[mastevano]

The load of the inductor is between 2-6 ohms and the inductance is about 10mH.