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# Second-order time constant

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#### mastevano

##### Newbie level 6
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).

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.

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.

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

Points: 2
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.

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

Points: 2