Consider this circuit taken from Over Voltage Protection Circuit for Automotive Load Dump

It provides over voltage protection when the input voltage exceeds 19.2V ( as set by R1 and R2).

The turn off delay is related to discharging C1.

enter image description here

Page 4 of the linked pdf says the following ( I have bolded the sentence I am interested in).

The circuit is designed to actuate quickly but reconnect more slowly. Capacitor C1 rapidly discharges to ground through the LMV431 when over voltage is detected. When conditions return to normal, reconnect is delayed by the R3•C1 time constant. Most loads (usually regulators) contain large input capacitors which provide time for the disconnect circuit to engage by limiting the transient slew rate. The nature of the expected transient along with the available capacitance will determine the required response time. The shut off action of this circuit occurs in about 12 µsec. Maximum transient rise times are limited in proportion to this time interval by Cload. This circuit was tested with a Cload of 1 µF. Larger Cload is allowed and recommended if fast rising, low source impedance transients are expected.

I am wondering if the time was measured or calculated ?
If it was calculated, how did you arrive ~12us ?


1 Answer 1


The shutoff time is determined by how long it takes the voltage on C1 to drop from 6.8V to the threshold voltage of Q2, plus some delay associated with shutting off Q1. Clearly, the only thing that can remove charge from C1 is D1, the LMV431AIMF.

Neither the threshold voltage of Q2 nor the current capabilities of D1 are well-defined, so the delay is probably measured (empirical) rather than calculated (theoretical).

You can assume, however, that it will scale linearly over some range with the value of C1. At the low end, it will be limited by the gate charge of Q2. At the high end, it will be limited by the dissipation limits of D1.

  • \$\begingroup\$ It's what I had assume that it was measured rather than calculated. Even my rough calculations are keeping me at ~28us (assuming ideal and using absolute Ik for D1 in datasheet). \$\endgroup\$
    – efox29
    Commented May 27, 2016 at 15:51
  • \$\begingroup\$ No, the dynamic impedance of D1 (from the datasheet) is well under 1 ohm. The absolute maximum current is given as 30 mA. If it saturates at that current, then the voltage on C1 would fall at a rate of 0.3 V/us. A drop of, say, 4V would require 13 us, which suggests that D1 probably sinks more current that. \$\endgroup\$
    – Dave Tweed
    Commented May 27, 2016 at 16:02
  • \$\begingroup\$ Ah. I understand. \$\endgroup\$
    – efox29
    Commented May 27, 2016 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.