I'm asking this because i'm trying to develop an active overvoltage protection circuit and unsure about how quickly it should react to overvoltage condition. Maybe someone can give me advice on that?
Thanks to everyone. I learned important things.
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6\$\begingroup\$ If you exceed the absolute maximum ratings in the datasheet, nobody will guarantee anything. So design you overvoltage circuit with that in mind. \$\endgroup\$– jippieDec 26, 2013 at 10:31
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\$\begingroup\$ @jippie From what you said it looks like any protection with delay is not protection at all, because overvoltage applied to protected element anyway. Is there protection device with instant action? \$\endgroup\$– user33393Dec 26, 2013 at 11:12
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\$\begingroup\$ @user33393, the trick is as per Andy's answer, to use inductors, caps and resistors to limit the rate of change to something you can handle. Remember also because of parastic capacitance a voltage will never increase with a zero delay, in practice they'll always be at least a little rise time. \$\endgroup\$– PeterJDec 26, 2013 at 12:48
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3 Answers
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An active over voltage protection scheme will be slower than such things as zeners. In any case you should ensure that any overvoltage condition is prohibited in rising to critical levels on what you are trying to protect more quickly than the protection device can respond. This can be achieved with passive components such as resistors, inductors and capacitors.
answered Dec 26, 2013 at 11:43
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\$\begingroup\$ So i place TVS diode first and assume that voltage across it never rises above some V, then i have my active protection, resistor, and capacitor. In fault condition capacitor charges from V, but, because of resistor,voltage can't rise above protected element normal operating range during active protection reaction time. And, if input transient is so powerful, that TVS diode can't handle it, there's nothing i can do. Right? \$\endgroup\$ Dec 26, 2013 at 13:05
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\$\begingroup\$ there is more to the active protection than R and C and I can't really agree or disagree without a circuit. \$\endgroup\$– Andy akaDec 26, 2013 at 13:53
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\$\begingroup\$ @user33393 I believe your LDO regulator is the vulnerable part but this will likely take 20+ volts in normal operation and by the time you've got to this sort of voltage the fuse will have popped so it'll be protected. The LDOR will also need capacitance on the input and this will ensure the rise time of the volts across the TVS is slowed so I don't see anything wrong with your circuit. If you are still concerned a small value resistor in series with the line to the LDOR will help the input cap slow things down significantly. If you're still concerned a zener across the o/p FET will also help. \$\endgroup\$– Andy akaDec 26, 2013 at 19:47
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\$\begingroup\$ I will connect LDO after Rlim. Regulator takes 50 V and TVS is 28 V rated. \$\endgroup\$ Dec 27, 2013 at 6:18
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There can not be a general rule. Some components increase the current exponentially in case of over-voltage and some increase the current just proportionality.
In addition some may have over voltage mechanism (like a clamping diode) and some may not so there is not a single answer that can cover every single component/circuit.
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It all depends on the mechanism of failure.
If the over-voltage damages the component by insulator breakdown, then no excess is allowed, no time is safe, pop! and it's dead.
If the over-voltage causes extra current to flow in a resistor, and heating is the problem, then you may get mS to minutes, depending on the thermal time constant of the circuit. If the supplier doesn't specify an \$I^2t\$ for the component, then it's down to you to do experiments, and make a suitable allowance for not all components being the same as the several you test.
