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I asked a question about protecting a circuit from current spikes from a piezo electric device: Protecting circuit from piezoelectric disc voltage spike. Someone graciously provided me with a solution in the image below. I have a follow-on question.

Assuming that the piezoelectric device is struck very hard, it's voltage can spike to 30v. That's going to generate current that splits and some goes through the 1M resistor, some goes through the zener diode, and some goes out as signal to an IC (an ATTiny85 in my case). It's my understanding that electricity tries to follow the path of least resistance. If the voltage/current is high enough it will overcome the zener diode and flow through that path. Why would any electricity ever take the 1Mohm path if that path provides such high resistance? Also, doesn't the chip provide for an "easier" path for the electricity to flow through as well so wouldn't it just want to flow through there and fry the chip?

enter image description here

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    \$\begingroup\$ Supplement to The Photon's answer: The zener diode starts conducting substantially slightly below its rated voltage and passes increasing current as Vz rises. As long as the zener is energy-dissipation rated for >= the amount of energy the piezo can ever deliver then it will NEVER be "overcome". In practice here Vz is unlikely to ever exceed 6V and probably much closer to 5.1V. \$\endgroup\$ – Russell McMahon Oct 21 at 23:06
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    \$\begingroup\$ if you have a bucket of water, and you punch a 1 cm hole and a 2 cm hole in the bottom, then does the water drain through the 2 cm hole only, since it is a path of lower resistance? .... the zener is like a hole punched near the top of the bucket \$\endgroup\$ – jsotola Oct 21 at 23:29
  • \$\begingroup\$ You misunderstand: you are unlikely to get 30V. You 'might' get 30 volt sometime with some circuit, but that's a rule-of-thumb number for a typical open-circuit piezo. The suggested circuit with 1M load is not expected to go above 5V. If it tries to, the zener will prevent it, but that's just protection. When it tries to go negative, the piezo will also prevent it. Design operation is with the piezo voltage less than 5 V, because the expected power output can't drive the 1M higher than that. \$\endgroup\$ – david Oct 22 at 8:32
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It's my understanding that electricity tries to follow the path of least resistance.

This is an oversimplification.

(DC) Current goes through every path available to it, in proportion to the conductance of that path. And conductance is just the flip side of resistance

$$ G = \frac{1}{R}$$

If the voltage/current is high enough it will overcome the zener diode and flow through that path. Why would any electricity ever take the 1Mohm path if that path provides such high resistance?

Ohm's law says, for a resistor

$$I=\frac{V}{R}$$

If you apply 5 V across a 1 megohm resistor, 5 microamps will flow through it.

If you apply 30 V across a 1 megohm resistor, 30 microamps will flow through it.

If it didn't behave that way, we wouldn't call it a resistor.

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    \$\begingroup\$ So assuming that everything is okay, if the piezo generates a tiny signal, the current-overload defense will act parasitically and drain off energy in all circumstances, correct? \$\endgroup\$ – mj_ Oct 22 at 1:45
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    \$\begingroup\$ @mj_ The zener is the overwhelming energy absorber once Vpiezo rises to about Vz \$\endgroup\$ – Russell McMahon Oct 22 at 2:15
  • \$\begingroup\$ @RussellMcMahon, so couldn't you just strike the 1Mohm resistor in this case? \$\endgroup\$ – mj_ Oct 22 at 13:55
  • \$\begingroup\$ @mj_ Re the 1M - note that they say "*experiment for best sensitivity" - I'm not sure of the exact mechanism (or if there is any) but they imply that a slight load on the piezo improves results. IF the main load is DC coupled the 1M would be much lighter load but if AC coupled the 1M MAY have some effect. But quite possibly not. \$\endgroup\$ – Russell McMahon Oct 22 at 20:11
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This is a supplement to The Photon's answer:

A zener diode starts conducting substantially slightly below its rated voltage and passes increasing current as Vz rises. As long as the zener is energy-dissipation rated for >= the amount of energy the piezo can ever deliver then it will NEVER be "overcome". In practice here Vz is unlikely to ever exceed 6V and probably much closer to 5.1V.

From this datasheet.

Here are voltage/current curves for a family of 500 milliwatt maximum dissipation zener diodes. It can be seen that even the 5V6 zener is conducting 50 mA by the time Vz reaches about 5.8V. It is unlikely that a piezo device will provide that much power.

enter image description here

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    \$\begingroup\$ Judging from the chart, it looks like zener diode's voltage top out. So a 5.6V stays around there and just the current spikes. In the piezo case, assuming you had a 30V input, would 30V be present across the zener diode or would it get stuck? \$\endgroup\$ – mj_ Oct 22 at 2:15
  • \$\begingroup\$ @mj_ Note what I said above. The curves are shown truncated due to their dissipation limits. In practice they keep going up until they melt - or you get a bigger one. || As above - I said that a 5V zener would be very (= immensely) unlikely to reach 6V with any usual piezo. Maybe a large large large one from a super sonar struck with a hammer would hive much more energy, but for typical sounders et the energy will be low. || "In practice here Vz is unlikely to ever exceed 6V and probably much closer to 5.1V." <<<< 30V \$\endgroup\$ – Russell McMahon Oct 22 at 7:06

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