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So I've been getting into the world of building circuits with an Arduino controller and a breadboard. I've been working in IT for years, but I'm not super familiar with electronics at this level.

While following a tutorial for building a "knock sensor" using a piezo speaker as the sensor, it connected the piezo speaker from ground to an analog input pin on the Arduino. Then, it connected a 1M ohm resister in parallel to that. All it said was that doing so protected the Arduino from the voltage spikes caused by the knock sensor.

After asking around, I learned that this resistor was called a "burden resistor," but I can't seem to understand how adding a resistor in parallel to the element that generates the voltage would protect the Arduino board.

Can someone explain this to a newbie like me?

This is the webpage that shows the burden resistor: http://arduino.cc/en/Tutorial/Knock

JYelton
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Gogeta70
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  • Possible duplicate of [What is the difference between burden resistor and a normal one?](http://electronics.stackexchange.com/questions/36341/what-is-the-difference-between-burden-resistor-and-a-normal-one) – alexan_e Jan 04 '14 at 00:54

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The piezo can generate very large voltages (e.g. 90V) above the operating range (6V) of the ATmega and would blow any input. That said the power is very little and any resistance will load the spike, so that current is absorbed in the 1Mg ohm resister, rather then the hi-impedance input of the ATmega.

Here is a link to Electrical Over-Stress and Electrostatic Discharge Protection of ICs by IT E2E group. It has link to a PDF presentation that does a great job explaining how the chips internals can tolerate or fail such voltages.

mpflaga
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  • the 1Mg takes the burden of the surge. – mpflaga Jan 04 '14 at 02:41
  • So the burden resistor going from the input pin to the ground provides a lower resistance pathway for the current to flow? In that case, why put a resister there at all? Why not just use a regular wire? Also, am i correct in assuming that the analog input is reading the voltage between the piezo and ground? Sorry if my questions are dumb. – Gogeta70 Jan 04 '14 at 02:46
  • @Gogeta70 A mere wire would short-circuit the input. I don't understand the suggestion. – user207421 Jan 04 '14 at 04:38
  • @EJP Well, if the idea is to create a low resistance pathway to ground, why use a resistor at all? I wire has a lot lower resistance than a 1M ohm resistor. – Gogeta70 Jan 04 '14 at 05:11
  • @Gogeta70 Because the idea isn't to 'create a low resistance path to ground'. The idea is to create a *lower* resistance to ground than the input impedance of the IC. A wire is a short circuit. You can't expect that to work across an input circuit. – user207421 Jan 04 '14 at 08:55
  • take a look at this web posting. http://www.openmusiclabs.com/learning/sensors/piezos/ It briefly explains piezo and how they work as pickups or speakers. Where figure 4 shows a pick up circuit with proper protection. – mpflaga Jan 04 '14 at 20:03
  • @EJP I guess i'm having trouble understanding the flow of current here. Current flow takes the path of least resistance, so the way i see it is that any current is going to flow to ground instead of into the analog input. How will the analog input detect the voltage generated by the piezo then? You said that simply using a wire creates a short circuit. How does adding a resistor to it change that? – Gogeta70 Jan 04 '14 at 21:52
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    kirchhoff's current law - it is not black and white, rather proportional. The input impedance of the IO pins is very large, Possibly 10MgOhms, not perfectly infinite. See http://www.ampbooks.com/home/tutorials/lesson-005/. So more of the charge is dissipated through the resistor then the HiZ input of the ATmega. – mpflaga Jan 05 '14 at 00:30
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    in response to above, a wire short instead of a resistor would be blank and white in that all the current would go through the shunt and none (even the very little needed) for the ATmega Input. Also not V=IR and if R=0 then V=0. And you don't want that as your input. – mpflaga Jan 05 '14 at 00:34
  • @Gogeta70 A wire is a zero resistance, which is a short circuit. A 1M resistor isn't. Surely you understand the difference between a piece of wire and a rather high-valued resistance? You're expected to. At least I expect it. – user207421 Jan 05 '14 at 05:28
  • @mpflaga I think i understand now. Basically, the burden resistor and the input resistance of the analog input are like 2 resistors in parallel. In which case, they both conduct an amount of current based on ohms law. And since the burden resistor has a much lower resistance than the analog input, it carries most of the current. Is that about right? EJP, i understand that a wire has nearly zero resistance. I just didn't understand why you needed resistance because i thought all the current went to ground anyway. – Gogeta70 Jan 05 '14 at 05:46
  • @Gogeta70 You need a resistance because you don't want to short-circuit the input. Otherwise there is no input. I don't know why this is so mysterious that it needs to be stated three times. – user207421 Jan 05 '14 at 12:09
  • @EJP i do understand that using a wire instead of a resistor would draw all the current to ground and act as a short circuit. I didn't understand how adding a resistor changed that, but after seeing the webpage that mpflaga posted i understand now. Thank you both for your help. – Gogeta70 Jan 05 '14 at 19:19