How to model piezo buzzer using LTspice - Page 1
Forum section: electronics / beginners

author: jj5
author_rank: Unknown
timestamp: April 18, 2024, 05:14:41 am
content: Hi there. I'm just getting started with LTspice and I want to model a circuit which contains a piezo buzzer. My FNIRSI DSO-TC3 component tester tells me that my piezo buzzer is a 21.6nF capacitor, so could I just model the piezo buzzer as a capacitor? I did try to search for information about how to do this on the web, but I couldn't find anything which I could understand.

author: jwet
author_rank: Frequent Contributor
timestamp: April 18, 2024, 02:56:13 pm
content: That's a pretty good first order model and the C is certainly dominant. A of E has some circuit example driving piezo's- kind of tough load.These buzzers are resonant, so a better model would be a resonant LC at the peak output of the buzzer. The C value would be you 21 nF (you read this at LF I'm guessing). The L would be whatever resonates to create the peak response. I would imagine the Q is about 10 so would inlcude a parallel R of whatever the peak reactance of the L or C is.This is similar to the model of a quartz crystal or ceramic resonator.

author: jj5
author_rank: Unknown
timestamp: April 19, 2024, 09:16:59 am
content: Thanks for your help with this.I found a datasheet for a similar component here:https://au.mouser.com/datasheet/2/1005/202005110dd2c-2325130.pdfIt lists the capacitance as 25nF, which stacks with my observations from my component tester. It also says the resonant frequency is 3.0 ± 0.5 KHz and the resonant impedance is 500 Ω (max).I found lib/sym/Misc/xtal.asy in my LTspice library. It models the piezo buzzer as a capacitor. It has settings for series resistance/inductance and parallel resistance/inductance, but no provision for nominating resonant frequency. If I know the frequency (3k, as above) can I calculate a series/parallel inductance? Do I put in the resonant impedance of 500 Ω as a series resistance?

author: jj5
author_rank: Unknown
timestamp: April 19, 2024, 09:43:42 am
content: Quote from: jj5 on April 19, 2024, 09:16:59 amApparently the series inductance (L) is given with: L = 1/((2πf)^2 x C); where f = 3,000 Hz and C = 0.000000025 F; so L = 112.58 mH. Does that sound right to you?

author: jwet
author_rank: Frequent Contributor
timestamp: April 20, 2024, 03:34:01 pm
content: i was just going to send something like that to you. Good luck.

author: jj5
author_rank: Unknown
timestamp: April 20, 2024, 03:35:37 pm
content: Thanks again for your help. One thing I still don't understand: do I put in the resonant impedance of 500 Ω as a series resistance?

author: jwet
author_rank: Frequent Contributor
timestamp: April 20, 2024, 03:53:27 pm
content: At 3000 Hz, the impedance of the cap and inductor are equal in magnitude but opposite in sign- they sort of a cancel. Since the L and C is in parallel, the combined impedance would go to infinity with no resistance in circuit. 1/Z = 1/z1 + 1/-z1, leads to a Z of infinity and a very peaky response. This doesn't happen due to R's in the circuit though quartz crystals come close with Q's of many thousands. The Q of a resonator is the fractional bandwidth that it is resonate over. It is a also equal to the loss in the resonator per cycle. With no R, Q would be infinite and the bandwidth would be zero and once you tickled the part, it would go forever. Think tuning fork. So if the Q of the resonator is about 10 - this is 300 hz wide 3 db bandwidth. The impedance of the cap is 1/2*pi*f*c= -j2123 ohms. The -j means that its complex, the inductor would have the same impedance but positive- they cancel. To make a Q of 10, add a parallel R of 212 ohms- my 500 was an approximation before knonwing much. You can sweep the circuit in spice and see all the magic. The R is a loss, it can go in parallel with both the L and C or in series with either component but in the loop. What a resonator does is slosh energy back and force between the two reactances. The R is a loss term and it burns some power on each "slosh" so has to be in the loop between the L and C, either parallel to both or more usually lumped with the L or C which is really the case.Hope this makes sense- have fun

author: jj5
author_rank: Unknown
timestamp: April 20, 2024, 04:45:48 pm
content: I had a go at doing a sweep, does this look right?

author: Zero999
author_rank: Super Contributor
timestamp: April 21, 2024, 08:31:03 am
content: Quote from: jj5 on April 19, 2024, 09:43:42 amI would expect the capacitance to be much smaller and the inductance higher.A crystal is equivalent to a large shunt capacitance, in parallel with an inductor, resistor and capacitor connected in series. It's the series circuit which determines the resonance, rather than the shunt capacitance, which is specified on the data sheet.https://www.analog.com/en/resources/technical-articles/modeling-of-quartz-crystal.htmlAs far as a piezoelectric transducer is concerned, I believe it's similar to a crystal, but with a lower resonant frequency and much lower Q as its purpose is to transfer the energy into the air. This is something I've looked into before, but didn't get very far, because it depends on how the transducer is housed and even the surrounding environment i.e. whether there are any reflections.

author: jwet
author_rank: Frequent Contributor
timestamp: April 22, 2024, 02:04:28 am
content: Your plot looks pretty correct to me. The db scale on the left is a bit arbitrary, you can make the peak 0 db by adjusting the amplitude of the source and then get more meaningful data related to the peak. These sweeps are also usually db on Y axis (a logarithmic unit) and log in frequency along X axis. If you're log log, you can easily pick out rolloff like 20 db/decade etc. These types of sweeps are called BODE plots- something for you to follow down the rabbit hole on... They're a very powerful tool and in a little more advanced form which includes a phase plot can give you an indication of stability, etc.As the last poster said- some of this analysis of acoustic devices in just spice don't give you the whole picture. When you mount a transducer and put it into an enclosure, all this nice RLC type analogy gets modified by what ever "resistance" the mounting does and the resonances in the cabinet- much like speaker design. Often piezo discs are mounted lightly with superglue at there edges and put in a cavity that acoustically enforces the sound. More to play with...