# Filter design for piezoelectric transducer

I'd like to use a piezoelectric transducer at one of its resonant frequency (4500 Hz). In order to get a clearer signal, I would like to use a band-pass filter.

First of all I calculated the impedance of the piezoelectric transducer. To achieve that, I put a variable resistor at the output of the transducer:

simulate this circuit – Schematic created using CircuitLab

Then I generated an audio signal with a frequency of 4500 Hz, and measured for two different values of Ri the amplitude of the voltage V across the variable resistance.

For both values, I used the formulaRi/(Ri+Z)=V/V1, which allowed me to calculate Z (I found 12 kOhms). Is the method correct?

Now my problem is: I'd like to apply a band-pass filter, with a centre frequency of 4500 Hz, and a band-width of a few hundreds Hz. But according to the formulas I've found for RLC band-pass filters, it seems that I will need an inductance definitely too big to reach such a BW. What would be the best way to implement the band pass filter, preferably with passive components? Thanks!

• Is this for a sensor (probably) or an emitter in which a BPF isn't necessary. What Q do you want ? 100 is possible with a pot. and 2 Op Amps or less with 1 – Tony Stewart Sunnyskyguy EE75 Oct 4 '16 at 22:45
• I'd like a bandwidth of a few hundred Hz. Also I'd prefer to use passive components, as I need to minimize the energy consumption. – Vincz777 Oct 4 '16 at 22:50
• how many micro amps is acceptible – Tony Stewart Sunnyskyguy EE75 Oct 4 '16 at 22:55
• If your piezo has a decent resonance @ 4500 Hz, you may not need a band-pass filter. Some piezos are so high-Q that you may have to work hard to de-Q it. Do you need extra filtering to knock down adjacent interfering signals & noise? – glen_geek Oct 5 '16 at 1:39
• @TonyStewart.EEsince'75 A few uA would be a maximum. – Vincz777 Oct 5 '16 at 7:29

## 1 Answer

• I would use this basic circuit with rail to rail Op Amp single supply but move all the grounds to Vbat/2 with split R's ,
• then use a trimpot for the output divider to change Q
• Q = f/BW. you indicate few hundred Hz or a Q=15
• then choose input R > 15k to 100k
• easy formula exist for this active BPF.
• Choose a low noise Op Amp with low current .
• wide range of supply voltage options are available

any questions?

Yes LC BPF will need large inductors, not feasible or accurate due to large tolerances.

If you want wide bandwidth but steeper out of band rejection and high sensitivity, you can use two stages cascaded with any gain that you want. This site has an auto-design input. 1% values can be selected to the nearest part or changed on the linked site. But again for a single supply, dc gnd must be changed to Vbat/2.

• Thanks! But how should I choose the Op-Amp? The voltage amplitude of the resonant frequency across the piezo is only 2-3 mV. In the link you provided, I didn't find any op-amp with large GBW, low voltage offset (but is it necessary?), and low power consumption. – Vincz777 Oct 5 '16 at 7:54
• digikey.com/product-detail/en/microchip-technology/MCP6144-E-ST/… These 4 IC's use much< 1uA... What will excite this resonator so precisely? – Tony Stewart Sunnyskyguy EE75 Oct 5 '16 at 12:04
• Thanks! Is the +/3 mV voltage offset of MCP6141 a problem, as my signal is only 2-3 mV wide? The resonator will simply be excited by an audio soundwave. Then I'd like to plug a comparator to the output of the filter, and the comparator will trigger an interrupt to an MCU. – Vincz777 Oct 5 '16 at 12:23
• You would be wise to use a diode peak integrator with hysteresis like a Schmitt trigger gate. A very loud bang can still contain spectrum in all frequencies. Use the extra OpAmp as a comparator with sufficient positive feedback >10% and BPF's reject DC. Choose gain wisely. With this design, you can choose any frequency and use an electret mic instead. – Tony Stewart Sunnyskyguy EE75 Oct 5 '16 at 12:50
• Not sure of what you mean by diode peak integrator. The comparator I use (MCP6541) has already a 3 mV hysteresis. I discarded using electret mic because of the energy consumption. – Vincz777 Oct 5 '16 at 13:18