I am trying to determine the specifics (windings, wire gauge, core material) for an impedance-matching transformer for a piezo pickup on an acoustic guitar. If I can arrive at a 2 megohm input, 20 kilohm output 100:1 ratio design, that should do it. I have found 100:1 ratio transformers online, but for an entirely different purpose, and no impedance specs given. The idea to is to replace an active, external-power consuming device (preamp) with a purely passive one. The frequency range should be about 50 Hz to 5 kHz; the lowest note (open 6th string E) is 82.4 Hz, and the highest (depending on design) is about 1000 Hz, so 5 kHz should give enough range for overtones/harmonics. Should I start based on a center frequency, or number of windings...or what? I know this is somewhat similar to a question asked in 2016, but I am not trying to boost a signal, just do appropriate impedance matching to an amplifier's input.
-
\$\begingroup\$ I didn't know that a guitar has a piezo pickup. \$\endgroup\$– Marko BuršičCommented Nov 4, 2021 at 18:33
-
1\$\begingroup\$ Why freq. range 50 - 5k? It is only base frequency, there is lot more higher ones. I think you want 20 kHz or you are going to lose whole instrument's timbre. I think you will get very low impedance on your transformer, even big industrial ones have low impedance. Next stage should have big impedance to work (amp ins have it). \$\endgroup\$– JomongerCommented Nov 4, 2021 at 18:51
-
1\$\begingroup\$ What is the voltage amplitude? Perhaps it could be used a JFET opamp directly, else you would need a transformer with quite lot of turns, that could even self resonate wit piezo, also the core IMO should be of soft iron. Much easier to make an attenuator for the opamp. \$\endgroup\$– Marko BuršičCommented Nov 4, 2021 at 18:55
-
1\$\begingroup\$ Minimum primary inductance to achieve 50Hz should be around $$ 2 \,\text{M}\Omega / 314 \,\text{rad/s} \approx 6000\,\text{H} $$ I don't think it's going to be doable \$\endgroup\$– carlocCommented Nov 4, 2021 at 19:00
-
1\$\begingroup\$ "The idea to is to replace an active, external-power consuming device (preamp) with a purely passive one." No way. \$\endgroup\$– Marko BuršičCommented Nov 4, 2021 at 19:08
3 Answers
The ratio of the transformer's primary and secondary impedances is not simply equal to the turns ratio, it is equal to the square of the turns ratio.
Turns ratio = Nprim/Nsec = sqrt(Zprim/Zsec).
For maximum power transfer to the load Zprim must be made equal to Rsource and so, for maximum power transfer, the following equation is used to calculate the transformer's required turns ratio.
Turns ratio = Nprim/Nsec = sqrt(Rsource/Rload).
So, when Rsource>Rload a step down transformer is required and when Rsource<Rload a step up transformer is required to achieve maximum power transfer to the load.
-
\$\begingroup\$ I'd like to point out that "For maximum power transfer to the load..." is not always the best choice where the goal is not to transfer energy but rather information. The information of the sound is in the charge generated in the piezo element, so 'the best' approach would be to use a charge amplifier like this: link but there are many more examples. \$\endgroup\$– HarryHCommented Mar 4 at 6:38
At low frequency, a piezo transducer can be modeled as a capacitor and a voltage source as shown in the schematic. Near resonance (one source said around 4.6 kHz), the impedance is complex and the amplitude response will peak. I couldn't find electrical data on piezo guitar pickups, but one source said the simple capacitance of typical guitar piezo pickup is around 500pF to 800pF.
simulate this circuit – Schematic created using CircuitLab
The amplifier input impedance will determine the low frequency response of the system and also equivalent input thermal noise. Higher Rin will make the high-pass cutoff frequency lower and will also lower the input noise. Assuming you want a low-cut frequency of 50 Hz, Rin would need to be 6.4 Megohms \$ (1/(2 \pi\;f\;C) = 1/(2\pi\;50Hz\;500pF)) \$.
Using a transformer at this high impedance is not practical. The magnetizing inductance calls for a primary inductance over 20k Henrys (magnetizing inductive reactance needs to be a few times higher than the transducer impedance). Plus, making a step down transformer with a 100:1 impedance ratio (10:1 voltage ratio) will attenuate the signal which will degrade the signal to noise ratio (not good).
Conclusion: Stick with the preamp approach and use a preamp with an input impedance above 6.4 M\$\Omega\$. If you use a lower input resistance amplifier, you'll need to pad the output of the transducer with a capacitor to flatten the low frequency response which will cause a loss in amplitude response.
I am on the road now, but a simple voltage divider is a capacitance transformer with and higher impedance load than source and a lower impedance output .
example 10k ohms at 100 Hz is 200nF and if piezo is >> 2nF , this could work for 100:1 V ratio. This may cause some low f attenuation. Although a FET buffer is better.
The same method is used for single f attenuation for HVAC to LVAC using C transformers on the grid.
