Piezo preamp impedance, my theory

The recent discussion of Acoustic Image Series III reminded me that I researched the impedance issue when I designed my DIY preamp. Here is what I learned...

It appears to me that a piezo pickup can be modeled as a voltage source in series with a capacitor. Equivalently, you could use a current source in parallel with a capacitor. The capacitance of the pickup, coupled with the input resistance of the amplifier, forms an RC high pass filter. For this filter to be benign to the overall frequency response of the system, the cutoff frequency has to be below the audio band. The cutoff frequency is equal to 1/(2 pi R C) [equation 1], so you can see that increasing R, or increasing C, will lower the cutoff frequency. Thus, if you know nothing about C, then you have your best chance with the highest possible value of R.

Another way of writing that equation is that if the cutoff frequency is fc, then the minimum value of R should be 1/(2 pi fc C) [equation 2]. With a capacitance meter, I measured 0.025 uF on my K&K Bass Max pickup. Assuming I want fc to be somewhere around 20 Hz, the minimum value of R would be roughly 0.32 MegOhms. Anything above 0.32 MegOhm should not significantly alter the sound of the pickup. Thus, the new Clarus head should be OK. Another calculation is that 1 MegOhm would be the bare minimum if the capacitance of the pickup is 0.008 uF.

Of course I am assuming that you don't want a high pass filter to affect your sound. It is entirely possible that some pickups sound better when they are "matched" to a particular load resistance. By trying different preamps, you may be inadvertently using that resistance to EQ your pickup. I am certainly making no value judgement against doing it this way.

Now we get into a bit more murky waters... during my research, I discovered that there is a way to "pad" the output of a pickup, in case its signal is actually distorting the preamp. A typical resistive attenuator is the wrong way, as it requires high resistance values that add noise. Instead, take advantage of the pickup capacitance. Pad the pickup by putting a capacitor in parallel with the pickup output. For instance, putting 0.025 uF in parallel with my Bass Max would pad it by 3 dB.

But when I modeled this circuit, I noticed that the padding capacitor does another thing... it renders the pickup less sensitive to the impedance of the preamp. If you think about it, putting 0.008 uF in parallel with a 1 MegOhm input would guarantee that no piezo pickup would be affected by the preamp impedance. Of course this would be at the expense of gain, but most piezo pickups have enough signal that you can give up some gain. You could hide the capacitor (preferably a polymer film type) in the output jack of the pickup. Let's make it 0.01 uF, which is a commonly available value. Indeed, an evil thought is to use much larger values of capacitance to match the lower input impedances of cheap electric bass amps.

Here is a thought about signal cable length... the fact that the pickup is a capacitor means that some additional capacitance introduced by the cable probably has negligible effect on the frequency response of the system.

A final note... there are some subtleties about raising the input resistance of the preamp. Under ideal conditions, increasing this resistance reduces the input noise of the preamp. However, this effect is buried under the noise floor of the input transistor or op amp on probably any commercial preamp. A much larger source of noise in an amplified bass is probably acoustical noise in the room coupled through the bass and into your amplifier.

And one more thing... it is my experience that really high input resistance amplifiers tend to be rather squirrely. Keeping the resistance to a moderate level may make good sense for the stability of the amplifier.

 

Oops, you're right. My brain is on ice.