Charge amplifier

A charge amplifier is an electronic current integrator that produces a voltage output proportional to the integrated value of the input current, or the total charge injected (wikipedia).

This is a particular useful circuit for a piezo microphone. The reason is that the typical circuit (using a FET or FET input OpAmp) needs a very high impedance to deal with the small amount of current created by the piezo microphone. A low impedance will result in loss of bass. High impedance circuit are very susceptible for noise, so shielding and a very short path between the piezo disc and the amplifier circuit is necessary.

The charge amplifier doesn’t have this problem. The gain of this amplifier depends on the ratio between a reference capacitor and the capacitance of the piezo disc. A the charge amplifier converts (a change in) charge into a voltage, there is actually not really something like “unity gain” (as this would imply voltage in = voltage out). Instead, we might want to go for a “typical” voltage RMS for pro-level audio (+4 dBu or 1.228V RMS). If this cannot be achieved directly from the charge amplifier, we might need extra gain via a (voltage) amplifier.

With regard to the simulation of this circuit: the piezo sensor is not modelled as a charge source, but a current source. (as explained in this article). This means that the frequency behaviour of this circuit is not what we expect. To get to the correct behaviour, we have to divide Vout by 1/frequency. This will give us a frequency plot with the expected high-pass filter at the cutt-off frequency.

The “gain” of the charge amplifier is fully based on \(C_f\), the feedback capacitor. A lower \(C_f\) will result in a higher gain and visa-versa. The noise amplification level will depend on the quotient \(C_f/C_p\) (\(C_p\) = the capacitance of the piezo). So you want \(C_f « C_p\). A typical value for \(C_f\) is 470pF to 10nF, depending on what the value of \(C_p\) is of the piezo. To obtain the optimal gain, some experimentation need to be done.

The circuit also contains \(R_f\), the feedback resistor. A feedback resistor is needed, because this will filter out any DC amplification. However, the combination of \(C_f\) and \(R_f\) creates a high-pass filter. To maintain low frequency sounds, Rf should be pretty large, 1M or even larger, depending on the value of \(C_f\). A smaller \(C_f\) means we need a larger \(R_f\). So “gain” and maintaining low frequencies are a trade off! The cut-off frequency for the high-pass filter is according to the following formula:

\[f_{cutoff} = \dfrac{1}{2 \pi R_f C_f}\]

As the circuit might also contain \(R_{in}\), the input resistor, we also have a low pass filter (\(R_{in}\) in combination of the capacitance of the piezo). The cut-off frequency for the low-pass filter is according to the following formula:

\[f_{cutoff} = \dfrac{1}{2 \pi R_{in} (C_p + C_c)}\]

In this formula \(C_c\) is the capacitance of the interface cable. If we have a very short cable, we can set this capacitance to zero.