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I'm having some difficulty understanding the frequency response of a charge mode amplifier, circuit shown in Figure 1 below.

                        Figure 1: Ideal charge mode amplifier circuit (1)

I believe I understand the principles behind an charge mode amplifier circuit, however I'm rather confused about the frequency response. Considering Figure 1 above, my understanding is that R1 and Cf will act as a low pass filter, where frequencies above a specific cutoff frequency (related to values of R1 and Cf) will be attenuated. However, having had a read through a Texas Instruments (TI) PDF (2) discussing this type of circuit, the frequency response doesn't appear to follow this rule (I was expecting a low pass filter response). The circuit and frequency response described by TI is shown in Figure 2 below.

            Figure 2: Practical charge mode amplifier and frequency response (2)

I'd really appreciate if someone could help me understand this.

References

(1): https://commons.wikimedia.org/w/index.php?curid=16867828

(2): https://www.ti.com/lit/an/sloa033a/sloa033a.pdf

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  • \$\begingroup\$ Those are two different circuits. The first has one pole in its transfer function, the other 2 plus a pole at \$\omega=0\$. First: \$i_1=-i_F => \frac{V_{in}}{R_i}=-V_o.j\omega C\$ so \$\frac{V_o}{V_{in}}=-\frac{1}{j \omega R_i C} \$: A pole at \$\omega=0\$. \$\endgroup\$ – HarryH Jun 9 '18 at 18:56
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Your first circuit is an integrator. It ideally puts the pole at zero, so there's no "corner frequency" as such...it's just 20dB/decade with (ideally) infinite gain at DC.

The second schematic adds Rf, which plays against Cf to give you a low-pass pole. I think they switched the high-pass and low-pass formulae on the frequency response chart.

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Fig.1 is a Miller integrater which integrates the input voltage.

In Fig.2 there should be no Ri.

That's the whole point in a charge amplifier, both inputs of the op amp, along with the output from the sensor, are held at Vcc/2 (virtual earth) and so the cable capacitance Cc and sensor capacitance Cp have no effect on the sensor's output signal.

When a charge amplifier is used, the output from the piezoelectric sensor is current which the charge amp integrates to produce an output voltage which is proportional to the charge generated by the sensor. It is the charge generated by the piezoelectric sensor which is proportional to the sensor's stimulating force. The integral of current is charge.

A voltage amplifier could be used instead to amplify the voltage output from the piezo electric sensor but only if the cable between the sensor and voltage amplifier is short. This is because cable capacitance will attenuate a voltage mode signal.

The charge amplifier's integrator has the response of a single pole, active, low pass filter with a pole at:

wc = 1/(Rf*Cf).

Rf must be kept very large in value so that the integrater integrates well down to low frequencies. The integrater only integrates properly when it is on its -20dB/decade (+90 degrees) part of its response.

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