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.


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

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

  • \$\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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.