Inverting OpAmp with voltage divider in feedback loop

Question

Last week I was looking for a programmable gain amplifier (PGA) with high gains (> 100V/V) and low noise. As far as I know most of the PGA integrated circuits out there are just (inverting) amplifier stages with the option to digitally switch feedback resistors. Please take a look at the block diagram of the LTC6912 for example:

In my opinion one of the problems you'll face when trying to minimize noise in a high gain standard (inverting) amplifier configuration, is the size of the feedback resitor. The high gain forces the resistor to get big, which results into a lot of noise.

So I started looking for other configurations where the feedback resistor doesn't have to be this big.

I found a very interesting article: https://www.electronicdesign.com/analog/digitally-programmable-amplifier-meets-sensor-gain-ranging-needs

In this article a PGA is built upon a sort of inverting amplifier configuration with a voltage divider in the feedback loop. Because I can't find anymore information about his type of circuit I started analyzing it myself and found the gain formula as follows:

The simulation result below shows the circuit with a large gain of 65.7dB with feedback resistors of only 39k. This seems very useful to me, however I've never seen it before...

When I started reading further, I found that the author used the same opamp configuration as a band pass filter, shown in the picture below.

Because I'm very excited about this topology I'm trying to find formulas for the cut-off frequencies of this circuit. However I really don't know how to do it. Is there anyone who can help me out? More information about these circuits is welcome as well! Thanks

Answer 1

To find the lower cutoff frequency and passband gain, you can use Miller's theorem to separate the \$100nF\$ into two capacitances to ground.

^{simulate this circuit – Schematic created using CircuitLab}

The schematic contains a simulation that shows you the similarity.

I determined \$A_{DC} \approx -16\ 443\$, resulting in an input capacitance of \$C_{in} = C\cdot (1-A_{DC}) \approx 1.6mF\$ and an output capacitance of \$C_{out} = C\cdot \left(1 - \frac{1}{A_{DC}}\right) \approx 100nF\$.

Doing so will allow you to calculate the lower cutoff frequency and passband gain.

$$A_{in\to B,miller} = \frac{R_8C_6s}{1+R_8(C_6+C_9)s}$$

There is a zero at f=0Hz, and a first cutoff frequency at

$$f_{p1} = \frac{1}{2\pi R_8(C_6+C_9)} \approx 0.1Hz$$

The passband gain is severely reduced by the capacitive divider circuit.

$$A_{pb} = A_{DC}\cdot\frac{C_6}{C_6+C_9} \approx 5.75 = 15.2dB$$

At high frequencies, the opamp can't produce high gains and the approximation breaks down. The higher cutoff frequency will be determined by the GBW of the opamp. If the opamp is considered ideal, then this is a high-pass filter.

Answer 2

No, the shown circuit is not a bandpass. It is a first-order highpass (inverting). Here is the transfer function (R2 and C2 are the feedback elements)

$$H(s) = -\frac{R_1C_2s}{1 + s(R_1C_1 + R_2C_2 + R_1C_2)}$$

In the given formula, the three resistors in the feedback path are replaced by R2. This resistor value can be simply found applying the star-to-triangle transformation.

As the result of this transformation you get three new resistors, however two of them play no role: One resistor appears as a load to the ground at the opamp output and the second one appears between the inverting opamp input and ground. Assuming an ideal opamp, both resistors have no influence on the gain.

Answer 3

For this type of circuit, you can apply the fast analytical circuits techniques or FACTs. These techniques will lead you straight to the answer expressed in a low-entropy format. Reduce the excitation to 0 V and determine the time constants involving capacitors \$C_1\$ and \$C_2\$. If you do that correctly, you find out that the \$b_2\$ term is 0: it leaves you with a 1st-order pole. For the zero, determine the high-frequency gain when \$C_1\$ is a short circuit: apply the extra-element theorem (EET) and determine the transfer function in this mode: you have \$H^1\$ and the denominator is equal to \$s\tau_1H^1\$.

What is needed now is to rearrange the expression so that a pole and a flat gain appear. This is what is done in the below Mathcad sheet where you clearly see the pole and the flat part properly defined:

As correctly pointed out by LvW, this is an inverting high-pass filter. For a tutorial on FACTs, have a look at the APEC seminar from 2016.