Electronic – Inverting OpAmp with voltage divider in feedback loop


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:

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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…

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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.

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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

Best Answer

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.