# T-network feedback of opamp [closed]

How are the output (Vo) and H(s) = Vo/Vin calculated in this schematic?

All you need to do it to replace the T-network with a single resistor.

How to do it can be found here:

Op-amp with T network in the feedback path

$$R_F = R_5 + R_6 + \frac{R_5 R_6}{R_7}$$ After doing all the simplification (C2 and C3 are connected in series, the R3 and R4 as well).

We end up with this equivalent circuit:

simulate this circuit – Schematic created using CircuitLab

And full analysis of this circuit can be found for example here:

Gain of filter is higher than calculated - why?

How does this filter work?

Deriving Bandpass Transfer Function

For example, the max gain will be equal to

$$A_{\text{max}}=\frac{R_FC_3}{R_FC_4+R_3C_3} \approx 55.54 \:[V/V] = 34.9dB$$

# The meaningless "T"

A circuit can be represented and named in different ways, but it is good that it has some meaning.

## T-network

"T-network" is an example of a meaningless name that does nothing to help its understanding. It is a way of drawing (grouping of elements) that resembles the letter "T" and that is it. It says nothing about the circuit's function.

simulate this circuit – Schematic created using CircuitLab

If we look carefully at the circuit diagram, we will notice that the two resistors on the right (R1 and R2) form a voltage divider, and the left resistor (R) is grounded, and can be considered as a load of the voltage divider. So the "T- network" is actually a loaded voltage divider. Usually its output resistance is much lower than that of the load so its gain is approximately R2/(R1 + R2). In the schematic below, I have adjusted (slightly increased) R2 so that the gain is exactly 1/10.

simulate this circuit

# Building the circuit

Many times I have had to explain this clever circuit trick, but now I will do it in a more different way - based on it, I will formulate a more general principle of virtual resistance modification. I will do it as usual in several successive steps.

## Real resistance

Let's take a 100 kΩ resistor but pretend we don't know its resistance and decide to measure it experimentally. For this purpose, we supply it with a 1 V voltage source Vs and measure the current I through it. As you can see, the resistance is the same - R = 1V/10μA = 100kΩ.

simulate this circuit

## Virtually increased resistance

Now let's apply the clever trick by reducing the voltage applied to the resistor R by a factor of 10. For this purpose, we insert an R1-R2 voltage divider between the supply voltage Vs and the resistor R. As above, we pretend not to notice the divider, and measure the resistance in the familiar way - R = 1V/1μA = 1 MΩ. However, what is our surprise when we see that it is 10 times higher!

The trick here is that we "do not see the divider" and use the previous 1 V voltage again... but the current is now 10 times smaller; so the resistance looks 10 times higher. Really clever trick!

simulate this circuit

## Virtually decreased resistance

And can't we (just out of curiosity) do the opposite - virtually reduce the resistance? Obviously, instead of a 1/10 attenuator we need to insert a 10 x amplifier (I have set the op-amp gain to 10 in the parameters window). As you can see from the gauge readings, both voltage and current increase 10 times, and the resistance "decreases" 10 times.

simulate this circuit

## Op-amp 4-resistor circuit

Finally, let's make an op-amp circuit of an inverting amplifier where the resistor R2 is virtually "increased" 10 times by the help of the R3-R4 voltage divider. Here the op-amp is "fooled". It does not "see" the R1-R2 voltage divider and "thinks" that the resistance R2 = 1 MΩ. That is why it increases its output voltage 10 times.

simulate this circuit

## Equivalent 2-resistor circuit

So the 4-resistor op-amp circuit above behaves as the classic 2-resistor inverting amplifier with R2 = 1 MΩ.

simulate this circuit