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How do I calculate the equivalent T network resistance of the inverting amplifier below? The answer is 10 MΩ, but I don't know how to calculate it. I'm not looking for just a formula, but an in-depth circuit analysis.

enter image description here

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  • \$\begingroup\$ Just do current mesh analysis with a virtual ground. \$\endgroup\$
    – Miss Mulan
    Commented Nov 10, 2022 at 14:39

2 Answers 2

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Your writing isn't entirely clear. But since you gave the answer you seek, I think I understand what you meant. You want to know how the output affects current entering the inverting input by way of the T-network. Your goal is to find the equivalent \$R_F\$ (if I do understand you) that would have a similar result as in:

schematic

simulate this circuit – Schematic created using CircuitLab

(Above, I've treated two resistors as being the same value, \$R_T\$, to help simplify things a little bit.)

Given that the inverting input can be taken as close to \$0\:\textrm{V}\$, for the purposes here, the question is essentially just asking, "How does voltage at the amplifier output affect current entering the inverting node?"

To figure that out, first figure out \$V_X\$:

$$\begin{align*} V_X &= \frac{V_O\cdot R_G\cdot R_T+0\:\textrm{V}\cdot R_T\cdot R_T+0\:\textrm{V}\cdot R_G\cdot R_T}{R_T\cdot R_T+R_T\cdot R_G+R_T\cdot R_G}\\\\ &= V_O\frac{R_G}{R_T+ 2\cdot R_G} \end{align*}$$

From that, you can easily compute:

$$\begin{align*} I_X &=\frac{V_X}{R_T}=\frac{V_O}{R_T}\frac{R_G}{R_T+ 2\cdot R_G}\\\\ &=\frac{V_O}{\frac{R_T\cdot\left(R_T+ 2\cdot R_G\right)}{R_G}}\\\\ \therefore R_F &=\frac{R_T\cdot\left(R_T+ 2\cdot R_G\right)}{R_G}\\\\ &=R_T\cdot\left(2+\frac{R_T}{R_G}\right) \end{align*}$$

The result is \$R_F\approx 10.004\:\textrm{M}\Omega\$


Just a note about T-networks, from my own personal experience with electrometers. (I was experimenting with circuits achieving below \$1\:\frac{\textrm{fA}}{\sqrt{\textrm{Hz}}}\$ input-referred noise levels and quite literally having to buy unpackaged dice and use wire-bonders and stable temps at \$-5\:^\circ\textrm{C}\$ [low, but not so low that my window would ice up] in tiny, sealed quartz-windowed modules in order to get there.)

These T-networks actually generate higher current noise (\$i_N\$) and the voltage divider at the output multiplies input offset voltage, drift, and amplifier voltage noise by the ratio of \$1+\frac{R_T}{R_G}\$. These input specifications are often pretty crappy anyway, so it quite quickly becomes insanely impractical to consider multiplying their already annoyingly high offset and drift (low input current FET opamps are almost always used here) by using a T-network instead of a big feedback resistor.

(Sure, the Johnson noise of the feedback resistor has to also be considered. But it's not nearly as much of a problem once you refer that back to the input as input current noise.)


NOTE to OP:

You wrote:

"Shouldn't the first \$R_T\$ and \$R_G\$ be in parallel, with the combination in series with second \$R_T\$? Resulting expression is \$\left(R_T\:\mid\mid\:R_G\right)+R_T\$. What is wrong with this expression?"

There is nothing wrong with that expression except that it's not shown within its proper context. Let's see where that goes.

Starting at the output and working backward:

$$\begin{align*} V_{TH} &= V_O\cdot\frac{R_G}{R_G+R_T}\\\\ R_{TH} &= \frac{R_T\cdot R_G}{R_G+R_T} \end{align*}$$

We now have to add the value of \$R_T\$ leading back to the inverting input node to \$R_{TH}\$ to get the total Thevenin resistance seen by the Thevenin voltage computed above. So, the current into the inverting node due to the output voltage of the opamp is:

$$\begin{align*} I_{V_O} &= \frac{V_{TH}}{R_{TH}+R_T} \end{align*}$$

But we are actually interested in the effective \$R_F=\frac{V_O}{I_{V_O}}\$ (the equivalent feedback resistance being determined by dividing the output voltage by the current it causes to enter into the inverting node, as if we had such a resistor there.) So:

$$\begin{align*} R_F &= \frac{V_O}{I_{V_O}}\\\\ &= \frac{V_O}{\frac{V_{TH}}{R_{TH}+R_T}}=\frac{V_O}{\frac{V_O\cdot\frac{R_G}{R_G+R_T}}{R_{TH}+R_T}}\\\\ &=\left(R_{TH}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right) \end{align*}$$

And now, here, you can see your factor present in the above equation. But notice that it is not alone! There is a second factor there.

We could just stop there. But let's follow through:

$$\begin{align*} R_F &= \left(R_{TH}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right)\\\\ &= \left(\frac{R_T\cdot R_G}{R_G+R_T}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right)\\\\ &=\frac{R_T\cdot\left(R_T+2\cdot R_G\right)}{R_G}\\\\ &=R_T\cdot\left(2+\frac{R_T}{R_G}\right) \end{align*}$$

Which is just the same thing as before.

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  • \$\begingroup\$ Shouldn't the first Rt and Rg be in parallel, with the combination in series with second Rt? Resulting expression is \$R_t||R_g+R_t\$. What is wrong with this expression? \$\endgroup\$
    – sarthak
    Commented Oct 6, 2017 at 19:22
  • \$\begingroup\$ @sarthak The only thing wrong is that you have taken a piece without putting it into its fuller context. If you do that, you get exactly what I wrote. Start with \$V_{TH}=V_O\frac{R_G}{R_G+R_T}\$ and \$R_{TH}=\frac{R_T\cdot R_G}{R_G+R_T}\$. Now, you need to compute the current into the inverting node, \$I=\frac{V_{TH}}{R_{TH}+R_T}=\frac{V_O}{R_F}\$. Therefore, \$R_F=\frac{V_O}{I}\$. What do you find for \$R_F\$? \$\endgroup\$
    – jonk
    Commented Oct 6, 2017 at 20:22
  • \$\begingroup\$ Between what nodes are you computing the Vth? \$\endgroup\$
    – sarthak
    Commented Oct 6, 2017 at 22:51
  • \$\begingroup\$ @sarthak \$V_{TH}\$ is the Thevenin voltage made up of \$V_O\$ and ground and the resistors, \$R_T\$ and \$R_G\$ on the right side of the schematic (not to include \$R_T\$ that ties to the inverting node.) \$\endgroup\$
    – jonk
    Commented Oct 6, 2017 at 23:07
  • \$\begingroup\$ @jonk ok now i understand how the feedback resistor is calculated , but i have another little question why R comp is 51k ? , it should be Ri || Rf thats equal to 100k not 51k \$\endgroup\$
    – user156489
    Commented Oct 7, 2017 at 6:26
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The simple answer is visible to the naked eye - another voltage divider inserted into negative feedback loop.

What is the idea?

T-network

If we look carefully at the circuit diagram of the so-called "T-network"...

schematic

simulate this circuit – Schematic created using CircuitLab

Loaded voltage divider

... 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 and it can be considered as unloaded; 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.

schematic

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 believing that building the circuit is the best way to explain it.

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.

schematic

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!

schematic

simulate this circuit

Op-amp 4-resistor circuit

After revealing the idea, 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.

schematic

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.

schematic

simulate this circuit

Another point of view

Negative feedback followers

The most basic negative feedback circuits, like negative feedback systems in life, are followers (basically they contain amplifiers but they are made to work as followers). These are the voltage follower and the voltage inverter.

Negative feedback amplifiers

To get them to amplify, we apply a well-known trick of life - disturb them. In an effort to compensate for the disturbance, they become amplifiers. So negative feedback amplifiers are actually "disturbed followers". These are the non-inverting amplifier and the inverting amplifier.

Implementation

In voltage amplifiers, disturbance means voltage attenuation and the attenuator we do it with is a voltage divider. So, to make an op-amp amplifier, we insert a "disturbing" voltage divider between the op-amp output and the input. Indeed, in the inverting amplifier it is a bit more sophisticated (there is also a second input for the input voltage), but it is still a divider.

Disturbed negative feedback amplifiers

When, for some reason, only one disturbance is not enough for us, we can disturb the amplifier with another disturbance (add another voltage divider). Such a "double-disturbed non-inverting amplifier" will contain two cascaded voltage dividers in its feedback network. Accordingly, a "double-disturbed inverting amplifier" will contain cascaded R1-R2 summing network and R2-R3 voltage divider in its feedback network... and this will be an "inverting amplifier with T-network" - a subject of this question.

It is as if in this circuit solution, the inverting and non-inverting circuits are merged into one.

Conclusion

From this point of view, an inverting amplifier with T-network is a "disturbed inverting amplifier".

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