2
\$\begingroup\$

I have encountered this topology. As you can see, there is a constant 3.5V voltage source connected to the non-inverting op-amp input. It's clear that this is an inverting op-amp ampifier but I couldn't figure out what exactly the R7 resistor is. Does this topology have a specific purpose (especially R7 resistor)?

enter image description here

\$\endgroup\$
7
  • 3
    \$\begingroup\$ "Op-amp topology" IMO usually refers to the circuit inside the op-amp, i.e. how many stages, output type, input type etc. \$\endgroup\$
    – tobalt
    Dec 16, 2022 at 4:53
  • 12
    \$\begingroup\$ This is just an inverting amplifier. The reference is 3.5 V instead of ground, but that doesn't matter. I'm not sure what the point of R7 is. \$\endgroup\$
    – Hearth
    Dec 16, 2022 at 4:56
  • 1
    \$\begingroup\$ Falstad link tinyurl.com/2lxsvxkz \$\endgroup\$
    – Electronx
    Dec 16, 2022 at 5:27
  • \$\begingroup\$ Probably R7 is to prevent a short circuit between 3.5V and the input voltage source.. \$\endgroup\$
    – ee_student
    Dec 16, 2022 at 8:01
  • \$\begingroup\$ The circuit does nothing without you providing more information like where that output is connected. \$\endgroup\$
    – Andy aka
    Dec 16, 2022 at 8:36

3 Answers 3

8
\$\begingroup\$

The challenge

It is always a big challenge to understand a circuit that is new to you. Let's then try to do it with the OP's "topology"...

Analyzing the circuit

I suggest that we analyze their circuit from left to right (as input signals flow). First we see two input sources - one is constant (3.5 V) and the other is adjustable.

Non-inverting amplifier

We can drive the non-inverting input of the op-amp through the first voltage source; then the circuit name would be "non-inverting amplifier".

Inverting amplifier

With the same success, we can drive the inverting input of the op-amp through the second voltage source; then the circuit name would be "inverting amplifier".

Differential amplifier

We can even drive both op-amp inputs through both sources; then the circuit name would be "differential amplifier". Indeed, it is quite imperfect since the two input gains are not equalized.

Reference voltage source

Once set, the "adjustable voltage source" becomes a "constant voltage source" like the other; so both can be considered as input sources. Thus both input voltages are constant and the op-amp output voltage is also constant; so the name of this circuit can be "buffered reference voltage source".

The role of bias voltage

The non-inverting configuration is more suitable for such an application because it is "self-biased". Let's consider the need of another (bias) voltage source here, in this inverting cobfiguration...

The op-amp is supplied only by one (7 V) voltage source. So the output voltage can be positive between 0 V and 7 V. To obtain it, the second (adjustable) voltage applied (through R3) to the inverting input has to be lower than the first (3.5 V) voltage applied directly to the non-inverting input.

So this is the role of the first (3.5 V) voltage source - to "lift" ("bias") the voltage applied to the non-inverting input above the adjustable voltage. If the non-inverting input was directly grounded, the adjustable voltage has to be negative (undesired).

Let's see what the "biasing" means here. We have actually added another voltage source in series and opposite to the input source so that their voltages subtract. The voltage sources are connected in series (regarding the op-amp differential input) and both they are grounded. In other cases (e.g., AC common emitter ampifier stage), a floating bias voltage (across a charged capacitor) is added to the voltage of the grounded input source. So the biasing is adding a positive or negative constant voltage to the input voltage.

The role of resistors

... R7

In my opinion, the resistor R7 is connected between the two input voltage sources for the case when the 3.5 V voltage source is disconnected. Then the op-amp will be saturated (maybe undesired). With this "protecting resistor", all voltages (including the output one) will be equal to the adjustable voltage.

There is no such a problem with disconnecting the adjustable voltage source since the inverting amplifier will become a follower.

... R8 and R10

This network acts as a load consisting of two resistors in series... but what the hell this is (still) I can't figure out. It practically does nothing because one of the resistors is too high (100 k). Perhaps the idea is to be able to short R10 in order to study the circuit behavior under heavy (330 ohm) load...

... R3 and R6

Since a good situation has been created here that stimulates and not suppresses original thoughts (rarely found in SE EE), I will allow myself to "philosophize" a little on this simple network of the two resistors R3 and R6 in series. I think this will be helpful for the OP (I know it from personal experience since the late 70s when I was frantically trying to figure out what the hell these resistors were).

R3-R6 network is a simple voltage summer that adds (subtracts) the circuit input and output voltages. It can be considered as a 2-input voltage divider. The op-amp adjusts its output voltage to make the difference (the summer output voltage) zero. As a result, the proportion between voltages (circuit gain) is the same as between the resistors... and it depends only on the latter.

It seems a little difficult to understand but it is actually all around us and we all operate according to this principle known by the trivial name "negative feedback".

Generalization

We saw a few powerful ideas here that we can add to our "collection of circuit principles":

Biasing: To match the input voltage to the input range of a device (amplifier, ADC, etc.), add an appropriate (in sign and value) constant voltage to the input voltage.

Voltage self-switching: To change the voltage of a real source driving a high-resistance load, connect an "ideal" source in parallel.

NFB inverting: Subtract the output from the input voltage in a parallel manner (through resistors) to obtain an inverted copy of the input voltage.

NFB inverting amplification: Subtract a part of the output from the input voltage to obtain an inverted and amplified copy of the input voltage.

NFB following: Use the result of parallel subtraction as an exact copy of the input voltage.

See also

How to Understand Circuits is a story from my Circuit Idea wikibook that reveals the philosophy of understanding. Two other stories reveal the philosophy of explaining and inventing circuits.

What is the idea behind the op-amp instrumentation amplifier? is my Codidact story about the philosophy behind the related differential and instrumentation amplifier.

\$\endgroup\$
8
  • 4
    \$\begingroup\$ (Downvoters please comment what makes this answer not useful.) \$\endgroup\$
    – greybeard
    Dec 16, 2022 at 16:37
  • 3
    \$\begingroup\$ I didn't downvote, but this answer doesn't get my upvote because it seems too general. The "adjustable voltage" on the schematic is clearly the input, and the 3.5 V just a reference, so adding analysis of it as a non-inverting amplifier, differential amplifier, and reference voltage source seems likely to just confuse the point. (okay, I can maybe see talking about the differential amplifier to explain how the addition of the 3.5 V reference affects things, but the other two seem nigh pointless.) \$\endgroup\$
    – Hearth
    Dec 16, 2022 at 16:49
  • \$\begingroup\$ @Hearth, I believe you will agree with me that besides specific explanations (abundant here), understanding and explaining circuit solutions needs general explanations (like mine) as well. About the second part of your comment: Once set, an "adjustable voltage source" becomes a "constant voltage source" like the other; so both can be considered as input sources. As a result, this circuit produces a constant output voltage; that is why I called it "reference voltage source". \$\endgroup\$ Dec 16, 2022 at 18:14
  • 1
    \$\begingroup\$ @Circuitfantasist I agree that general solutions are good, but I disagree with putting them on a specific question. A specific question should get a specific answer, and a general question should get a general answer--if you think a general answer should exist, but no suitable question is out there, make your own question and self-answer! \$\endgroup\$
    – Hearth
    Dec 16, 2022 at 21:53
  • 1
    \$\begingroup\$ @electronx, Yes, I do everything myself like this: First I have to come up with the idea for the illustration. Then I start sketching it casually with pencils on a piece of paper, rubbing a lot. Finally, I draw it (first with a pencil) more beautifully on squared paper... and finally repeat it with colored markers. Then I wait for daylight and take a picture of it with my phone. I'm very excited to put it in my next answer and wait for approval like yours... but I usually get downvoted with no explanation as to why. This puts me off for a while... and all over again… So for years... \$\endgroup\$ Dec 17, 2022 at 20:58
3
\$\begingroup\$

It’s an inverting amplifier, referenced to 3.5V. It’s a typical setup using a ‘virtual ground’ using a single-ended supply. It has a voltage gain of -3.32 (-R6/R3) for an input signal referenced to 3.5V.

R7 isn’t really doing anything as the (+) input is tied to the 3.5V voltage, and we’ll assume ‘adjustable voltage’ is low impedance. Perhaps the intention was to provide the input a DC path to the 3.5V virtual ground.

Otherwise, to understand it better consider that R3 and R6 form a voltage divider, and the op-amp will force the divider to be equal to 3.5V. What will the output voltage be if the input is 3.5V? 3.6V? 3.4V?

\$\endgroup\$
3
  • \$\begingroup\$ +1 for the final "voltage divider explanation". Just to remind you that it is not such a simple "1-input voltage divider". It has two inputs and it is controlled oppositely on its two sides - on the left by the input voltage, on the right by the output voltage. More precisely, it is a "voltage summer" with weighted inputs acting as a subtractor. \$\endgroup\$ Dec 16, 2022 at 19:27
  • 2
    \$\begingroup\$ However, if the non-inverting input is pegged at 3.5V, R7 does nothing and can be removed. Then it is just an inverting amplifier with offset. \$\endgroup\$
    – PStechPaul
    Dec 16, 2022 at 19:50
  • 1
    \$\begingroup\$ @PStechPaul, Exactly! The situation is: Two voltage sources - an "ideal" one (3.5 V) with zero internal resistance and real one ("adjustable") with R7 "internal" resistance are connected in parallel. What will dominate? Of course, the ideal voltage source. This is a powerful circuit idea that can be seen in many circuit solutions. It allows to switch between (commutate) two voltage sources by a simple 2-terminal switch (SPST). Diodes and transistors can act as such a switch. \$\endgroup\$ Dec 16, 2022 at 20:06
2
\$\begingroup\$

What is the name of this op-amp topology?

That is simply an "inverting amplifier".

What may make it confusing is that you drew it upside down. By convention, the - input is above the + input. If you redraw it in the conventional manner, and turn R7 so that it's vertical, then the function should become apparent.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Just to clarify that I was the one who added the tag "inverting-amplifier":-) \$\endgroup\$ Dec 16, 2022 at 18:42
  • 1
    \$\begingroup\$ @Circuit fantasist Thank you for doing so and thank for the clarification. \$\endgroup\$ Dec 16, 2022 at 18:55
  • \$\begingroup\$ I'm looking forward to finding out what the magic function of R7 is when drawn vertically :-) \$\endgroup\$ Dec 16, 2022 at 19:13
  • \$\begingroup\$ It sets the operating point if nothing is connected to the input. That becomes apparent when drawn conventionally. \$\endgroup\$ Dec 16, 2022 at 19:38
  • \$\begingroup\$ I.e., what I wrote in the section "The role of R7" of my answer? \$\endgroup\$ Dec 16, 2022 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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