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I will be quoting this article.

In a closed loop op-amp circuit the output pin of the op-amp is connected with the either of the input pin to provide a feedback. This feedback is called as the closed loop connection. During closed loop an op-amp works as an amplifier, it is during this mode an op-amp finds many useful applications like buffer, voltage follower, Inverting Amplifier, Non-Inverting amplifier, Summing amplifier, Differential amplifier, Voltage subtractor.

And

... but op-amp when used in open loop (without feedback) will have a very high uncontrolled gain which is practically not useful.

It says something about "very high and uncontrolled gain". I assumed they used a closed op-amp so they could weigh inputs and outputs.

Is it necessary to use a closed-loop op amp if I wanted a precise, but unweighted output?

Also, if it is necessary, do I need any resistors, for an unweighted-output?

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    \$\begingroup\$ The article explains closed-loop and open-loop very well. What does unweighted mean in this context? \$\endgroup\$
    – Mattman944
    Sep 10 at 17:09
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    \$\begingroup\$ The feedback is what makes the op amp even linear; you don't use an op amp open-loop unless you need a comparator (and don't need a high performance one, at that). \$\endgroup\$
    – Hearth
    Sep 10 at 17:25
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    \$\begingroup\$ @Walter If you close the loop with no resistors (or other elements) involved, you just get a voltage follower. A useful circuit, but not a subtractor. \$\endgroup\$
    – Hearth
    Sep 10 at 17:35
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    \$\begingroup\$ That quote looks poor enough not to have been transposed verbatim. \$\endgroup\$
    – Andy aka
    Sep 11 at 10:48
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    \$\begingroup\$ Then it's not good grammar for an article \$\endgroup\$
    – Andy aka
    Sep 11 at 15:32

3 Answers 3

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Time for a history lesson!

It turns out that it is difficult to make vacuum tubes and transistors of a precise and stable gain, but quite easy to make them of a "very high gain".

Op-amps created from transistors or vacuum tubes thus share this property. It is difficult to create an op-amp of a precise and stable gain but very easy to make op-amps of a "very high gain".

Story has it that Harold Black was sitting on the train home from work from Bell Labs and it suddenly came to him that negative feedback could be used to obtain mass producible, accurate and stable gains in spite of this problem.

Negative feedback lets you take advantage of a "very high" but not very precise gain and puts all the gain accuracy and stability onto the external components used. It is much easier to make precise and stable resistors. Not to mention it is much cheaper to make a wide range of resistor values than it is to make an semiconductor amplifier and every possible gain that might be needed.

So in theory you could use make an amplifier out of transistors for a particular gain without resistors. But it will be difficult to make and that gain will not be very stable over over a wide range of input voltages, temperatures, or supply voltages.

You cannot do this with an op-amp since op-amp are specifically designed to offload all the gain accuracy and stability from the transistors inside the opamp to the resistors (and other passive components) on the outside.


So to make a subtract voltages, use opamps and resistors. This is what they were made for.

You may need to do jump through some hoops doing things that seem ackward and strange compared to the neat and tidy math on paper.

For example, summing involves running each input voltage through a resistor to a shared node. And the relative resistances used contribute a weight to each input voltage, so there are always weights. There is no such thing as unweighted but you can have equal weighting.

But these also form a voltage divider of sorts so the final result is also smaller than what you know it should be at which point you must use an opamp+resistors to re-amplify it to offset this (or larger or smaller, whatever you want really).

You're getting into things that are best off understood using KVL circuit analysis which if you don't know, can learn, or you can just find canned circuits with equations online such as this: https://www.electronics-tutorials.ws/opamp/opamp_4.html

If you want a negative input you going to need negative voltages. If you have a positive input you want to convert to a negative input, you can do this by multiplying by -1 which is what an inverting amp with a gain of 1 will do.

However, note that opamps can only accept voltage inputs and produce output voltages within their voltage supply range. Often they can't even do this as most op-amps will not have their input and output range cover the entire supply range. It often falls short by about 3V on either side. Thee so-called "common-mode input range" and "output range" on the datasheet. That means that if you want to use something like -2V as an input or have -2V as an output then you will need an to provide a supply voltage that includes -2V (plus aforementioned overhead) in its range.

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    \$\begingroup\$ I am suggesting you use an opamp with resistors. This is what they were made for. You may need to do some things that seem ackward and strange compared to the neat and tidy math on paper though. For example, summing involves running each input voltage through a resistor to a shared node. And the relative resistances used contribute a weight, so there are always weights. There is no such thing as unweighted but you can have a weight of one. \$\endgroup\$
    – DKNguyen
    Sep 10 at 17:45
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    \$\begingroup\$ But these also form a voltage divider of sorts so the final result is also smaller than what you expect on paper at which point you must use an opamp+resistors to re-amplify it to offset this (or larger or smaller, whatever you want really). \$\endgroup\$
    – DKNguyen
    Sep 10 at 17:45
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    \$\begingroup\$ @Walter You're getting into things that are best off understood using KVL circuit analysis which if you don't know, can learn, or you can just find canned circuits with equations online. electronics-tutorials.ws/opamp/opamp_4.html And if you want a negative input you going to need negative voltages. You can turn a positive number into a negative number by multiplying by -1 which is what an inverting amp with a gain of 1 will do. But opamps can only accept and output voltages within their voltage supply range. \$\endgroup\$
    – DKNguyen
    Sep 10 at 17:45
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    \$\begingroup\$ Just for the record, Harold Black came up with the concept of the negative feedback amplifier in a flash when he was crossing the Hudson river on the Lackawanna Ferry on his way to work. The ferry was crossing from the Hoboken terminal to Manhattan where Harold Black worked at Bell Labs which was then situated on West Street. The date was Tuesday August 2nd 1927. Whilst on the ferry he sketched a simple diagram of a negative feedback amplifier plus equations on a copy of The New York Times and signed it. When he reached the lab it was witnessed and signed by Earl C. Bleassing. \$\endgroup\$
    – James
    Sep 10 at 18:45
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    \$\begingroup\$ @DKNguyen The first commercial op amp was the K2-W, which consisted of two 12AX7 dual triodes, a neon glow tube, and a whole bunch of discrete resistors and capacitors in a bakelite shell. It was released in the late 1940s. I'm sure the design had been previously used in less integrated form, too. \$\endgroup\$
    – Hearth
    Sep 11 at 6:27
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Yes, you need closed loop on an op-amp for any kind of precise output.

The only configuration that requires no resistors and has a precision output is the voltage follower. In that case output voltage = input voltage (ideally).

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There is a packaged device that can more or less do this, it is called an "instrumentation amplifier" (internally, these are a network of buffers and subtractors). These still usually need some configuration resistors to set the gain.

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