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How would I design an op amp with a gain of less than 1? Is this even possible? Also would it matter if I used a non inverting or an inverting op amp?

I've designed a non-inverting op amp with a gain of 2 like so. Can I modify it to have a gain of say 1/2?

1 + R1/R2

schematic

simulate this circuit – Schematic created using CircuitLab

I'd appreciate some help here.

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    \$\begingroup\$ You don't specify any requirements besides the attenuation. So no op-amp and just a voltage divider will give you 0.5 gain (non-inverting). \$\endgroup\$ Commented Nov 20, 2013 at 3:50
  • \$\begingroup\$ I'm studying transfer functions at the moment so I think at the minimum a buffer is required to separate the stages of my function. \$\endgroup\$
    – codedude
    Commented Nov 20, 2013 at 3:58
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    \$\begingroup\$ Then you can just put a unity gain buffer at the output of the voltage divider. \$\endgroup\$ Commented Nov 20, 2013 at 3:59
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    \$\begingroup\$ nit-picking : you would never design an opamp with gain less than 1. If you did, it wouldn't be an opamp. You might use one in an amplifier with low gain : the simplest way would be a voltage divider followed by a buffer. \$\endgroup\$
    – user16324
    Commented Nov 20, 2013 at 11:24
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    \$\begingroup\$ Where did I put my negative resistors? \$\endgroup\$
    – Mike
    Commented Apr 6, 2017 at 15:19

7 Answers 7

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An op-amp is not required. Given the minimal requirements given, a voltage divider will do. Gain = 0.5.

schematic

simulate this circuit – Schematic created using CircuitLab

If you need to buffer the output, you can just place a unity gain op-amp at the output of the voltage divider.

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  • \$\begingroup\$ I'm a little confused on how this gives a gain of .5. The voltage divider produces an equivalent resistance of 5kΩ. Why is this a gain of .5? \$\endgroup\$
    – codedude
    Commented Nov 20, 2013 at 4:03
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    \$\begingroup\$ @codedude Well, the resistors will divide Vin in two, so Vout = Vin/2. For example, the input current will be Iin = Vin/(R1+R2), and Vout = IinR2. If R1=R2=R, then Vout = VinR/(R+R) = Vin/2. \$\endgroup\$ Commented Nov 20, 2013 at 4:07
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    \$\begingroup\$ @codedude BTW, you should really make sure you understand these simpler circuits before moving up to analyzing op-amps and their transfer function equivalents. \$\endgroup\$ Commented Nov 20, 2013 at 4:11
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    \$\begingroup\$ @apalopohapa Exactly! Why ask about op-amp circuits with a given gain, if a voltage divider makes you stumble. \$\endgroup\$
    – Kaz
    Commented Nov 20, 2013 at 4:14
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    \$\begingroup\$ :) Definitely need to brush up on some stuff. \$\endgroup\$
    – codedude
    Commented Nov 20, 2013 at 4:23
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The non-inverting amplifier circuit cannot produce a gain of less than 1. Inverting amplifiers on the other hand can be built for gains less than 1 because there is no "1+" in their gain equation.

The best approach to making a circuit with a noninverting gain of 0.5 depends heavilly on what your other conditions are. In particular what are the requirements for input and output impedance of the circuit?

Options may include.

  1. If you can live with the implications regarding input and output impedance just use a voltage divider.
  2. If you need a high input impedance then use a unity gain buffer followed by a voltage divider.
  3. If you need a low output impedance than use a voltage divider followed by a unity gain buffer.
  4. If you need both a low input impedance and a high output impedance (what a non-inverting amplifier normally offers) then use a unity gain buffer followed by a voltage divider followed by another unity gain buffer.

Someone else suggested (in a now delted answer) using two inverting amplifiers. I don't see much merit in that approach.

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    \$\begingroup\$ Two inverting amplifiers would likely have lower audio distortion and some other subtle advantages. The defined input impedance could be a feature in some cases. See Jim Williams' (RIP) directive : "a little known tenet of precision op amp circuits... always invert (except when you can't)" \$\endgroup\$ Commented Mar 31, 2016 at 13:45
  • \$\begingroup\$ What about using a diff amp configuration? It's basically using two matched voltage dividers to get whatever gain you want, right? ( en.wikipedia.org/wiki/… ) \$\endgroup\$
    – jacobq
    Commented Sep 12, 2016 at 11:28
  • \$\begingroup\$ I don't see any advantage of the diff amp configuration over the voltage divider followed by unity gain buffer configuration. \$\endgroup\$ Commented Apr 6, 2017 at 14:56
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To have a non-inverting operational amplifier configuration with a gain of less than one you can do the following:

  • Place a voltage divider at the input of the non-inverting amplifier. Calculate R1 and R2 to obtain the gain you want. In this example, as the two resistors have the same value, the gain of the voltage divider is 0.5
  • Short the feedback resistor (0Ω). You want the amplifier to have a gain of 1, so the overall gain is the same as the voltage divider alone.
  • The R3 can be any reasonable value, I've just used 270kΩ so it's the same as the others.

schematic

simulate this circuit – Schematic created using CircuitLab

The advantage of this set up over a simple voltage divider is that the amplifier provides enough current to feed the next step (represented by the U1 module, which can be whatever device). Therefore you can use much higher resistors R1 and R2, so there is very little current drained from the source.

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If you want the advantages of the non-inverting amplifier with gain \$\ge\$1 (high input impedance, low output impedance) you can use two op-amps as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

Here R1/R2 are a voltage divider that provides that actual gain of 0.5 and OA1 and OA2 are simply voltage followers.

One limitation of this configuration is that the voltage follower OA1 must be able to follow the input over the full range, so if the input approaches the supply rails too closely it may not work accurately.

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See if this reply helps. You drew the schematic for a non-inverting op-amp with a gain of 2; GAIN equals 1 + (ResistorTOP / ResistorBOTTOM). If ResistorTOP = ResistorBOTTOM then overall gain is 1 + 1.

If you use the schematic for the yes-inverting op-amp, then you can have a gain equals (ResistorTOP_FEEDBACK divided by ResistorBOTTOM_INPUT), and this can be 1/2 = half, but the signal is inverted or (times -1). You can use two op-amps in the yes-inverting config. First op-amp achieves -1/2, and the second op-amp achieves times -1 (for an overall gain of positive 1/2).

The last suggestion is: if the input signal is not very delicate, then use a RESISTOR voltage divider that cuts the INPUT signal in half, then apply it to a non-inverting OPAMP circuit of times 1. Voltage at the divider comes from : voltage IN x RESISTOR BOTTOM divided by (RESISTOR BOTTOM + RESISTOR TOP).

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Connect a second op amp to replace the voltage divider . This will increase the feed back and to stabilize output with equal inputs, the gain will be less than 1 . If feed back is with gain 2 the output would be exactly 0,5 .

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    \$\begingroup\$ A schematic is better than words. There's a button on the editor toolbar that's easy to use. Double-click a component to edit its properties. 'R' to rotate. "Save and insert" when done. You don't need a CircuitLab account if you use the button on the SE editor toolbar. \$\endgroup\$
    – Transistor
    Commented Oct 17, 2021 at 13:42
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To connect a op amp with gain bellow 1 is easy. Main principle is to understand the operation of op amp . A op amp is in princip a comparator. It will switch on or off when reference voltage is higher or lower than signal. A positive gain is achieved by leading a propotion of output voltage in to refernce input by a voltage divider . The op amp is in balance when both inputs are equal, which means that when 1 / 1000 propotin of output is lead to reference input the output signal will be 1000 times input signal.

I guess You are able to figure out the rest ? All what is needed is to gain the output signal which is lead to the input reference . If - input is for example 2 volts and the output is amplifyed with 2 to the input + , the balance will be found when output is 1 V . A gain of 0,5 .

Tested and it do function .

Best regards Torsti Forsman

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