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These days I'm looking at operational amplifiers; from what I've seen, implementing them in a circuit is quite simple, at least when they are connected as "non-inverting". Determining the gain/amplification is possible by doing a calculation of two resistors, R1 and R2 (should R2 be called a "feedback resistor"?)

Non-inverting operational amplifier

(The image is taken from http://mustcalculate.com/electronics/noninvertingopamp.php.)

Let me do a practical example to explain where my questions are:

In my example I choose to implement an op-amp (for example, the TLV272, which is also "rail to rail") as "non inverting amplifier". Then I want to increase a voltage of 10 volt to 15 volt (to be sure, I will feed the op-amp with a power supply of 15 volt). Well: by the equation I have to choose a value of 20 kΩ for R1 and a value of 10 kΩ for R2, which is equal to an amplification of 3.522 dB (voltage gain 1.5).

OK, but I could also do the same by choosing R1 as 200 kΩ and R2 as 100 kΩ, or increase these values until R1 of 200 MΩ and R2 of 100 MΩ (or at the totally opposite: R1 of 2 milliohm and R2 of 1 milliohm): in all these cases I will still have a gain of 1.5, but with totally different ranges of resistors, in terms of values.

I can't understand the criteria (in terms of range) how these resistors should be choosen. Maybe this criteria is related to the kind of signal which the op-amp will have to manipulate on his input? Or what else? And in practical example, which will be the difference if I increase a signal using "R1 = 2 kΩ R2 = 1 kΩ" and "R1 = 200 MΩ R2 = 100 MΩ"?

EDIT: I've seen that my question has been edited, also to correct my grammar: thank you. I'm sorry for my misspellings, but english is not my main language. Next time, I will do an attempt to be more accurate in my grammar.

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    \$\begingroup\$ I know someone's going to write a good long detailed answer to this, but short & sweet: your opamp has to source/sink the current through those resistors, so low values = high current. But, resistors cause noise - and that noise is proportional to the value of the resistance. So trade-off. I'm sure there are other considerations, but those are the first which spring to mind. \$\endgroup\$ – brhans Jun 3 '15 at 13:05
  • \$\begingroup\$ Don't you worry: thank you, anyway, for your simple answer :) \$\endgroup\$ – Mister D Jun 3 '15 at 13:11
  • \$\begingroup\$ Also, with higher value resistors the circuit might become unstable and could oscillate. You can prevent that by adding a small capacitor across R2. In practice, the resistors will be between a few hundreds of ohms upto 1 Mega ohm. \$\endgroup\$ – Bimpelrekkie Jun 3 '15 at 13:12
  • \$\begingroup\$ @Rimpelbekkie I can't understand, in that application, when a value should be considered "higher". 100 ohm compared to 10 ohm? 10Kohm compared to 1Khom? And so on. \$\endgroup\$ – Mister D Jun 3 '15 at 13:15
  • \$\begingroup\$ By higher I mean that the chance of having oscillations increases as the value of the resistors in crease. The actual value of resistor above which oscillations can occur depend on the opamp so there is no absolute value. It depends on the properties of the opamp you find in the datasheet. \$\endgroup\$ – Bimpelrekkie Jun 3 '15 at 14:07
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As you have figured out, the gain is only a function of the ratio of the two resistors. Therefore, at first glance, 2 kΩ / 1 kΩ, and 2 MΩ / 1 MΩ are equivalent. They are, ideally, in terms of gain, but there are other considerations.

The biggest obvious consideration is the current that the two resistors draw from the output. At 15 V out, the 2kΩ/1kΩ combination presents a load of 3 kΩ and will draw (15 V)/(3 kΩ) = 5 mA. The 2MΩ/1MΩ combination will likewise only draw 5 µA.

What does this matter? First, you have to consider whether the opamp can even source 5 mA in addition to whatever load you want it to drive. Perhaps 5 mA is no problem, but obviously there is a limit somewhere. Can it source 50 mA? Maybe, but probably not. You can't just keep making R1 and R2 lower, even keeping their ratio the same, and have the circuit continue to work.

Even if the opamp can supply the current for the R1+R2 value you picked, you have to consider whether you want to spend that current. This can be a real issue in a battery operated device. 5 mA continuous drain may be a lot more than the rest of the circuit needs, and the major reason for short battery life.

There are other limits too at high resistances. High impedance nodes in general are more susceptible to picking up noise, and high-value resistor have more inherent noise.

No opamp is perfect, and its input impedance not zero. The R1 and R2 divider form a voltage source of impedance R1//R2 driving the inverting input of the opamp. With 2MΩ/1MΩ, this parallel combination is 667 kΩ. That needs to be small compared to the opamp's input impedance else there will be significant offset error. The opamp input bias current must also be taken into account. For example, if the input bias current is 1 µA, then the offset voltage caused by the 667 kΩ source driving the input is 667 mV. That's a large error unlikely to be acceptable.

Another problem with high impedance is low bandwidth. There will always be some parasitic capacitance. Let's say for example that the net connected to the two resistors and the inverting input has 10 pF capacitance to ground. With 667 kΩ driving it, you have a low pass filter at only 24 kHz. That may be acceptable for a audio application, but a serious problem in many other applications. You might be getting a lot less gain at high frequencies than you expect from the gain-bandwidth product of the opamp and the feedback gain.

As with everything in engineering, it's a tradeoff. You have two degrees of freedom in chosing the two resistors. The gain you want only nails down one degree. You have to trade off the current requirements and output impedance to decide the second.

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  • \$\begingroup\$ A very exhaustive and comprehensive answer. Thank you. I have understood the fact about the current that the two resistors will draw. Let's take the case that I want to increase again this current from the output of the op amp, to the input of a transistor (eg a 6 amp transistor connected as "emitter-follower"). I can feed that transistor with 5 mA ( (15 V)/(3 kΩ) ) or would be better drive the transistor with more current, so with lower values of resistors for the op-amp? \$\endgroup\$ – Mister D Jun 3 '15 at 14:31
  • \$\begingroup\$ You stated: "What does this matter? First, you have to consider whether the opamp can even source 5 mA" Can you provide me a "real" example with the given op amp (TLV272)? So: can it source 5 mA? How much (max) current can it source without exceeding his features, without damaging it? I ask this so I can do a verify with the datasheet, because I'm not very able to interpret the various value for the op amp in the object. So, with your explanation I will be more able to understand a datasheet. Thank you. \$\endgroup\$ – Mister D Jun 3 '15 at 14:53
  • \$\begingroup\$ @Mist: Unfortunately the output current is poorly specified, although what is there is quite clear and obvious. See the bottom section of the chart on page 6, clearly labeled "Output current". With 10 V power is can typically source 13 mA, which is really no spec at all. You need to derate heavily. If this matters, get a better-specified opamp. \$\endgroup\$ – Olin Lathrop Jun 4 '15 at 11:29
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As mentioned above, low value feedback resistors have relatively high current which the amplifier must drive. In an inverting amplifier, Rin sets the input impedance, so it is best not to have too low a value because the signal source must drive this.

At the other end of the scale, very large resistors not only generate noise (thermal or Johnson noise), but due to the natural capacitance* of the part, they form a filter in the feedback loop, which at worst can undermine the loop stability of the amplifier. Quite apart from changing the ac response of your circuit in interesting and hair-pulling ways, this effect gets worse at lower gains, and at gains of below 4 (typically, depends on the specific amplifier) can bite quite painfully. Indeed, there are numerous amplifiers designed specifically to have a minimum gain and are unstable below this gain (the advantages include better transient specifications).

As a general rule, I limit feedback resistors to no more than ~220k for either inverting or non-inverting configurations. If this does not yield sufficient gain, use an extra gain stage.

There are tricks one can do (a T network of resistors in the feedback loop is a well known one) to raise the gain of a single stage, but amplifiers are cheap and take up negligible space.

In inverting topologies, the choice of feedback resistor is primarily driven by the requirements of the signal source which sets the input resistor (usually minimum) size.

  • This becomes clear when one defines capacitance as existing between any two points of differing electrical potential

HTH

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To give a really short answer: something in the range of tens ofs will probably be good (with most OP-amp models and for most applications). Try 40 kΩ for R1 and 20 kΩ for R2.

This is of course not ideal in all circumstances, but it should usually work fine with a reasonable tradeoff between power consumption and noise level. Olin Lanthrop and Peter Smith have explained in detail what disadvantages you get with too high or too low resistance values.

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    \$\begingroup\$ +1 - although I think this answer would fare better as a comment, IMO you deserve an upvote for clearly stating what most of us already know, but what is seldom said - that most op-amps can usually go nicely with 10k-ish values. I would even go a bit further, to state that the usual range of R value for most of the common EE applications is 100 - 100k, and op-amps are devices which work best with values near the upper boundary. I myself found some cheapo op-amps that didn't want to work good with 47k-100k res, going down to 10k-33k range made them work perfectly. \$\endgroup\$ – vaxquis Jun 3 '15 at 23:03

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