According to Sedra & Smith's microelectronics textbook, in order to avoid a loss of signal strength, voltage amplifiers are required to have a high input resistance, which I agree with. However, in case of the inverting op-amp, why does \$R_1\$ also have to be high? I don't understand why a voltage divider is necessary in this case, considering an ideal op-amp has infinite impedance in its input and all the current is going to go through \$R_2\$ anyways. From my understanding, what we want is for \$R_1\$ to be low, since that's what's going to result in a high \$v_o\$ given that \$v_o = -\frac{R_2}{R_1} v_I \$.

enter image description here

For reference, I'm referring to the text in section 2.2.3 of Sedra & Smith's 7th edition of "Microelectronic Circuits".

  • \$\begingroup\$ Why would you want the current through R1 to be high? The output voltage is \$v_o=\frac{R_2}{R_1}v_i\$ \$\endgroup\$ – Chu Apr 28 at 19:40
  • 1
    \$\begingroup\$ I think I didn't express myself properly. What I mean by the current through \$R_1\$ being high is that I want \$R_1\$ to be low, since in that case the amplifying effect would be greater. \$\endgroup\$ – rafaelfp Apr 28 at 19:44
  • 2
    \$\begingroup\$ The voltage gain is \$\frac{R_2}{R_1}\$. For a voltage amplifier, the input current is normally low, so \$R_1\$ would be typically in the \$k\Omega\$ region. \$\endgroup\$ – Chu Apr 28 at 21:03
  • \$\begingroup\$ My respect for the Sedra&Smith's bestseller... but using the voltage divider principle to explain the role of R1 is inappropriate and misleading here. If Rs and R1 were a voltage divider, then we would take the voltage drop across R1 as a voltage output (Vin for the inverting amplifier). Instead, we use the current through R1 as an output, which we then convert to a voltage drop across R2. Finally, the op-amp creates a "copy" Vout = VR2 of this voltage at its output. Rs and R1 are simply two resistors in series... so Rs is added to R1 thus decreasing the input current, the total gain and Vout. \$\endgroup\$ – Circuit fantasist Apr 30 at 10:30

10 Answers 10


The problem the author is trying to avoid is overloading of the source signal.


simulate this circuit – Schematic created using CircuitLab

Figure 1. A voltage source with it's source impedance, Rs feeding an inverting amplifier with input impedance Ri.

Remember that the inverting input of an inverting amplifier is at virtual ground.

  • Let's say my Vs is 1 VRMS and has an output impedance of 1 kΩ.
  • Our amplifier has a gain of -1. We expect an inverted version of Vs to appear on the output with a reading of 1 VRMS.
  • We're disappointed! With the components I've drawn we can see that the voltage at Vi is half of Vs. As a result the amplifier output will be 0.5 VRMS.

If we replace Ri with a 10 kΩ resistor we'll be much closer to the expected value, 10/11 VRMS. The higher the input impedance the less it loads the source.

Note that with some systems such as 50 Ω signal generators the source has a 50 Ω impedance and is designed to give the nominal voltage on the output when driving a 50 Ω load. If measured using an oscilloscope while there is no load the reading will be double the value set on the output control.

| improve this answer | |
  • 1
    \$\begingroup\$ @rafaelfp. Thank you for accepting my answer despite a couple of typos near the end. I've fixed them. I hope it helps. \$\endgroup\$ – Transistor Apr 28 at 22:12
  • \$\begingroup\$ "The problem the author is trying to avoid is overloading of the source signal." If the input source is overloaded by sourcing/sinking unacceptably high current, the op-amp output will be overloaded by sinking/sourcing the same current since the four elements - Vi, R1, Rf and Vo are connected in a loop. If Ri = Rf, both Vi and Vo will be equally loaded with the same resistance. If Ri < Rf, Vo will be loaded with lower resistance than Vi. \$\endgroup\$ – Circuit fantasist Apr 29 at 20:10
  • \$\begingroup\$ Not necessarily. I'm thinking of devices such as a hi-Z microphone or guitar pickup being fed into a lo-Z amplifier input. \$\endgroup\$ – Transistor Apr 29 at 20:13
  • \$\begingroup\$ In the last sentence of my comment above, read "Ri > Rf" (a typo). In the case of a "hi-Z source", there is no need of R1. The source will act as a current source and the combination of Rf + op-amp will act as a transimpedance amplifier. \$\endgroup\$ – Circuit fantasist Apr 29 at 20:50
  • 1
    \$\begingroup\$ @Circuitfantasist I'm glad you addressed this, because that's what was making me confused in the first place. I accepted the answer because I came to terms with the fact that ideally we would like \$R_1\$ to be high because then I wouldn't have to worry about the source's resistance, since the gain would be defined only by \$R_1\$ and \$R_2\$. But I'm glad you understood my initial concern that \$R_1\$ wouldn't really be necessary, considering we're just using the current through \$R_1\$ as the output. \$\endgroup\$ – rafaelfp Apr 30 at 17:00

If VI is an ideal voltage source then it will have zero output resistance in which case R1 doesn't have to have a high value.

In the real world VI will have an output resistance (RI) which forms a potential divider with R1. Thus if RI has a high value and R1 has a small value most of the signal will be lost across RI.

As a rule of thumb, R1 should be at least 10X the value of RI so that most of the source voltage is transferred to R1.

Remember, the inverting input of the op amp (point 1) is held very near to ground. That is to say it is a virtual earth.

| improve this answer | |

The choice of R2/R1 determines the gain. The choice of R1 (or R2, since they're proportional to each other) is determined by a number of factors. You may wish the input impedance to be high, so you'd want R1 to be high, however bias current and other factors limit how high R1 and R2 can be.

So if you pick R1 =100K\$\Omega\$ and the gain should be -10, then you need R2= 1M\$\Omega\$. If the input capacitance of the op-amp is too large, the stability and cutoff frequency will be affected. Even at DC a mere 10nA bias current will cause an output offset of 10mV.

| improve this answer | |
  • 1
    \$\begingroup\$ Even though this choice is determined by a number of factors, one of the factors mentioned in the textbook for choosing a high value for \$R_1\$ is to avoid loss of signal strength through the voltage divider principle. My problem is that I don't see how this principle applies to this case. \$\endgroup\$ – rafaelfp Apr 28 at 19:49
  • \$\begingroup\$ Assume the source vi is not ideal, but has a series resistance... for example 50\$\Omega\$. If you make R1 500\$\Omega\$ you'll lose about 9% of the signal. \$\endgroup\$ – Spehro Pefhany Apr 28 at 20:13

Op-amps have a very high input impedance. Almost no current enters through the input terminals.

Say the input voltage is 10 volts and the input resistance is 1 ohm. As the lingering input acts as a virtual ground, the current through the resistor will be 1 amp. If feedback resistance is also 1 ohm then the output voltage will be -10 volts.

But if the input resistance was 1k ohm then the current through the input resistance would be 10 mA. If the feedback is also 1k ohm then the output will be -10 volts.

So, in both cases, the output voltages are the same, but the current is reduced on a large amount. So, less power loss.

| improve this answer | |

Your problem is that you are assuming your signal source has a zero impedance and can drive any load impedance. In practice, neither of these assumptions are true. If the source impedance is R, then the input signal, as seen by the amplifier, will be divided because of the voltage divider action of R and R1. This will reduce the overall gain. Furthermore, a real signal source is current limited and thus cannot drive any impedance.This is why DKNguyen implied that choosing R1 = 1 ohm is not practical since most voltage sources cannot drive 1 ohm loads.

| improve this answer | |

If I fully understand your question, you always want your amplifier to have the highest input impedance possible, so that it minimizes the effect of the source input. A low input impedance would swamp the source.

| improve this answer | |
  • \$\begingroup\$ Welcome to EE.SE, Jeff. I think you are trying to say "have the highest input impedance possible, so that it minimizes the effect on the source" but that's not always true as explained in the 50 ohm devices in my answer. "Swamping" the source suggests filling the source up with something. The problem is draining the source or overloading it by taking out more than it is designed for. This is also explained in my answer. \$\endgroup\$ – Transistor Apr 29 at 18:06

The inverting and non-inverting inputs of an ideal opamp are virtually shorted. The other assumptions for an ideal opamp would be infinite gain, banwidth and output current no current draw on it's inputs and zero output resistance.

So R1 is the input resistance of the inverting topology and this is the load that the previous stage sees. R1 sets also the current that will flow through R2 and the output of the inverting opamp. Now in reality it all depends from the previous stage, the output stage of the opamp and the load of the inverting opamp. Can the previous stage drive R1? Can the inverting amp output stage handle the current that R1 value dictates?

| improve this answer | |
  • \$\begingroup\$ "The inverting and non-inverting inputs of an ideal opamp are virtually shorted." No, while this is true in standard inverting and non-inverting amplifier configurations it is not always true. Comparator and Schmitt trigger applications deliberately allow the inputs to be at very significantly different potentials by design. Otherwise your answer is correct. You can edit to clarify. Welcome to EE.SE. \$\endgroup\$ – Transistor Apr 30 at 15:11
  • \$\begingroup\$ "The inverting and non-inverting inputs of an ideal opamp are virtually shorted" (if there is a negative feedback as Transistor noted). And how are they shorted? Maybe they are connected by a piece of wire? It would be interesting and useful to explain... Your thoughts in the second paragraph were intriguing for me... and they made me think... Regarding "Can the previous stage drive R1?", it always can drive it since the virtual ground is not firmly fixed. If the current through R1 becomes too high (> VOUTmax/R2), the virtual ground will "move" and this way it will protect the previous stage. \$\endgroup\$ – Circuit fantasist Apr 30 at 18:47

tl; dr version:

  • If Vin has zero source impedance (an impossible unicorn), then the values of R1,R2 or anything else in the op-amp don't matter.
  • If Vin does have source impedance (like all voltage sources in the real world) then the op-amp R1/R2 circuit will load Vin and affect the overall gain.


The ideal op-amp has two defining characteristics:

  • (1) infinite gain (strongly amplifies the difference between the inputs)
  • (2) infinite input impedance (no current flows into the inputs)

Because of (1) and (2), R1 and R2, they can be any value so long as their ratio R2/R1 achieves the desired gain. Why? Because of (1) above: Vo will swing to the voltage to bias the R2/R1 voltage divider so that it cancels the difference between the inputs. And because of (2), there's no bias current to account for so the resistors can be near-infinite (except that R1 can't be zero, because, well, dividing by zero.)

That is:

  • (3) Vo = -Vin * R2/R1

Which you may recognize as the equation for an ideal inverting op-amp.

In the inverting configuration, Vo is working via R2/R1 to cancel the difference on its inputs, such that the (-) input is the same voltage as (+) input, that is, the same as ground. Sometimes this is called a virtual ground: it's the same voltage as ground, but not actually ground.

The upshot is, Vin sees a Thévenin load that is, effectively, R1 to ground.

Back to that Vin unicorn. If Vin has zero impedance, as it would by definition if it were an ideal voltage source, then the gain is strictly set by R2/R1, as in (3).

So, yes, with that ‘ideal’ unicorn Vin voltage source, you can ignore R1 and the the op-amp loading it. You’re technically correct - you don't care about 'the input resistance of the inverting op-amp being so high.'

But also, you're wrong, and that is what your instructor is leading you towards. Which is...

The Real World

Brush off that unicorn glitter. Ideal voltage sources don't exist. There is nothing in the universe that can deliver infinite current. So Vin impedance is always a factor.

Since we know Vin always has impedance; this reduces the gain. That is,

  • (4) Vo = -Vin * R2 / (R1 + Zo[Vin])

The bigger the Vin impedance, the less the gain.

We can minimize the effect of it, compensate for it, or even nullify the effect by using a buffer. The choice of solution depends on how strongly Vin can drive the signal.

  • Very low Vin impedance: just make R1 as big as practical.
  • Low Vin impedance: adjust R1/R2 to correct the gain based on (4) above.
  • high Vin impedance: add a buffer to Vin.

Here's what that buffer setup looks like:


simulate this circuit – Schematic created using CircuitLab

I've introduced a useful op-amp circuit called a unity-gain follower. The unity-gain follower hides Vin from R1 loading by using its ability to source and sink (theoretically infinite) current in its job of making an exact copy of Vin.

The follower's gain is:

  • (5) Vo = Vin * (1 + 0 / infinity), that is, gain is 1.

Meanwhile, the follower (+) input impedance is high (also, theoretically infinite) so presents no load to Vin. Now we truly don't care about R1 and R2, and we can go back to that unnatural unicorn frolic as we've created a copy of Vin that has infinitely low output impedance.

Can this without the addition buffer? Yes, by using the non-inverting connection (ground R1, connect Vin to (+).


simulate this circuit

This works if you don’t need the inversion, and can live with the gain never being less than 1:

  • (6) Vo = Vin * (1 + R2 / R1)

For non-inverting gain.

| improve this answer | |
  • 1
    \$\begingroup\$ I realize that they can have any value in order to achieve a desirable gain, but what I don't understand is why does a high \$R_1\$ help to avoid loss of signal strength. \$\endgroup\$ – rafaelfp Apr 28 at 19:53
  • \$\begingroup\$ R1 presents as a load from Vin to the (+) voltage, which in this circuit, is ground. That is because the op-amp Vo is strongly driving R2 and R1 so that (+) and (-) are the same voltage. \$\endgroup\$ – hacktastical Apr 28 at 19:59
  • 1
    \$\begingroup\$ But suppose that we didn't have an ideal power source and there was a small resistance \$R_I\$ from \$v_I\$ in series with \$R_1\$, how would that affect the amplifier? From my understanding, there wouldn't be a loss of signal strength, since the entirety of the signal in \$v_I\$ would still be amplified, regardless of the value in \$R_1\$. \$\endgroup\$ – rafaelfp Apr 28 at 20:05
  • 1
    \$\begingroup\$ But your answer doesn't address the point that I'm making. You wrote that the lower you make \$R_1\$ the more it loads down \$v_I\$ (which from my understanding means that some of the signal strength is lost), but I don't see how that is true. I understand that \$R_1\$ is the load from \$v_I\$ to ground, but my point is that this load doesn't need to be large to preserve signal strength. \$\endgroup\$ – rafaelfp Apr 28 at 20:34
  • 1
    \$\begingroup\$ That is, if you assume Vin is a zero-impedance (ideal) voltage source, the value of R1 is indeed irrelevant, because you can assume that Vin can supply infinite current regardless of its load. So, yes, you're right, but also, wrong, as this is never the case in the real world. Look at eq. 4 in my answer to see the relationship. \$\endgroup\$ – hacktastical Apr 28 at 20:46

Besides all the mentioned loading aspects - one of the most important reason for the desired high input resistance is the closed-loop gain Acl. Only in this case (no remarkable current into the opamp input nodes) the closed loop gain is - with very good accuracy - determined by the external feedback path only (as another precondition: Very large open-loop gain Aol and very small ouput resistance).

Otherwise, the closed-loop gain expression (Acl=-R2/R1) would contain (and would also depend on) the neglected input resistance values.

| improve this answer | |

The traditional approach for explaining the inverting configuration, particularly "R1 phenomenon", is by using the virtual ground concept. My observation is that this is a rather formal and abstract concept and it is not well understood what it really is. For example, when we say that "the inverting and non-inverting inputs of an ideal op-amp are virtually shorted" (tonal_123), an obvious question arises: And how are they shorted? Maybe they are connected by a "piece of wire"? It should be explained in details what and where this "wire" is... since beginners often think that op-amp inputs are internally connected through it:)

That is why, I decided to explain the phenomenon of this connection without mentioning any ground - neither real nor virtual (both they still exist but not used in the circuit diagrams below). Thus my explanation will be valid even in the case of a "floating" circuit; it is universal. In short, my idea is:

Since the question is about the value of resistance R1, we can use more direct way of explaining it - in terms of resistances.

1. Ideal input voltage source (Rs = 0). Let's replace the specific op-amp circuit with a simpler equivalent electric circuit - Fig. 1a, where the op-amp output is represented by a variable voltage source with voltage VOA = VR2.

Inverting amplifier - R1

Fig. 1. An equivalent electric circuit of the op-amp inverting amplifier driven by an "ideal" voltage source

The result is surprising - it turns out the four elements of this configuration (VIN, R1, R2 and VOA) are connected in a loop; the same current flows through them and the sum of voltages across them is zero (KVL).

Trying to grasp the idea of this connection, we can distinguish three pairs of elements here:

> Composed voltage source (VIN + VOA). First, we can see the two sources VIN and VOA at the lower part of the circuit diagram. They are connected in series, in the same direction, so that their voltages are summed up. So we can think of them as of one composed voltage source (circled in yellow) with effective voltage Veff = VIN + VOA.

> Composed resistor (R1 + R2). Also, at the upper part of the circuit diagram, we see two resistors in series that can be thought as of one composed resistor with resistance Reff = R1 + R2. Thus the current is I = Veff/Reff.

> Composed zero voltage source (VR2 - VOA = 0). Then, we look to the right of Fig. 1 where there is another intriguing pair of elements - R2 and VOA (circled in blue). The current flowing through R2 "creates" a voltage drop VR2 = I.R2 across it. The op-amp creates, by the mechanism of the negative feedback, equivalent voltage VOA = VR2. These voltages are summed in series so that their sum is zero. Thus the voltage drop VR across R2 is compensated by the additional voltage VOA and the current is I = VIN/R1 as though there is no resistor R2 connected. Figuratively speaking, the combination of R2 and the op-amp output is equivalent to a "piece of wire" (outlined in pink).

> Composed zero resistor (R2 - R2 = 0). To make our (and OP's) life more interesting... and to show that circuitry can be not only a boring craft but also a romantic art... we can explain the same phenomenon through the mystic concept of negative resistance. Thus, besides understanding something very important in circuits, we will have fun too. We can reason in the following way:

The current I flows through the resistor R2 and, as a result, a voltage drop VR2 = I.R2 appears across the resistor. The op-amp "copies" this voltage so the voltage VOA = VR2 appears at its output. Both voltages are proportional to the current (Ohm's law) but while the first is subtracted from the input voltage VIN (since it is a drop), the second is added to the input voltage (since it is a voltage). So, we can make a conclusion that the op-amp output behaves as a kind of "inverted resistor". It is accepted to say that it is a resistor with negative resistance... in short, a negative resistor.

Figuratively speaking, the combination of the positive resistance R2 and the negative resistance -R2 of the op-amp output is equivalent to a "piece of wire" with zero equivalent resistance - Fig. 1b... and we can think of this network as a sort of "active superconductor" (outlined in pink). Another interesting thing about this artificial superconductor is that we can make its resistance not only zero, but even more than that...

This magic will continue until the op-amp is saturated. From this moment, the positive resistance R2 will come to picture. For example, if R1 = 0, should a short circuit to become since the total loop resistance is 0 + R2 - R2 = 0. But the op-amp will saturate and the current will be limited by R2 (0 + R2 - 0 = R2).

Of course, some input resistance (R1, Rs or both) is still needed to decouple the input voltage source from the op-amp inverting input and this way, to provide a negative feedback. If you connect an "ideal" voltage source directly to the op-amp input, the op-amp output will not be able to confront it through R2 and the negative feedback will not function. The gain will be maximum op-amp open loop gain... and the op-amp will be saturated. So both Vin and Vout have to impact the negative input through resistors, in opposite directions.

2. Real input voltage source (with Rs). When the input voltage source has some internal resistance Rs - Fig. 2, it is added to the other resistances. So there are a total of four resistors in the loop - three positive (Rs, R1 and R2) and one negative (-R2). But the last two are mutually destroyed and the current in the circuit is determined only by the first two - I = VIN'/(Rs + R1).

Inverting amplifier - Rs+R1

Fig. 2. An equivalent electric circuit of the op-amp inverting amplifier driven by a real voltage source

3. Without op-amp (VOA = 0). To appreciate something in life, you have to remove it for a moment. So let's imagine what the circuit would look like if the op-amp was gone and the right end of R2 was directly connected to ground - Fig. 3.

Inverting amplifier - Rs+R1+R2

Fig. 3. Equivalent electric circuits of the op-amp inverting "amplifier" without op-amp (VOA = 0)

The negative resistance -R2 disappears and the positive resistance R2 comes to picture. The total resistance is a sum only of positive resistances - R1 + R2 or Rs + R1 + R2. As a result, the current decreases. This situation will happen when the op-amp reaches the supply rails (saturates).

But still... what is the idea behind these four elements in a circle (Fig. 1a)?

The idea is to put the input and output voltages in a steady proportion VOUT/VIN (= 1, > 1 or < 1) set by the ratio R2/R1 between two constant resistors R1 and R2. How is this acomplished? Here is another fresh explanation:

Think of the combination VIN and R1 as of a current source (sourcing voltage-to-current converter)... and of the combination VOA and R2 as of a current sink (sinking voltage-to-current converter). The current source and sink are connected in a loop so the same current should flow through both. The source is a "master" and the sink is a "slave" in this "tug of war" named inverting amplifier. So, when the input source increases its voltage to "push" more current, the op-amp output sink decreases its voltage to "suck" the same current...

The OP question was an illustration of how a delusion can be useful. It was obvious that an ideal voltage source (Rs = 0) cannot be connected directly to the inverting op-amp input (R1 = 0)... but the persistence with which OP defended this thesis made us think of the circuit idea... which was the most important thing. I hope my thoughts have helped you to realize the problem in depth.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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