# Infinite output impedance in amplifier, instead of 0?

The question asks: "Exercise 14.5 Assume an ideal op amp and use the summing-point constraint to find an expression for the output current io in the circuit of Figure 14.14. Also find the input and output resistances of the circuit".

I'm confused about the answer for output impedance. The solutions manual says that "because io is independent of RL, we conclude that the output impedance of the amplifier is infinite." Isn't the output impedance always 0 in an ideal op amp? What do they mean by current independent of load? Wouldn't that always be the case unless there was a controlled source?

• Nitai Michalski, I don't know who wrote this but it's the last place you can find out anything; don't take it seriously. Commented Jul 17 at 19:31
• @Circuitfantasist I recognize where this problem is from. It is from Electrical Engineering; Principles and Applications by Hambley. This is a well-renowned book used by universities and it is not right to say "don't take it seriously" in my opinion.
– Carl
Commented Jul 17 at 21:02
• @Carl, books of this kind (I would also include Sedra's book) are sterile, scholastic and dogmatic. They analyze ready-made circuit solutions as if they were given by God, and do not show the meaning of all this, the ideas, principles, concepts on which they are built. The funny thing is that the title ("Principles and ...") claims to do this, but it doesn't. If it did, the OP wouldn't be asking this question here; and in general, these books raise many of the questions asked here. I personally only use them as a basis for reasoning. Commented Jul 18 at 8:57

Isn't the output impedance always 0 in an ideal op amp?

The ideal op amp itself has zero output impedance, yes. But the output impedance of the amplifier you build with it can be anything but zero, depending on the configuration, of course. Take a look at the marked text in your question again. It clearly says "the amplifier has infinite output impedance", not the op amp.

What do they mean by current independent of load?

Basically a current source.

The circuit in your question is a form of voltage-controlled current source. As you already know, an op amp with negative feedback keeps the input terminal voltages equal. The negative feedback loop is closed by the load resistor, RL. Since the voltage across the RF is equal to the non-inverting terminal voltage (or the input/control voltage), the current through RF and therefore RL will be kept constant. So you have a current source and its magnitude is set by the input (control) voltage and RF.

And remember, a current source has theoretically infinite output impedance.

This opamp is being used as a transconductance amplifier, so it doesn't work the way you normally think of an opamp working.

Assume $$\V_{in}\$$ is a fixed voltage 500 mV, the opamp wants to make both inputs have the same voltage so it will output enough current to cause there to be a 500 mV drop across $$\R_F\$$. Since the output current depends on $$\V_{in}\$$ and $$\R_F\$$, it is independent of $$\R_L\$$. Since $$\\Delta R_L\$$ has no effect on the current it's like the output impedance of the opamp is infinite.

Imagine a current source made with a very high voltage and a very high resistance in series with a relatively low resistance load.

simulate this circuit – Schematic created using CircuitLab

With a 0 Ω load the current is 1 uA, with a 1k load it's 0.999999 uA, very little change. If you could make the voltage and resistance of the source infinite the current wouldn't change at all with a change in the load. So if the load doesn't affect the current the impedance must be infinite.

• but isnt $i_o = \frac{v_o - v_F}{R_L}$ also ? or do we only use that Load resistance to convert the generated current to voltage , but doesnt that mean the output impedance is load-dependent? Commented Jul 17 at 18:08
• @KnowledgeSeeker Yes, but Vo will be dependent on RL so the current will stay the same with changes in RL. RL affects the output voltage, not the output current. Commented Jul 17 at 18:16
• Ah yes your point, i would like to see a rigourous mathematical proof about this phenomenom . Commented Jul 17 at 18:20
• @Circuitfantasist Nah, it's ideal, it can handle anything! Commented Jul 17 at 19:20
• @Circuitfantasist I'm trying to show that with a source with infinite impedance the current would be independent of the load in an easy to understand way. Commented Jul 18 at 17:10

Let's treat the op-amp and its periphery as a "black box", leaving only the load resistor $$\R_L\$$ outside:

simulate this circuit – Schematic created using CircuitLab

The "black boxes" above each have two terminals, X and Y, and you have no idea what's inside the box. Each box has some internal resistance, which we aim to measure.

We could measure it by changing the external resistance $$\R_L\$$, or by applying some external potential difference between X and Y, or even by forcing some current through the box. Whatever method we use, we shall measure the resulting change in voltage $$\\Delta V_{XY}\$$, and change in current $$\\Delta I_O\$$. Effective resistance $$\R_{XY}\$$ between X and Y inside the box will be, according to Ohm's law:

$$R_{XY} = \frac{\Delta V_{XY}}{\Delta I_O}$$

For box B, you'd find that there's no change in voltage, since its internal voltage source will strongly oppose any change in $$\V_{XY}\$$, but current $$\I_O\$$ is able to rise and fall freely:

$$R_{XY(B)} = \frac{\Delta V_{XY}}{\Delta I_O} = \frac{0}{\Delta I_O} = 0\Omega$$

In other words, box B (or any ideal voltage source) is said to have zero impedance.

Box C is a current source, and we would expect it to strongly oppose changes in current through it, $$\\Delta I_O=0\$$, while freely permitting $$\V_{XY}\$$ to change:

$$R_{XY(C)} = \frac{\Delta V_{XY}}{0} = \infty \Omega$$

Ideal current sources, therefore, are said to have infinite source impedance.

Box D will have an impedance somewhere in between zero and infinity, because in this case it would be possible to impose any arbitrary voltage between X and Y, and it's also possible to inject an arbitrary current through the box. With either "perturbation" method, if you plotted $$\I_O\$$ on the horizontal axis, vs. its corresponding $$\V_{XY}\$$, the slope of that graph is $$\R_{XY}\$$. Just by inspection, you could predict that this slope would be:

$$R_{XY(D)} = \frac{\Delta V_{XY}}{\Delta I_O} = 100\Omega$$

For more information, and a graphical treatment of using voltage and current changes, and the slope of the characteristic V-I graph to represent impedance, see this answer.

To answer your question, it remains to be determined which of the boxes B, C and D has behaviour which most closely resembles the behaviour of box A.

We are permitted to attach any load $$\R_L\$$ we desire and you could even apply some arbitrary potential difference between X and Y, and it would still keep $$\I_O\$$ constant. The op-amp will adjust $$\V_{XY}\$$ to whatever value is necessary to keep $$\I_O\$$ fixed. Thus $$\\Delta I_O = 0\$$, so:

$$R_{XY(A)} = \frac{\Delta V_{XY}}{0} = \infty \Omega$$

This is consistent with the behaviour of the ideal current source in box C.

It's true that the op-amp is indeed a voltage source, using voltages to obtain this behaviour, but in the context of a two-terminal "black box", for which we only have access to the exposed terminals, nodes X and Y, any external "observer" circuitry would perceive it to behave like a current source, for which the impedance inside the box is effectively infinite. No amount of voltage applied, and no impedance of load attached (op-amp constraints notwithstanding, such as limited output voltage range and current limits) will result in a change in $$\I_O\$$.

• A nice, if a bit stiff, presentation... It brings to mind the four-terminal network theory that's common in transistor analysis. Either way, it's a nice addition to the mix of answers. Commented Jul 21 at 6:23
• @Circuitfantasist "stiff" LOL. You mean missing some of my classic humour and hilarious rhetoric, that I'm famous for? Commented Jul 22 at 12:20
• No, Simon, I just meant this more schematic way of presenting it as circuit blocks. Maybe "stiff" is not the best word for it but what to do as English is not my native language. Otherwise, I welcome your willingness to break away from your usual detailed way of presenting working CircuitLab simulations and use this compact way of laying out schematics. I don't have the strength to give up "live" simulations, maybe I'm overdoing it (I wish every schematic was "live"). But this works very well by supplementing the answers... Commented Jul 22 at 14:25
• I also experiment with different techniques. Today I wasted a lot of time talking to Gemini trying to get it to come up with circuit solutions but it turned out to be harder than I thought. Apparently it is better at analyzing other people's ideas than generating its own... But overall the discussion was very useful to me even though it left me mentally exhausted… Commented Jul 22 at 14:25

You are correct, the output impedance of an ideal op-amp is zero.

However, the output impedance of the whole op-amp circuit that drives the load resistance Rload is not zero.

You can think that the load resistance Rload can be any arbitrary value, and still the current only depends on the value of Rf and Vin.

You can use Kirchhoff's laws or golden rules of ideal op-amps to prove that voltage over Rf equals Vin and thus current is only defined by Rf.

So no matter how many teraohms or micro-ohms the Rload is, there will be up to teravolts or whatever is required at the op-amp output, and over Rl to make enough current flow that voltage over Rf equals Vin.

• Justme, I think it's just a kind of a non-inverting amplifier. Commented Jul 17 at 19:44
• @Circuitfantasist Now that you said it, it is. However, I simply viewed it as an ideal non-inverting unity gain buffer amplifier, with the non-ideal output impedance labeled as Rload, and the feedback cancels effect of Rload out so it simply drives the Rf with the input signal. Commented Jul 17 at 20:47
• Justme, I love this explanation where this circuit is presented as a "disturbed follower". The load resistance (voltage) is the disturbance that the op-amp overcomes. Depending on what we take as an output, we get various circuits: 1) the voltage across Rf - a "disturbed follower", 2) the op-amp output voltage - a non-inverting amplifier, 3) the current through Rf - a constant current source. Commented Jul 18 at 8:13
• Perhaps it should only be specified that for this circuit to be a non-inverting amplifier, the load must be a linear resistor. Commented Jul 18 at 17:04
• @Circuitfantasist I don't agree. The ideal opamp can do literaly anything to make Vin- and Vin+ equal, so it would not even be an amplifier but an unity gain buffer. Commented Jul 18 at 18:13

When they say "output impedance of the amplifier" they mean output impedance of the closed loop amplifier.

The circuit represented by the symbol below is an op amp, which you typically put feedback around to make various kinds of amplifiers. In other words, the op amp is an open loop amplifier and, after placing feedback around it, you create a closed loop amplifier.

simulate this circuit – Schematic created using CircuitLab

Depending on the feedback network, you can make 4 types of closed loop amplifier: voltage, current, transconductance, and transimpedance. They will have either infinite or zero input impedance, and infinite or zero output impedance.

So, you are right, an ideal op amp does have zero output impedance, but depending on the kind of feedback network, you can create a closed loop amplifier with either zero or infinite output impedance.

• And how is the negative feedback implemented in this case? Commented Jul 18 at 16:44
• It is a series-series feedback circuit: Rf senses the current through the load and applies a voltage proportional to that current at the input of the open loop amplifier. This scales both the input and output impedance of the open loop amplifier up by the loop gain. Commented Jul 18 at 16:47
• Perhaps more precisely, "applies a voltage proportional to that current at the input of the open loop amplifier (in series and oppositely to the input voltage)". Commented Jul 18 at 17:00

The circuit presented by OP is a constant current source (op-amp and current-setting resistor RI) that supplies a "floating" load RL. Through negative feedback, the voltage across RI is maintained equal to the input voltage (Vin). As a result, the current flowing through both RI and RL depends only on Vin and RI, following the equation IL = Vin/RI.

# Beyond IL = Vin/RI

Such "textbook explanations" using established terms that have become verbal clichés do a good job in many cases but not when the goal is to understand the circuit solution. Understanding is different from knowing and doing. It requires you to understand the idea, the concept, the principle... Apparently, that is why the authors of the famous book from which the OP's question was born showed this necessity in the book title, but for some reason they left it to us to do it. Then let's take advantage of this unique opportunity and discover the principles on which this class of circuits is built. This will give the OP a good foundation in their future encounters with similar circuit solutions.

## Classic negative feedback solution (OP's circuit)

Basic idea: To set a current of a certain value through the load according to the principle of negative feedback, we need to convert it to a voltage (because our devices work with voltage) and compare it with the input voltage. Then we vary the current until the difference between the two voltages becomes zero. As a result, IL = VRI/RI = Vin/RI.

Implementation (conceptual circuit): For this purpose, we (the "op-amp" ) apply a variable voltage VOA to the RL-RI network. Then we subtract in series manner the voltage VRI from Vin, and properly apply the result to the op-amp differential input. Finally, we vary the voltage until reach equilibrium (VNI = 0 V).

simulate this circuit – Schematic created using CircuitLab

Op-amp non-inverting current source (OP's circuit): We can make an op-amp do this routine work. For simplicity, I have replaced the resistors with meters of the same internal resistance.

simulate this circuit

Let's investigate the current source at various loads (from 0 to 10 kΩ) by the help of the CircuitlLab DC Sweep Simulation. As you can see, the curve of the current (in blue) is absolutely horizontal.

Transistor current source with emitter degeneration: This circuit solution has long been known in its transistor implementation. To make the circuit more perfect, I have added a bias voltage.

simulate this circuit

Problem: In the configurations above, both input voltage source and current-setting resistor are grounded but the load is "floating" which is in most cases inconvenient.

## Negative feedback solution with floating input voltage source

Conceptual circuit: We can swap the resistors so that the load is grounded but now the input voltage source is "floating". As you can see, Vin, RL and the null indicator NI are connected in a loop.

simulate this circuit

Op-amp circuit: Vin, RL and the op-amp differential input are connected in a loop.

simulate this circuit

## Both negative and positive feedback solution (NIC)

Conceptual circuit: The input source may be grounded if it is a current source.

simulate this circuit

Op-amp current mirror: This is the odd negative impedance converter (NIC). The op-amp simultaneously changes the voltages at its inputs to reach equilibrium (VRI = VR).

simulate this circuit

## No negative feedback solution (bootstrapped RI)

Conceptual circuit: Constant current can also be obtained without using the principle of negative feedback. If we are observant enough, we will notice that the voltage at the upper end of the current-setting resistor RI follows the voltage at its lower end. So we can add a constant offset voltage to the load voltage and apply it to the upper end of RI.

simulate this circuit

As you can see in the graphs beluw, the curve of the upper end is lifted by 1 V above the curve of the bottom end.

Improved Howland current source: This idea is implemented in the so-called "Improved Howland current source". The load voltage variations are amplified 1x and appear at the top RI end.

simulate this circuit

## Inverting current source

Conceptual circuit: Finally, let's see how the same idea is implemented through an inverting op-amp configuration.

simulate this circuit

Op-amp inverting current source: Here, the voltage across the current setting resistor RI is subtracted from the input voltage and the result is zero (the so-called virtual ground). The load is virtually grounded which is better than to be floating. I have used an op-amp with supply ends to show the load limitations.

simulate this circuit

When RL exceeds 10 kΩ, the op-amp output voltage reaches the supply rail and stops changing.

# What "infinite output impedance" means

The main property of a constant current source is that its current does not change when we change the resistance of the load (the voltage across it). But if we don't want them to understand us :-), we say that "its output impedance is infinite". What is actually going on in the OP's circuit that this is the case?

Regardless of the specific implementation, the same thing happens: When we start increasing the load resistance RL, the op-amp current source starts increasing the load voltage VRL by the same amount so the ratio VRL/RL = IL does not change. So the main idea in these constant-current circuits is dynamic voltage.

The same effect can be achieved with infinite "static" (constant) voltage and infinite constant resistance but the losses will be great.

So the end result for both implementations is the same - infinite resistance, but the nature of that resistance is different:

• In the first case it is not physical resistance but dynamic voltage that creates the same voltage opposition as the physical resistance (aka "bootstrapping").

• In the second case it is physical resistance.

The output impedance of a two port network can be found by replacing the load resistor by a current source with known current $$\I_\text{TEST} \$$, calculating the voltage drop across it, and dividing the two. In this case it would look like this: -

simulate this circuit – Schematic created using CircuitLab

No current flows into the opamp terminals so $$\V_\text{TEST-} = -I_\text{TEST}R_f \$$. The output of the opamp is equal to the open loop gain $$\A_0\$$ multiplied by the voltage difference between the terminals, so $$V_\text{TEST,+} = A_0(V_\text{in}-V_\text{TEST-})$$ Assuming that $$\V_\text{in} \neq V_\text{TEST-}\$$ we can without loss of generality set $$\V_\text{in} = 0 \: \text{V} \$$, causing

$$V_\text{TEST,+} = -A_0V_\text{TEST-} = A_0I_\text{TEST}R_f$$

The voltage drop across $$\I_\text{TEST}\$$ is

$$V_\text{TEST,+} - V_\text{TEST,-} = A_0I_\text{TEST}R_f - I_\text{TEST}R_f = I_\text{TEST}R_f(A_0-1)$$

Finally , the output impedance

$$Z_\text{out} = \frac{V_\text{TEST,+}-V_\text{TEST,-}}{I_\text{TEST}} = \frac{I_\text{TEST}R_f(A_0-1)}{I_\text{TEST}}$$

$$Z_\text{out} = {R_f(A_0-1)}$$ For an ideal opamp $$\A_0 \to \infty\$$ so clearly $$Z_\text{out} \to \infty$$

Two things to note here: First, is that the circuit is simply a non-inverting amplifier. The interesting part, is that the feedback resistor is considered as the load resistor $$\R_L\$$ which I haven't seen before. Secondly, in the analysis above note that the principle of a virtual short doesn't apply to the opamp's input terminals because the ideal current source determines the voltage across $$\R_f\$$ independently of everything else. In a general setting with a regular load resistance the virtual short principle will apply, however.

• Carl, you cannot connect a constant current source at this place because a conflict occurs between two current sources connected in series (op-amp and Rf is the other current source). Instead, you need to connect a voltage source and vary its voltage by reading the current. In your circuit, the op-amp output voltage will reach the supply voltage, and the op-amp will simply become a "battery". Also, not only RL is the feedback resistor; Rf and more precisely, the combination (voltage divider) Rf and RL forms the negative feedback. Next, "the principle of a virtual short" is always valid... Commented Jul 18 at 9:15
• ... and should be applied even in your case; otherwise, as I have said above, the op-amp is not an op-amp but a "battery"... Commented Jul 18 at 9:17