# How does an op amp amplify things in respect to its electron flow?

I'm a newbie learning about electronics for fun. While I was doing some, research I bumped into operational amplifiers. Unfortunately, no matter how extensively I looked at different resources I could never find the answers to my problems, so I am here to consult some experts.

Let me get this straight. An op amp takes the voltage difference between two inputs and multiplies the output by a certain gain. In an ideal op ap there is no current going through the inputs meaning the voltage for the two inputs is zero. What's even more confusing is the electron flow of things because some other post mention that "The general rule is that the current flows from the power supply pins to each other and the output. That means (approximately) none flows via the input pins."

Does that mean there are no electrons flowing out of the inputs? I get that the impedance for the inputs is supposed to be infinite so no current gets through but at that point you might as well just cut off the inputs and have only the power supply pins and output if you have zero voltage across both inputs relative to ground because it seems like the inputs are doing nothing. If there are no electrons* flowing out of the inputs then how is the rest of the circuit supposed to be powered? I get that you can use a feedback loop but does that mean an op amp cannot work without a feedback loop?

Moreover, are you really comparing a voltage difference? How are you supposed to compare the difference between two voltages that are zero and amplify it? It seems to me like I'm taking an electron flow from the output and dividing it by 100,000 times and spiting it back out from the two input pins.

Looking at this image, all I see are a bunch of transistors, resistors and one capacitor.

Do you mean to tell me that the voltage difference is somehow multiplied by 100,000 after going through a bunch of resistors? In my mind it makes sense for voltages to drop after going through a resistor and even if you have a power source that can provide energy to amplify the input difference, how is it strong enough to amplify it by 100,000 times if I'm only providing like 9 volts across the power supply?

Like I said in the beginning of the post, I am a newbie and this may be a completely incorrect interpretation of how an op amp works. You don't have to answer every question I listed but I would appreciate if you could explain this to me without using too many convoluted terms.

• 1. You don't need current (electron) flow to have a voltage. Why do you think you do? 2. The voltage difference between a real op amp inputs is small but not zero. It equals the output voltage divided by its open-loop gain. Commented Aug 12 at 13:55
• well, I get that you don't need current (electron) flow to have a voltage because I guess something like a batter does not and yet still has a voltage difference. But does that mean an op amp is basically a voltmeter that can amplify the voltage difference? Commented Aug 12 at 13:58
• @Saisama what else do you expect your opamp to amplify? Commented Aug 12 at 14:01
• Most people spend their entire first class in electronics building up to the ideal opamp, and then do not get to your second picture until a later class. Since there is so much to unpack here, you're not going to get a good understanding from 1-2 paragraph answers on StackExchange. Considered picking up a used textbook and working through it? Can probably find one for next to nothing online. Commented Aug 12 at 14:05
• Re, "it makes no sense to me that the difference between the two inputs is supposed to be zero." An op-amp, by itself, does not cause the voltage between its '+' and '-' inputs to be zero. But, in many useful op-amp applications, the external circuit surrounding the op-amp feeds the the op-amp output back to the inputs in a way that drives the difference toward zero. Commented Aug 12 at 14:42

In an ideal op ap there is no current going through the inputs meaning the voltage for the two inputs is zero.

This isn't how you should reason about the op-amp.

There is very little (maybe as low as picoamps) current flowing in or out of the inputs because they have very high impedance.

The voltage between the two inputs is (very nearly) zero (if it is) because you have built a negative feedback circuit around the op-amp and it is working in its normal operation regime.

The high impedance of the inputs is not the reason for there being zero voltage difference between the inputs --- the negative feedback design is.

Does that mean there are no electrons flowing out of the inputs?

Yes. If there is zero current into the inputs, then there are zero electrons flowing out of the inputs.

Note that typically the inputs have some capacitance, so when the voltage changes a small current will flow in or out of the inputs, and therefore some electrons will flow out of or into the inputs.

I get that the impedance for the inputs is supposed to be infinite so no current gets through but at that point you might as well just cut off the inputs

No, this isn't correct. You need to connect the inputs because the input pins are connected capacitively to other components within the op-amp. The potential on the inputs can still affect the operation of the op-amp even though there is no current flowing.

If there are no electrons* flowing out of the inputs then how is the rest of the circuit supposed to be powered?

It's an amplifier. The output emits more power than the input takes in, by definition. You should never have expected the power for the op-amp (or any other amplifier) to come from the input pins.

I get that you can use a feedback loop but does that mean an op amp cannot work without a feedback loop?

Even without the feedback loop the op-amp will do its basic job of producing an output voltage that is proportional to the potential difference between its input pins (unless limited by saturation when the output gets too near the supply rails).

The feedback loop only helps to limit and control the gain so that this operation is useful for building linear amplifier and filter circuits.

How are you supposed to compare the difference between two voltages that are zero and amplify it?

The difference you are comparing is not actually zero, but it may be very small, on the order of microvolts. Since this is small enough not to noticeably change the behavior of the surrounding feedback circuit we usually approximate it as zero, even though it must actually be non-zero for the op-amp to function.

It seems to me like I'm taking an electron flow from the output and dividing it by 100,000 times and spiting it back out from the two input pins.

This makes no sense to me. The signal goes in to the input pins, not out.

Do you mean to tell me that the voltage difference is somehow multiplied by 100,000 after going through a bunch of resistors?

The signal mainly goes through the transistors, not the resistors. The resistors are there to prevent signal being shorted to the power rails.

That said, you are just going to have to learn a great deal about circuit analysis before you will be able to look at this schematic and analyze it correctly. This typically takes at least half a semester in a university course on circuit analysis.

For a start you should stop worrying about electron flow when analyzing circuits. Electron flow is important when analyzing the operation of individual diodes or transistors, but only makes things confusing when studying complete circuits. Learn to do your analysis in terms of conventional current and node voltages and your life will get much easier.

how is it strong enough to amplify it by 100,000 times if I'm only providing like 9 volts across the power supply?

Remember that the feedback loop is keeping the differential input voltage tiny. If the input voltage is 1 uV, then multiplying it by 100,000 only gives an output voltage of 100 mV, which is perfectly easy to produce from a 9 V supply.

• Thanks for clarifying the reason for 0 voltage difference between the two inputs. Most of the resources I looked at stated this fact but did not explain why but I understand now. Loved your explanation on the other questions I raised but I wanted to clarify a question you responded to because I asked "If there are no electrons* flowing out of the inputs then how is the rest of the circuit supposed to be powered?" but what I meant was if something was connected to the input pins and needed electron flow to power it, it would seem counterintuitive to restrict the flow of electrons. Commented Aug 12 at 18:55
• But you mentioned that there is signal going through the input pins and I'm inferring that signals are different from voltage and current is that correct? Commented Aug 12 at 18:59
• @Saisama remember current goes THROUGH, voltage ACROSS something. You are using the word 'powered' incorrectly, the power is from the power pins, from them the energy to amplify the signal at the input pins is drawn. You are not powering it with the input signal. The inputs are however sensitive to voltage, and voltage doesn't go THROUGH those inputs, it is just an 'electrical potential' on them. Think of them like sensors for now maybe? They sense what voltage is there and amplify the difference, no (or really very very little) current flows through the input pins, but the signal is 'read'. Commented Aug 12 at 19:26
• So, I just need to think of them as like voltmeters that take the voltage difference and amplify it? Commented Aug 12 at 19:35
• @Saisama well, almost, not quite voltmeters as they have their own characteristics, but think of them as a 'voltage reading' on each of the input pins. They have such high impedance almost nothing will flow but the reading will be there. The difference will be amplified by the opamps gain. Without negative feedback this output will be next-to-unusable as the opamps natural gain is so high, but by subtracting a percentage of the output from the input (negative feedback) we can control it and set the gain where we want. Commented Aug 12 at 19:43

The word "transistor" is a condensation of two words - transfer resistor. At its heart, a transistor is a current-controlled variable resistor. Starting with an NPN device, as the current into the base increases, the effective resistance between the collector and emitter decreases proportionally. The "constant" of proportionality is called the gain of the transistor. I put constant in quotes because if there is anything about a transistor that is not constant it's its gain. But that gain is predictable within operating ranges, which is what all of those charts on a datasheet describe.

Thus, in the most strict, pedantic sense, a transistor does not amplify anything. It attenuates. It is a variable resistor, and that's what resistors do - they limit electron flow (current). A resistor restricts current, and this has the effect of reducing voltage in certain circuit arrangements.

In the 741 schematic, the current into the base of Q14 modulates the current moving from pin 7 (V+) to pin 6 (Vout). But at no time can the total power available at pin 6 be greater than the total power available at pin 7. Even in a power opamp, where microamps of input current cause amps of output current, the device does not create energy and does not, technically, amplify the input. It attenuates the available power in a way such that output mimics the input but at a higher power level.

Do you mean to tell me that the voltage difference is somehow multiplied by 100,000 after going through a bunch of resistors?

Yes. And not just resistors, but also "transfer resistors." There are approximately 20 transistors in the 741 schematic, but at the concept level it takes only three. First transistor, 1 uA into the base causes 100 uA to flow. That 100 uA is directed to the second transistor, where it causes 10 mA to flow. That 10 mA is directed to the third transistor, where it causes 1 A to flow. To get a more linear response, low noise, low distortion, low everything, it takes a lot more parts, external feedback circuits, etc. But an open-loop gain of 100,000 is actually pretty easy.

You don't need a current to flow, to have a voltage. Cells do this, as do photovoltaic cells and charged capacitors. It's true that to obtain a voltage across certain components, you must push a current through them (or vice versa), such as a resistor, whose voltage and current obeys Ohm's law, but that's just a characteristic of those particular elements. Cells and capacitors, though, can have a voltage across them even when no current flows through them. That's not to say you can't have current through them, just that you don't need a current to have a voltage.

Voltage is just the presence of an electric field, and it's possible for an electric field to alter the electrical characteristics of other nearby elements, without a current flowing. Field effect transistors (FET) work on this principle. No current flows when you apply a voltage to a FET's gate (except momentarily, due to the presence of parasitic capacitances everywhere), and yet that voltage is able to modulate conductivity between its drain and source. This property of FETs permits you to combine it with a resistor to form a potential divider:

simulate this circuit – Schematic created using CircuitLab

The resistance (or conductivity) of the FET's channel between X and Y changes depending upon the potential difference between A and B. By altering the potential difference (voltage) between A and B, you can obtain a much larger change in potential difference between X and Y (voltage amplification), and yet no current flows into or out of the gate (except momentarily, to charge parasitic capacitances).

For a FET-based op-amp, no current flows into or out of its inputs, except leakage and capacitive charging/discharging currents, but for a BJT based model, there is some current, although it's tiny. The big op-amp schematic you showed uses BJTs, and so there is some current into/out of the inputs.

It is not correct to make a blanket claim that current or voltage is "lost" through a resistor, this is just a misunderstanding of a resistor's role in the bigger picture. Take the following resistor potential divider, for example:

simulate this circuit

On the right I increased the resistance of R2, and yet the voltage across it increased, in contradiction of that claim. Moreover, the voltage across R1 decreased, so whether voltage is "lost" or "gained" even depends on where you measure it.

You have to dispense with your existing understanding of how current and voltage work, as propounded by countless "explain like I'm 5" articles and videos on the internet, and replace it with the real-world tried-and-tested mathematical model called the "Lumped Element Model" (LEM), from which we obtain Ohm's law, Kirchhoff's Voltage Law (KVL), and Kirchhoff's Current law (KCL).

These laws (and others relating to capacitors and inductors) take away the pain of dealing with quintillions of electrons, and their motion, and distill the physics down to simple algebra in terms of voltage, current and impedance (resistance). Of course you can try to understand the underlying physics, but understanding the function of the op-amp is not a physics problem, it's an engineering problem, for which you have the LEM, and for which you never consider charges like electrons, or their motion.

To understand how a resistor resists, or a transistor amplifies, then delve into the physics, but if you want to understand the behaviour of a circuit containing resistors and transistors, then forget about the electrons, and consult the LEM.

• Thank you for responding to my question. After reading all the replies I noticed a lot of people mentioning that I should focus more on current flow to help me understand circuits and I feel like you gave a good explanation on why. But you mentioned that I should consult the LEM and I just want to clarify, do you mean the company LEM? Commented Aug 13 at 16:21
• @Saisama No, I mean the Lumped Element Model! In particular KCL, KVL and Ohm's law, which are part of the LEM as applied to electronic circuits. You'll notice in my answer these all have links to useful web pages about them. Commented Aug 13 at 17:09

Looking at this image, all I see are a bunch of transistors, resistors and one capacitor.

Do you mean to tell me that the voltage difference is somehow multiplied by 100,000 after going through a bunch of resistors?

But you just said that it was a bunch of transistors and resistors. So yes, the voltage difference is multiplied by 100k after going through a bunch of transistors and resistors. Exactly right.

Transistors are the key difference. They can amplify signals. And there are different kinds of transistors - the ones you show are bipolar junction transistors. They need a constant input current to do their work, even if that current is minuscule. As soon as the input current ceases, and subsequently the stored base charge is depleted (if any), the output current ceases as well.

Today a lot of op-amps use insulated gate field effect transistors, a.k.a. MOSFETs. The MOSFET literally has an extremely good insulator between its "input" (gate) and the conductive channel (its ends called source and drain). A MOSFET is as close to an electrometer as you can get. Its inputs do have some capacitance, so to change their potential some charge must flow while the potential changes. But once the potential has been set, no further current is flowing. In practical discrete mosfets, the input current flows almost entirely through the impurities and contamination in/on the device package and on the PCB. When the mosfet still on its wafer in the ultra-clean FAB environment, the gate will retain its charge for days, months, years or decades - depending on the details of how it was fabricated.

Perhaps the most famous example of "practically zero" input current is the mosfets in an EPROM. The only way an EPROM retains its data is via stored electrical gate charge of the MOSFETs it's built with. Any current flow will alter the charge and make the EPROM forget its contents. The best EPROMs have minuscule gate charges and it still takes a century or more for them to lose their contents when stored in a basic domestic freezer. The gate leakage currents in them are low enough that it's not a continuous current anymore, but single electrons "jumping ship" every once in a while.

Other common "zero current gate" devices are CCDs (charge-coupled devices) used in specialty cameras. The charge loss as the voltages are passed down the sensor row is measured in single electrons. A row readout will often lose zero electrons. So yes, zero current is flowing then, even though the pixel voltages are shifted across the transistors in the row.

So, precisely zero currents - with zero electron flow - are not some exotic thing. Good CCD camera sensors do just that. You can hold them in your hand and take pictures with them :) Good CMOS image sensors also lose so little charge during readout that on some pixels the readout current is not just approximately zero, but exactly zero - every electron accounted for.

In an ideal op ap there is no current going through the inputs meaning the voltage for the two inputs is zero.

An ideal op-amp not only has infinitely high open-loop gain, but also infinitely high impedance on the inputs. So no current flows into the inputs independently of whether the voltage between the inputs is zero or not.

For example, with an ideal op-amp powered from +/-15V, the following circuit gives you 15V on the output, with no current flowing anywhere. The voltmeter of course has to be ideal as well, i.e. with infinite input impedance:

simulate this circuit – Schematic created using CircuitLab

That means (approximately) [that] none [current] flows via the input pins.

That's correct. Voltage can be measured with no current flow using ideal devices. Non-ideal real-world devices can do that "in the limit". There are op-amps with so little input current that just measuring it requires a special lab setup with extreme attention to cleanliness of the dielectric surfaces, shielding from external noise/interference, etc.

In very practical terms, there are op-amps out there with input conductance lower than the mutual conductance of the input traces on the typical pc-board they would be mounted on. You could put the circuit on a "perfect" insulator such as oxide-coated wafer inside of a modern chip fab, touch it with your (dry) finger, and the minuscule amount of stuff transferred over would conduct better than the input of the op-amp :)

I get that the impedance for the inputs is supposed to be infinite so no current gets through but at that point you might as well just cut off the inputs

That is, in the limit, what electrometers are supposed to do :)

Moreover, are you really comparing a voltage difference?

No. Just amplifying it. Voltage is a potential difference by definition, so your use of the word difference is, while common, technically superfluous. An op-amp amplifies the potential difference between its inputs.

How are you supposed to [amplify] the difference between two voltages that are zero [...]?

The difference is not zero, it is infinitesimal. That's the critical difference. EE textbooks often throw around infinitesimal and zero interchangeably, but mathematically those are quite different terms! Zero is a number. An infinitesimal value is not a number, just a description of a number - a description of a process that has to be used to analyze the value is-if it was a number. Just like the words "a blue car" are not a blue car, just a description of one.

• Thanks for the clarification I appreciate you taking your time to break down my questions and provide a clear and concise answer. I will take a look at MOSPFETs as it seems to be in the realm of what I am looking for. Commented Aug 13 at 16:17

# The three basic operations

This is one of three basic operations – attenuation, following, and amplification. Their purpose is to produce a proportional replica of the (input) voltage. It is interesting to note that both attenuation and amplification are fundamentally the same operation, but they operate on different quantities: attenuation acts on the input voltage, while amplification acts on the supply voltage. Attenuation and following can be achieved without an additional power source, while amplification requires one. Let's see why.

# The attenuating amplification

Amplification means to increase the input voltage in some way proportionally. Since we cannot do this directly, we proceed in the opposite way - we take a much larger (supply) voltage and reduce it, but it still remains larger than the input.

Simply speaking, to amplify something means to make it bigger. In electronics, we cannot just make the input signal bigger, so we use a bigger power source and then make it smaller, but still bigger than the original signal.

Attenuation itself is somewhat absurd because in practice it means "throwing away" energy. In life, we never do this (reduce by discarding). For example, if it is too warm in a room during winter, we set a lower temperature on the heating device rather than opening a window (without turning off the heater); we do the same with the air conditioner in the summer. To lose weight, we do not throw away food products but simply do nott buy as much, etc. However, here we allow ourselves to "throw away" excess voltage in order to obtain the desired "amplified" voltage.

# Implementation

In electronics, attenuation and, consequently, voltage regulation is achieved using the so-called voltage divider. It consists of two elements (transistors) that, generally speaking, possess the property to resist current. By changing the “resistance” of one, the other, or both simultaneously and in opposite directions, we can regulate (amplify) the voltage.

Obviously, the voltage limits are zero (when the bottom resistance is zero) and the supply voltage (when the top resistance is zero). This is why the output voltage of a real amplifier cannot exceed the supply voltage.

# CircuitLab experiments

I will now illustrate these concepts using simple electrical circuits. For the sake of concreteness, let's assume we will vary the input voltage twice.

## Attenuator

To attenuate the input voltage Vin by a factor of two, we connect a voltage divider with a gain of 0.5 between the input source and the load (voltmeter). It consists of two equal resistors (R1 = R2) in series. Half of the input voltage is lost across R1, and the other half constitutes the output voltage Vout.

simulate this circuit – Schematic created using CircuitLab

Here we assume that the load is high resistance ("ideal" voltmeter). So, the role of resistor R2 is to complete the circuit, allowing current to flow and creating a voltage drop across resistor R1. If the load had a low resistance, R2 would not be necessary.

## Follower

Therefore, if the load has a high resistance and we remove R2, the output voltage will follow the input voltage.

simulate this circuit

## Amplifier

To create an amplifier, we need a voltage-controlled voltage source with twice the input voltage.

Conceptual circuit: In CircuitLab, we can achieve this using a behavioral voltage source.

simulate this circuit

"Resistor amplifier:" In practice, we can implement an amplifier using a voltage divider supplied by a voltage source with a sufficiently high voltage V > Vin. The voltage division ratio of the voltage divider, primarily dependent on resistance R2, is controlled by the input voltage, Vin.

simulate this circuit

Transistor amplifer: In a genuine amplifier, a transistor acts as the variable resistor R2 in the conceptual circuit above.

simulate this circuit

The values are for illustrative purposes only.

# Applications

For example, in the circuit diagram of 741 op-amp internal structure, such “voltage dividers" are Q3-Q5, Q4-Q6, Q9-Q10, Q13-Q19 and Q14-Q20 pairs. Really, they are more sophisticated (using the so-called "dynamic load"), and to understand why, you need to learn more about op-amps.

You should really just forget that op-amps have a property which multiplies the input voltage difference by very large number, which in case of ideal op-amp is infinite.

It does not help understanding ideal or practical circuits which are covered easily by Kirchhoff's current and voltage laws applied to the feedback path.

You simply have to remember that an ideal op-amp outputs any voltage which makes the inputs equal, and no current flows into any of the inputs.

You should also forget thinking in individual electrons. It does not help either. Just use standard current flow.

If the Opamp doesn't have a feedback loop, the output voltage is going to be clamped at the voltage of power supply rail, which is +9V in your diagram. If you amplify an input signal with feedback, the maximum positive and negative value of output is going to be just under the voltages of positive and negative voltage supply rails. If the op amp has 100k open loop gain, the output voltage with feedback is going to be under ± 9v in the above circuit diagram.

Regarding electron flow, very low current/low number of electrons might flow into/out of input pins of the op amp. If you are driving a load, most of the current flowing out of the Opamp at the output comes from the power supplies. Electrons are flowing into or out of the power supply rails.