Trying to understand how operational amplifiers work

1. Nov 19, 2011

Femme_physics

Last edited by a moderator: May 5, 2017
2. Nov 19, 2011

yungman

The model looks reasonable except the other side of $Gv_{in}\;$ don't come out. It is only generalizing that no current flow in or out the input. It is not exactly like $R_{in}\;$ in your drawing. It is an input stage of either a BJT or FETs. The input current range from 1uA down to 0.01pA depending of the technology.

The definition of op-amp is it has an inverted and a non inverted input. The amp has very high gain of whatever input is presented ACROSS the +ve and -ve input ( called differential input). Therefore in a closed loop feedback where the output is connected back to the -ve input with a resistor R2 in your case. any voltage at the +ve input will cause a differential voltage across the two inputs. The amp has infinite gain, so the output react immediately and swing to the direction of the +ve input. It will swing until the voltage at the -ve input equal back to the +ve input, then the output stop rising.

Let's use your drawing, the voltage at the input is the voltage divider voltage of the output by R1 and R2. so if R2 is twice the value of R1, then the voltage at the -ve input is 1/3 of the output voltage according to the divider, so to every 1v input, you get 3v output in the same direction. So your amp is a non inverting configuration with a gain of 3.

Hope this help.

Alan

3. Nov 19, 2011

I like Serena

Hey! Did you trace the links of my pictures back to their origin (which is wikipedia)?

Yep! An op-amp is a complex piece where the current is drawn from the additional power supplies.
In your picture the current is drawn from Vs+ and Vs-.
This is done in such a smart way that V+ and V- do not draw a noticeable current.

Last edited: Nov 20, 2011
4. Nov 19, 2011

waht

An ideal op-amp is defined to have:

1) infinite input resistance: Rin -> infinity
2) zero output resistance: Ro -> 0
3) infinite gain: G -> infinity

$$V_{in} = V_{+} - V_{-}$$ And $$V_{out} = G\cdot V_{in}$$
This is the starting place. Every op-amp circuit can solved by applying these conditions.

Of course, in reality nothing is ideal. A real op-amp integrated circuit is a non-ideal approximation of an ideal op-amp. As an example, it may have parameters like Rin = 100K, Rout = 4 or G = 20,000.

But solving circuits such as the non-inverting amp above, it kind of assumes that you have to use the ideal op-amp model, unless it's stated otherwise and you are given op-amp parameters.

So what's really going with the non-inverting op-amp circuit, and why has the ideal op-amp been defined the way it is?

The answer to that comes from control theory which is a huge subject that deals with feedback systems. In such a system, some of the output is fed back to the input of some element with defined input and output.

In this case, the defined element is the op-amp. You can see, the voltage divider formed by R1 and R2 samples the output voltage and feeds a fraction of it to the input terminal V_.

From a voltage divider set up,
$$V_{-} = V_{out} \cdot \frac{R_1}{R_1+R_2}$$ And $$V_{+} = V_{in}$$
As to not confuse those variables used in the schematics above that use V_in in two different places for different things.

Let's call
$$V_{d} = V_{+} - V_{-}$$
$$V_{out} = G\cdot V_{d}$$

Having established this, you can follow this calculations to obtain the gain of the amplifier which is defined as

$$A = \frac{v_{out}}{v_{in}}$$ and it can be seen that as you take the limit of the op-amp gain G to infinity things nicely cancel and simplify where A converges to a finite value.

This is the long way of solving the problem, but it gives insight into what really happens. There are also shortcuts you can do that are derived from the op-amp definitions and control theory. From condition 1, infinite input resistance will block input currents so that I_ = 0 and I+ = 0. And from control theory under negative feedback conditions, the output will try to cancel the input, so that v_d = 0.

With
$$V_d = 0$$
It follows that $$V_{+} = V_{-}$$
And $$V_{in}=V_{+}=V_{-}$$
By applying KCL, Kirchhoff's Current Law at V_:
$$\frac{V_{in}}{R_1} = \frac{V_{out} - V_{in}}{R_2} + I_{-}$$
With I_ = 0, if you carry this out, the gain comes out to:
$$A = 1+\frac{R_2}{R_1}$$
The shortcut doesn't always work with op-amp circuits, so you have to be careful. But just in case, long way will always work.

Last edited: Nov 19, 2011
5. Nov 19, 2011

vk6kro

Here is a diagram of the insides of an opamp. This one is a fairly old type and there are much better ones around now.

The complex arrangement of transistors is typical of opamps, although the input devices (Q1 and Q2 in this case) could be FETs in some other types of opamps.

Fortunately, you don't have to know the exact internal circuit of each opamp as the external circuits used compensate for differences in the internal design.

Notice that this opamp has an internal capacitor of 30 pF. Capacitors inside integrated circuits are expensive to produce so you won't see this very often.

6. Nov 20, 2011

technician

3 essential things to know to sort out op amps
1) V at the - input = V at the + input ( this is a good assumption because the gain of an op amp is so high)
2) No current flows into or out of the +input or -input (this is a good assumption because the input resistance is so high)
3) The output voltage cannot be greater than the +supply voltage or less than the -supply voltage
In your circuit Vin connected to the +input must also be the voltage at the -input
This means that there is a voltage = Vin across R1
This means that a current is flowing through R1 and this current must come from Vout of the amplifier.
This current flows from the output through (R1+R2) in series.
So Vout = I x(R1 + R2) and Vin = I x R1
The voltage gain is Vout/Vin = (R1+R2)/R1
You see how this is subtly different to the gain of an inverting amplifier (-R2/R1)

7. Nov 20, 2011

Femme_physics

So the inner workings of an op-amp is beyond the scope of understanding to a beginning electronics student who just started with op-amps? I appreciate the replies, everyone. It does seem too complex.

8. Nov 20, 2011

technician

Don't be discouraged!!
You do not need to know all the details of the inner workings of an op amp to be able to use them and design circuits using them.
The applications you have met recently can be sorted out in a fairly straight forward way

9. Nov 22, 2011

jim hardy

find this old document from Texas Instruments

"Op Amps for Everyone"

it has a document number starting with SLOD
yu should be able to find it from google.
it is a good text. I think Bassalisk found it mqybe he still has link

opamps sound crazy at first

the secret is to realize this:::
because opamps have such preposterous parameters (gain of millions, input resistance of giga-ohms) it is incumbent on designer to surround them with a circuit that keeps them operating within limits of power supply.

so you as designer are in control.
Opamp is just a triangle with Vout = 10^6 X (Vin+ - Vin-)

around 1967 i had a difficult time transitioning from tubes, where gain of 40 might be good, to Fairchild ua709 opamp with its gain of a hundred thousand.....

10. Dec 3, 2011

Femme_physics

I know I am a little late, but I shall take your advice into considerations now

Yes sir!
.

What smart way? What additional power supplies?

I don't get where does this 10^6 comes from. It appears like massive energy made out of the blue, and we all know that's impossible.

Fair enough, it's complex. But nowhere in all this do I see how it pumps up energy out of the blue.

11. Dec 3, 2011

technician

FP
We get hung up on all the details of what an op amp does and how it does it.
Don't forget the energy/power ultimately comes from the power supply. It does not come from the op amp....this controls how the energy is delivered.
As you have seen we are so concerned about the details that most of the time the power supplies are not even shown on the circuit diagram. We take it for granted.
Think about a car, the engine and controls are complicated but the energy comes from the fuel....hope that analogy does not sound too trivial.

12. Dec 3, 2011

Femme_physics

Fair enough, it doesn't really matter where the energy comes from. It's the amplification process that forever befuddles me. How can you amplify a given source of energy? Going with the fuel example, it's like putting fuel in a machine that eventually through a complicated process creates more fuel than was originally put , just by having gone through devices, without actually adding extra fuel.

You see what I mean?

13. Dec 3, 2011

technician

I would say that the energy of the fuel is converted to a different form of energy but not more energy.
In a similar way an op amp enables the voltage of a battery and the charge contained in the battery to be manipulated to produce what you want.
Small electrical signals can be used to control large signals. That is what amplification means

14. Dec 3, 2011

pantaz

But you are adding extra fuel. It's just coming via a different path.

Amplifiers still obey Ohm's Law. A basic NPN transistor amp is a good way to envision it -- a small signal into the base picks up additional power from the collector, thus appearing amplified at the emitter. It's a similar mechanism with any amplifier.

15. Dec 3, 2011

I like Serena

Vs+ and Vs-.

Your Vin signal itself is not amplified.
It only opens a kind of gate for the large current to flow, drawn from Vs+ and Vs-.

It only takes a little energy to open the fuel valve, after which a lot of fuel can start flowing.

16. Dec 3, 2011

Femme_physics

So it doesn't really amplify the current? And can you give an example of "Small electrical signals can be used to control large signals", please?

Ahhhhhhhhhhhhhh........so Vs+ and Vs- are the REAL big current, whereas Vin is just the gate current, like "base current" of transistors, yes?

Ah. But now, why do you even need this analogous electrical valve? I thought amplifiers are meant to amplify current, not to CONTROL it. That's what resistors and switches are for.

17. Dec 3, 2011

I like Serena

Yep!

Huh?
An amplifier does control current (like you said, you can't create energy out of nothing).
Just like a switch (but an advanced switch that is controlled electrically).

18. Dec 3, 2011

sophiecentaur

Not many electronic devices work without an external Power Supply!
An amplifier 'amplifies' the low power of an input signal by controlling power from the power supply to produce a vastly increased version of the signal power. The analogy with a valve or a tap is a good one. The crudest way of controlling power is an ON/OFF switch. It is, in effect, a very simple non-linear power amplifier. The power from your finger is enough to control the multi kW power feeding a massive heating system. Also, the low power involved in moving the slider on a rheostat (big variable resistor) can control, in a linear fashion, the high power delivered to a motor; this is also a form of amplification. An electronic amplifier is achieving the same sort of thing only in a more sophisticated way.
With an OP Amp, the Power Gain is so huge that you can equate the input power to zero for some purposes. I know that sounds daft but, in comparison with the output power, it is negligible during calculations. You always 'tame' this enormous gain using Negative Feedback. At the negative input to the amplifiers in your diagrams, the current flowing 'in' through the series input resistor is almost exactly the same as the current flowing 'out' through the feedback resistor. A miniscule current flows into the Op Amp input and it can be treated as virtually zero. (Electronics is Engineering - in which assumptions like that are constantly being made, in one way or another)

The overall performance of the Amplifier is defined by that feedback and depends very little on the Op Amp characteristics (as long as the device gain is high enough at the highest operating frequency).

19. Dec 3, 2011

technician

FP
If you take an audio system as an example of amplification the input could be from a microphone or a record player pickup. These are usually very low power and only provide small voltages (mV) and small currents (μA or mA).
We want to reproduce this small signal through a loudspeaker and require POWER of maybe 10Watts. This could be achieved by a voltage of 10V and a current of 1A (or some other combination!) through the loudspeaker.
This would be called power amplification and needs amplification of voltage and current.
Amplifiers are usually designed in 2 sections: A Voltage amplifier followed by a current amplifier to give the required power output. The current amplification tends to be fairly simple ...a single transistor could be enough.
The voltage amplification is a little more involved because signal processing is usually incorporate in the voltage amplification.
Tone controls, volume control etc.
If you start with maybe a small signal of 10mV from a microphone and you want an output of 10V to drive the loudspeaker you can see that a voltage gain of 1000 is needed.
This is the area we are in with op amps.
I hope this gives you some background that lies behind the implementation of amplification!!!
The tiny input power controls the large output power which comes from the power supply.

20. Dec 3, 2011

Femme_physics

So why is it called an amplifier and not a controller?

In fact, I'd go so far as to say an amplifier doesn't really "AMPLIFY" the original current but rather REDUCE to control it. It should be called a regulating-reducer if anything!

Is that because it stores current and then relaunches it, or something like that? Like a capacitor with a regulator?

Again, flow "REGULATION", not flow "amplification".

So a very small current flows FROM to mike to the loudspeaker, and suddenly turns into a big current due to the op-amp? But how? But we go back to the same point-- if op-amps are like valve, how can it increase the current?