Op-Amp Saturation: Why Use an Op-Amp with 15V Supply?

[╔(σ_σ)╝]

I have been studying op-amps in my 2nd year circuits class and I had a question to ask.

From what I understand the output voltage is supposed to be < power supply voltage.

If this is the case, why do we use op-amps ?

I mean if I have 15V and I need to power something why can't I just use the 15V instead of using an op-amp only to realize 13V as my output voltage ?

For example if I have an ideal inverting opamp powered by 15V, -15V dc my output voltage is approximately

$$V_{o}$$= -($$R_{f}$$/$$R_{1}$$)* $$V_{i}$$

Where Rf is my feedback resistance, Vi is my input voltage connected to my non inverting terminal and R1 is my resistor connected to my inverting terminal.

Why can I just put Vi = 30 V ( supposing Rf > R1) and get |Vo| > 30V ? I know it would "saturate" theoretically but what is wrong with that ?

When we saturate the circuit does this mean that Voutput= Vpowersupply or does that it mean I can get more voltage but my op-amp would not be working in a "safe" mode ?


In summary I thought and op-amp would amplify voltage but it doesn't seem to be true. I mean it take a input voltage and then spits out a bigger value which is limited by the power supply. I thought all circuits would function properly as long as the power balance equation is satisfied ?

If I have 15V to waste by should I plug it into an op-amp instead of using it directly on my device ?

 

[Bob S]

Hi Pi-

All an op-amp does is very accurately amplify the difference of V_pos and V_neg by the factor A, where A is maybe 10,000 or so. Furthermore, the inputs have an infinitely high impedance and no input bias current.

V_out = A(V_pos- V_neg)

If you had a 15 volt supply and wanted say a 10 volt source output with a very low source impedance, you could either use 3 0ne-ohm resistors , or use 3 1k resistors and an op-amp. Which would draw less current?

[added] Saturation occurs when either the op-amp inputs or output are within a volt of either power supply voltage (or less depending on the op-amp).
Bob S

 

[KingNothing]

You seem to be misunderstanding the use of op-amps. Op-amps are used to multiply the amplitude of the input voltage. They are not intended to provide a constant DC output - like you've said, that would be useless.

For example, say I am singing into a microphone. That microphone produces a sine wave of 20 millivolts peak-to-peak. Obviously 20mVpp cannot power a speaker. To power a speaker, you might use an op-amp to multiply the 20mVpp signal into a 10 or 20Vpp signal, which can power the speakers.

In review, Op-Amps are used to multiply signals, not provide DC voltage.

 

[Bob S]

KingNothing-
As long as you follow the rules of an op-amp, like I stated above, you can amplify, not multiply, an input signal up to the voltage limits I stated above, within the gain-bandwidth product of the op-amp.
Sketch me an op-amp circuit that can multiply two input signals V₁ and V₂.
Bob S

 

[╔(σ_σ)╝]

Hi Pi-

All an op-amp does is very accurately amplify the difference of V_pos and V_neg by the factor A, where A is maybe 10,000 or so. Furthermore, the inputs have an infinitely high impedance and no input bias current.

V_out = A(V_pos- V_neg)

If you had a 15 volt supply and wanted say a 10 volt source output with a very low source impedance, you could either use 3 0ne-ohm resistors , or use 3 1k resistors and an op-amp. Which would draw less current?

[added] Saturation occurs when either the op-amp inputs or output are within a volt of either power supply voltage (or less depending on the op-amp).
Bob S

Okay I understand what you are trying to say. I think I have a better understanding now.


My second question was, what happens when the circuit is saturated ? Does it still work ? Why is saturation "bad" ?

Btw as per your questions the 3 1k ohm resistors would draw less current.


KingNothing

I do not recall anywhere in my post where I stated that " op-amps are used to provide a constant dc voltage".

 

[KingNothing]

KingNothing-
Sketch me an op-amp circuit that can multiply two input signals V₁ and V₂

Why? No one ever said anything about multiplying two input together. The difference is multiplied by a gain.

 

[Darryl]

Op amps are a very very powerfull tool for electronics engineering. They can be applied to all sorts of great applications.

From memory, an idea op amp has infinate gain, zero input impedance and infinate output impedance and infinate Ft.

The Key to op-amp design is in the feedback loop, and input circuit.

Is a saturated Op amp still working correctly, yes it is.

If you design the input circuit and feedback circuit correctly, you can use an op-amp to turn a small noisy pulse into a clean large pulse, in that case the op-amp will turn on with it sees the small pulse, and it's output will swing to near it's rail voltage for the duration of the input pulse, producing a clean output pulse at the size of the saturated output of the op-amp.

threshold detectors, integrators, gyrators, switches, amplifiers, buffers all manner of mathematical functions.

Practically: your biggest problem with op-amp design due to their high gain, and high I/P Z, low O/P Z they will oscillate at the drop of a hat. It's not unusual to get several Mhz oscillations from op-amp.

Most famous op-amp is the classic 741.

 

[vk6kro]

My second question was, what happens when the circuit is saturated ? Does it still work ? Why is saturation "bad" ?

Saturation is only bad if it results in distorted output.
If you have a 2 V peak to peak input signal and a gain of 10 the output will be distorted unless the opamp can produce 20 volts peak to peak output. An opamp with + and - 12 volt supplies may be able to do this, but one with + and - 9 volts supplies will not be able to do this.

Saturation is just a normal consequence of the internal construction of the opamp. The opamp can't produce more output than its supply voltage and it usually can't output as much as the supply rails either.
Also, unless the output is exactly centred between the supply rails, one direction of swing will saturate before the other. This might result in "flat topping " of a sinewave, for example and distorted output at less than the saturation limits.
If it happens, you just need less signal, less gain or better adjustment of the DC conditions in the circuit.

I found this confusing:
I mean if I have 15V and I need to power something why can't I just use the 15V instead of using an op-amp only to realize 13V as my output voltage ?
then...
I do not recall anywhere in my post where I stated that " op-amps are used to provide a constant dc voltage".

 

[455_F4C3]

I understand what ╔(σ_σ)╝ is trying to say, and I have the same question a long time ago, but could never figure it out myself. If anyone here want to enlighten me, and hopefully help our friend ╔(σ_σ)╝, I would be most grateful.

Here is what I never understood:

If op-amp is use to amplify a circuit, and its output is limited by its Vcc+ and Vcc-, why do we even need it?

For example, Vcc+ and Vcc- is +9V and -9V, and the voltage source provides 20V, why bother with an op-amp at all, when you can just power the load with a constant DC from the voltage source... that is what I am confuse about.

Please forgive me, if this seem like a REALLY STUPID question. lol.

Thank you.

 

[vk6kro]

No, it isn't a stupid question.

The point is that opamps are used in many ways, but one of them is to take an input and make the output exactly, say, 5 times bigger. The input may be varying and we want the output to vary as well, but be 5 times bigger.

An example is when an opamp is used as a microphone amplifier.

The output of some microphones is very small. It can be 0.01 volts AC. If you wanted to have a signal big enough to drive the line input on a computer (which needs about 1 volt AC) then you need to amplify the microphone signal.

But you don't want to just hope it will be big enough. You want a gain of 100 so that the output will be 1 volt AC. (100 times 0.01 volt = 1 volt)

An opamp let's you do this and you could make lots of them and they would all have a gain of 100. Opamps give you the precision you need to make results predictable.

Notice that the output of the opamp will never be more than 1 volt AC in this example and that it can never be driven to saturation.

 

[mahdiye88]

Hi Pi-

All an op-amp does is very accurately amplify the difference of V_pos and V_neg by the factor A, where A is maybe 10,000 or so. Furthermore, the inputs have an infinitely high impedance and no input bias current.

V_out = A(V_pos- V_neg)

If you had a 15 volt supply and wanted say a 10 volt source output with a very low source impedance, you could either use 3 0ne-ohm resistors , or use 3 1k resistors and an op-amp. Which would draw less current?

[added] Saturation occurs when either the op-amp inputs or output are within a volt of either power supply voltage (or less depending on the op-amp).
Bob S

i want to know saturation level just depend on internal circuits of the op-amp?can't we say anything?cant we generalize ?

 

[uWave_Matt]

Here is what I never understood:

If op-amp is use to amplify a circuit, and its output is limited by its Vcc+ and Vcc-, why do we even need it?

For example, Vcc+ and Vcc- is +9V and -9V, and the voltage source provides 20V, why bother with an op-amp at all, when you can just power the load with a constant DC from the voltage source... that is what I am confuse about.

Thank you.

I think the confusion the OP and yourself have is in what signal is amplified. The DC voltage is used to power the op-amp, and is not the signal that is amplified. Refer to the image below:

The signal to be amplified will be an AC signal and will be fed into the op-amp through pin 2 or 3. Your Vcc +/-9V will be attached to pins 7 and 4. The amplified signal will come from pin 6 and will be the same as the signal that was input at 2 or 3, but with a greater magnitude.

In order to amplify the signal, you have to provide some form of power from pins 7 and 4. You don't get to create a bigger signal out of nothing. But if you figure out how to do that, let me know and we'll go get rich.

 

[Bob S]

I wonder if you are really looking for a voltage multiplier. There are many vendors that sell multipliers on the web. There are also do-it-yourself designs. See

http://en.wikipedia.org/wiki/Voltage_multiplier

Bob S

 

[Proton Soup]

I understand what ╔(σ_σ)╝ is trying to say, and I have the same question a long time ago, but could never figure it out myself. If anyone here want to enlighten me, and hopefully help our friend ╔(σ_σ)╝, I would be most grateful.

Here is what I never understood:

If op-amp is use to amplify a circuit, and its output is limited by its Vcc+ and Vcc-, why do we even need it?

For example, Vcc+ and Vcc- is +9V and -9V, and the voltage source provides 20V, why bother with an op-amp at all, when you can just power the load with a constant DC from the voltage source... that is what I am confuse about.

Please forgive me, if this seem like a REALLY STUPID question. lol.

Thank you.

you know, I'm not even sure i'd want to think of an op-amp as a general purpose amplifier. sure, for very small signals with modest amplification requirements, it's true. but not generally as a power amplifier like for your stereo.

the real magic of op-amps is that they do a wonderful job of analog signal processing. and the first thing you want to get right is to not change the signal you're examining. if you load the signal with your input circuitry, you change it. op-amps overcome this by having a very high input impedance. next, they are fast and accurate, adding very little noise to the processed signal. and if you've ever taken an electronics course where you've designed multistage amplifiers using discrete transistors, you will quickly appreciate the handiness of an off-the-shelf component that does this better than you ever could do on your own.

now, once you've done some signal processing/preamplifiction with your op-amps, you may feed this into some power amplifier stage if say you want to drive speakers on a sound stage. and there you will be using discrete transistor output stages. but these will also be limited by some driving voltage, even though it's higher than +/-9V. that's just life, you will always have a limit. and that's where design comes in, managing all your constraints.

as for the question about a constant DC voltage source... let's say the load is a pair of ear buds for listening to music from you ipod. delivering a constant DC source to that load would mean that no sound waves are generated from the speakers. the signal has to be dynamic (variable AC). and the signal has to be lower than the voltage that would damage the speaker, which makes it important to have an amp that actually does saturate - it provides protection.