Why does op-amps gain decrease at high frequency

In summary: Specifically, an ideal opamp is one in which the transfer function is an infinitely high-pass type. That is, the output voltage remains constant as the input frequency is increased. However, this is an impossible requirement to achieve in practice, as the open-loop gain of an opamp decreases with frequency. Nevertheless, the opamp's internal circuitry attempts to maintain the close-loop gain as best as possible.In summary, the opamp's close-loop gain decreases as frequency is increased, causing the output voltage to drop off.
  • #1
PianoDentist
7
0
Ive read on loads on places that tell me this but don't explain why.

They explain the closed loop gain drops off because the open loop gain in their 'real' equations is no longer infinite/very high so the assumpstions that lead to ideal equations fail.

but nowhere can I find an explanation of exactly why as frequency increase gain would decrease
 
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  • #2
At high frequency and high gain the output has to change voltage very quickly - this means very large internal currents and is ultiamtely also limited by the capacitance and inductance of the internal connections
 
  • #3
Opamp maintain the close loop gain because it has high open loop gain at low frequency. The differential voltage across the two input needed to get the correct voltage is very small. Therefore the close loop gain is accurate and maintain steady gain. But as you increase the input frequency, the open loop gain decrease, you need more input differential voltage to produce the same output, this will cause error and the gain start to roll off.
 
  • #4
it has somthing to do with "slew rate "...its analogy in physics is velocity...juz as there's a limit to velocity. (velocity of light)...so there is with op-amp...slew rate is the rate of change of o/p w.r.t. time ...
Slew rate of an op-amp is constant...that means the rate of change of o/p is constant ...now increasing the freguency of i/p signal implies we are increasing the rate of change if o/p...which is in turn constant...so on increasing freq. gain decreases... :)
 
  • #5
Slew rate and frequency response are totally different thing. Slew rate is the speed it step from one voltage to another usually specified in 1V step. An op-amp can have very high frequency response( GBW) but slower slew rate. This is particular prominent at large output swing where it just cannot swing fast enough. But if you lower the amplitude, you can see the -3dB frequency is much higher.

It is very common to have slew rate limited. You can easily see if you sweep the input frequency with sine wave. Adjust the output to be...say two to three volts or higher. When you look at the output while increase the frequency, the sine wave at the output all of a sudden become triangular wave without any roll off in amplitude. That is slew rate limited.
 
  • #6
You can think of an opamp as a low pass filter.
All opamps have a limit on upper frequency.

In a LPF, At low frequencies, the output amplitude is equal to input. But as the frequency increases, the capacitive reactance decreases and the output amplitude starts to decrease.
 
  • #7
Actually I asked this question my professor not so long ago.

You see op amps can be internally frequency compensated and not internally frequency compensated.

The one that ARE internally compensated have a capacitor inside. Now why they are compensated, and what are they compensating for?

Normally op amps go beyond 1MHz frequency, but after this, they are not very stable. Meaning, it can go into oscillation very easily. This can lead into a thing called "ringing". You might want to google that.

But if you put that small capacitor inside, you cut off frequencies over 1MHz, effectively. Which makes it a low pass filter in a sense.

But even if you don't have that capacitor, your frequencies will go down after a certain frequency, because of parasitic capacities. [between two transistors, diodes, etc. you get the idea :D]

All in all, frequencies go down because of capacitors, capacitors are low-pass filters, because their reactance:
[itex] X_c =\frac{1}{\omega C} [/itex]
[itex] \omega -> \infty[/itex] [itex] X_c -> 0 [/itex].
 
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  • #8
There are so many things to look at the opamp, that's the reason they have so so many of them AND the data sheet is so long on every one of them. They don't print it for the fun of it. Best way to learn opamp after study books is to get a few data sheet and read through it. Look at the specific condition where they obtain the data. Books can only teach so much, only the basic in how to set the gain. But the data sheet is where the rubber hit's the road.
 
  • #9
This is a question that's never overly easy to answer.

The simplistic, overly simplistic answer, is to point out that nothing occurs instantly. If you turn on a lamp, the filament takes time to heat before it radiates light. If you step on the accelerator of a car, it takes time for additional fuel and air to increase the revs of the engine. Likewise, if you change the input to a transistor, it will take time to respond.

The next ingredient to explaining your problem is to explain how to make what's termed a phase shift oscillator.

Imagine if you will, that you have a number of transistors, each one's output going into the next one's input. There's two things you might expect to happen:

1. When you apply an input, it takes longer for it to appear on the output as you increase the number of transistors, because each adds to the delay.
2. Depending on how you connect them, as you continue to add transistors, the output will become more sensitive to the input. You'll have more gain.

If you combine ingredients from 1 and 2 and input a sine wave, it's possible that you can find a frequency where the output has not only amplified the sine wave, but due to the delays, it's also delayed the sine wave by 180 degrees.

If your amplifier is inverting, then it already shifts the phase by 180 degrees. Tie it's output to the inverting input, and as long as the gain is greater than one at the magic frequency, you'll get oscillation.

The simplest op amps handle this by deliberately making one transistor stage much much slower than the others. Usually, this stage works as a simple filter and starts decreasing the gain at a few tens of Hz.

At most, this single stage cannot shift the phase more than 90 degrees, so if it starts working sufficiently low in frequency, the op amp's open loop gain will be less than 1 for higher frequencies, where the other transistor stages begin to cause additional phase shift.

I hope this helps,

Best Wishes,

Mike
 
  • #10
first of all, all transistors and devices like that crap out when the frequency gets high enough. the electrons and the holes can slosh back and forth up to a certain rate and no faster. for a transistor, injecting current into the base causes an effect with the amplified current at a slightly delayed time, just because of the finite nature of the physics. this is also why computers have finite clock speeds.

because op-amps are "monolithic", there are capacitances and leakage throughout the circuit that is burned onto the chip. every part is connected to the "substrate" via a reversed-biased diode (that is supposed to be "off" or non-conducting). at low frequencies these leakages are not so bad, but at high frequencies, these same leakages of current affect the operation of the circuit. capacitors at high frequencies become short circuits, so at high enough of a frequency every internal component (the little transistors and such) is shorted out to every other component. that's when the op-amp is good for nothing.
 

1. Why does the gain of op-amps decrease at high frequency?

The gain of op-amps decreases at high frequency due to the presence of parasitic capacitances in the circuit. These capacitances, which exist between different components and within the op-amp itself, act as low pass filters and cause the gain to decrease at higher frequencies.

2. How do parasitic capacitances affect the gain of op-amps?

Parasitic capacitances cause the op-amp to behave like a low pass filter, allowing low frequencies to pass through with minimal attenuation but attenuating high frequencies. This results in a decrease in gain at high frequencies.

3. Can the decrease in gain at high frequency be compensated for?

Yes, the decrease in gain at high frequency can be compensated for by using external components such as capacitors or resistors to create a frequency-dependent feedback network. This allows the op-amp to maintain a constant gain over a wider frequency range.

4. Are there any other factors that can contribute to the decrease in gain at high frequency?

Aside from parasitic capacitances, other factors such as the internal frequency response of the op-amp and the load connected to the output can also affect the gain at high frequency. These factors should also be taken into consideration when designing a circuit.

5. How can the decrease in gain at high frequency impact the performance of a circuit?

The decrease in gain at high frequency can limit the bandwidth of a circuit and affect its ability to accurately amplify high frequency signals. This can result in distorted or attenuated output signals, impacting the overall performance of the circuit.

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