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I am bit confused about the cut off frequency and the max frequency above which the output response starts to distort which is slew rate frequency. Are these two things same?

Cutoff frequency (f) = GBP / Av

Slew Rate Frequency (f) = SR/(2*pi*Vp)

Is cutoff frequency = slew rate frequency?

Thanks

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They are not at all the same thing. Assuming we are talking about an op-amp amplifier, the cutoff frequency is the point where the small-signal gain is down by 3dB.

The slew rate frequency is a non-linear large signal phenomenon. It has to do with how fast the output can change in response to a large input signal. It's due to the finite amount of current available to charge/discharge the dominant pole compensation capacitance.

Slew rate limitations will make a sine wave with high enough amplitude and frequency distort to start to look like a triangle wave.

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  • \$\begingroup\$ So does this means we will always have its (cut off) value around 20 KHz, when 3db is standard frequency ? \$\endgroup\$ – Dhiraj Dhakal Feb 10 '17 at 2:20
  • \$\begingroup\$ Please forgive me, the cut off frequency value would be around 1.41Hz when converted from 3db, and i want to ask do we always set cutoff frequency = 1.4hz while solving the problem? \$\endgroup\$ – Dhiraj Dhakal Feb 10 '17 at 3:00
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There is a third cause of distortion. And it also is frequency sensitive.

The Input DiffPair also affects distortion. Bipolar diffpairs, without emitter resistors to linearize, have predictable IP2 and IP3. These numbers, somewhat different for IP2 versus IP3, are near 0.1voltsPP across the Pin+ to Pin-.

That means 0.1volt at 1KHz would produce 0.1volt at 2KHz. This is IP2. Dropping down by 10:1, to 0.01volt across Pin+ to Pin-, would produce 0.001 volt at 2KHz. In dB, the 2nd order distortion would drop dB_for_dB.

For the 3rd order, given 1KHz and 1,100Hz of level 0.1 volt each, the behavior of 2*F1 +- 1*F2, and 1*F1 +- 2*F2 will produce "shoulders" at 900Hz and 1,200Hz of level 0.1volt also. Dropping the level of 1Khz and 1,100Hz by 10:1, thus 0.01vpp for each, will produce "shoulders" at 900 and 1,200 at 0.0001vpp. In db, the 3rd order distortion would drop 2dB_for_db.

Why is this important? At low frequencies, the huge open loop gain is your friend, and causes voltage of Pin- to Pin+ to be tiny, implementing "virtual Ground". At high frequencies, the dropoff in open loop gain causes the voltage of Pin- to Pin+ to grow 10:1 as the frequency increase 10:1. And large voltages between input pins is what causes the diffpair (bipolars or fets) to distort.

Example: opamp UGBW is 100MHz, your circuit is unity-gain, your frequency is 10MHz, and your input voltage is 1voltPP. The gain at 10MHz is only 10, thus the input error (how much the virtual Ground differs from Zero Volts) is Vout/Gain = 1vpp/10x = 0.1voltpp. Thus the distortion behavior we predicted in our first paragraphs is our distortion situation here.

And as I cautioned, IP2 level is not same as the IP3 level, but they are only a few dB (4:1?) apart.

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Short answer: Primarily, the slew rate plays a role in high-gain amplifiers (opamps) with feedback. The amplifying capability of such amplifiers has frequency limitations which are characterized by two different bandwidth definitions:

(1) Small-signal bandwidth - determined by the small-signal open-loop gain response (gain-bandwidth product)

(2) Large-signal bandwidth - determined by the slewing capabilities of the amplifier with feedback: B=SR/(2*Pi*Vmax). Within this bandwidth sinusoidal input signals will be amplified without any remarkable distortions for amplitudes not larger than Vmax.. Larger frequencies will cause gain reduction, triangular distortions and additional phase shifts.

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Is cutoff frequency = slew rate frequency?

Just look at the formula for the slew rate frequency: -

Slew Rate Frequency = \$\dfrac{S.R.}{2\pi\cdot Vp}\$

It contains an amplitude value Vp. This means it is dependent on the amplitude of the output signal. The cut-off frequency is not dependent on anything other than how the device is designed/built.

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  • \$\begingroup\$ OK i have one problem where cutoff frequency has to be made equal to the slew rate frequency in order to get the answer, can you explain me why and if possible solve it? Ex. If an opamp is wired to yield a voltage gain of 100 whose Unity Gain Bandwidth is 2 MHz and Slew Rate 1.3V/μs, and i/p signal is sinusoidal. Find the max. i/p signal that can be applied that produces distortioless o/p. [Answer Vip=0.127V] \$\endgroup\$ – Dhiraj Dhakal Feb 10 '17 at 2:22
  • \$\begingroup\$ Equate both formulas and solve for Vp. \$\endgroup\$ – Andy aka Feb 10 '17 at 10:08
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They are different. For example, if an op amp is outputting a small waveform within the gain-bandwidth requirements, all is well. Then, you try to output a bigger signal that exceeds the slew rate, and the output gets very distorted.

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