# Step Response Info in Frequency Response Plot?

• posted

I was wondering, if I have a Bode Plot of the Frequency Response of an op-a= mp, if I can characterize the op-amps step response from this info, or simi= larly the op-amps slew rate?

I'm thinking yes, I think yes because I could see the bandwidth, and if I m= ultiply the slew rate time I need by 3, then take the reciprocal of that to= get a frequency, as long as that frequency falls into my bandwidth I will = be ok for a step that has that rise time.... so as long as f =3D 1/(3*rise = time) falls into my bandwidth then I'm ok. Is this true?

thanks!

PS - I'm using 3*rise time, as a rule of thumb.

• posted

if I can characterize the op-amps step response from this info, or similarly the op-amps slew rate?

multiply the slew rate time I need by 3, then take the reciprocal of that to get a frequency, as long as that frequency falls into my bandwidth I will be ok for a step that has that rise time.... so as long as f = 1/(3*rise time) falls into my bandwidth then I'm ok. Is this true?

OpAmp slew rate IS related to bandwidth, but not necessarily in a straight forward way.

Examine the equations in...

Slew rate is a function of available current to drive the compensation capacitor.

BTW, the standard rule-of-thumb is TR = 0.35/BW, accurate only for passives. ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
• posted

If you take the Bode plot (mag and phase), normalize it, and write it in th= e form of a Fourier transform (or sample it and write the complex FFT coeff= icients), then if you take the inverse Fourier or iFFT, that will give you = a pretty good indication of the impulse response and you can make some reas= onable claims about the step response from that.

I've used Excel (with the analysis datapak add-on) to do this, many times.

Also... the rule of thumb of Tr =3D 0.35/BW applies only to Gaussian system= responses (H(t) =3D Gaussian function) which, as Mr. Thompson points out i= s generaly limited to functions formed from RC or RL networks.

Tom P. near Albuquerque

• posted

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

I was under the impression the derivation just assumed a single pole, and since most, e.g., op-amps are designed with dominant pole compensation anyway, it's reasonably accurate for them?

• posted

-amp, if I can characterize the op-amps step response from this info, or si= milarly the op-amps slew rate?

multiply the slew rate time I need by 3, then take the reciprocal of that = to get a frequency, as long as that frequency falls into my bandwidth I wil= l be ok for a step that has that rise time.... so as long as f =3D 1/(3*ris= e time) falls into my bandwidth then I'm ok. =A0Is this true?

Frequency response and step response can be derived from one another so long as the system is linear. Slew rate limiting is a nonlinear phenomenon - things become more complicated. If you're "far enough away" from SR limiting (i.e. risetime >> slew rate * amplitude), you can just use the values from the linear calculation. If you're inside the SR limiting, that will give you the (approximate) risetime.

For a single pole, the 10-90% risetime is ~2.2x the time constant. For an amplifier with "gaussian response", the risetime is 0.35/ bandwidth. Otherwise the ratios vary depending on how crappy the transient response can be.

Hope that helps...

• posted

OpAmp models that use only a single pole don't model slew rate adequately. ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
• posted

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

Understood; they're not directly related.

Do op-amp designers have some rule-of-thumb regarding what their full-power bandwidth (effectively, slew rate) and the gain-bandwidth product is? Some full-power bandwidths are almost shockingly small compared to GBW, but of course it's usually when you're talking micropower parts.

---Joel

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No. Bandwidth without an amplitude doesn't say anything. That is where the slewrate comes in. You should always look at both the bandwidth and slewrate.

Several months ago I needed an amplifier which could output 24V p-p /

200mA at 140kHz. I found some power opamps at TI with a bandwidth of 2Mhz (IIRC). That seemed okay until I looked at the slewrate!
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Failure does not prove something is impossible, failure simply
indicates you are not using the right tools...```
• posted

It take current consumption to slew rapidly.

Distortion-free full-power bandwidth IS a slew rate limit.

In my spice model I play some math games and set some current sources so you can somewhat independently set GBW and SR. ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
• posted

if I can characterize the op-amps step response from this info, or similarly the op-amps slew rate?

multiply the slew rate time I need by 3, then take the reciprocal of that to get a frequency, as long as that frequency falls into my bandwidth I will be ok for a step that has that rise time.... so as long as f = 1/(3*rise time) falls into my bandwidth then I'm ok. Is this true?

In theory yes, in practice, sort of. As long as the amplifier stays in its linear range, its impulse response is the inverse Fourier transform of its closed-loop transfer function, and its step response is the integral of that.

However, hitting the amp with a step of more than a few tens of millivolts will make it go nonlinear during slew, and there will be some additional transient as it recovers linearity at the end.

What's more, any little whoop-de-doos in the open loop transfer function will cause settling artifacts. That's why the settling behaviour of a composite amp is usually so disappointing.

So if you need to know the rise time, a first-order guess is that it's the output step height divided by the slew rate. For small signals, it's roughly 0.37/f_3dB for 90 degrees phase margin. For underdamped circuits, the rise time will be a bit shorter than this but the settling time will be longer.

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal```
• posted

if I can characterize the op-amps step response from this info, or similarly the op-amps slew rate?

multiply the slew rate time I need by 3, then take the reciprocal of that to get a frequency, as long as that frequency falls into my bandwidth I will be ok for a step that has that rise time.... so as long as f = 1/(3*rise time) falls into my bandwidth then I'm ok. Is this true?

Sorry! Slew rate is NOT dependent on bandwidth. Bode plot of frequency response can give one an idea to SMALL SIGNAL step response and vice-versa. Slew rate is a *LARGE* signal response - usually completely different.

• posted

They're not identical, but they're closely related for a given circuit topology.

The ratio of slew rate to bandwidth depends on the frequency compensation and the input stage transconductance. Interestingly, a lower transconductance means you can have higher slew rate for a given bandwidth and phase margin.

The classic National Semi AN-A (James E Solomon, IEEE J. Solid State Circuits, V SC-9, #6 (Dec 1974,

shows that a unity gain stable bipolar op amp with a plain-Jane differential pair input stage has a maximum slew rate that depends only on temperature and closed-loop bandwidth. It comes out to

SR

• posted

...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
• posted

Fair enough, the paper was limited to pole splitting compensation as in a 741 (possibly the last recorded useful application of root locus theory).

The claim that GBW and SR are unrelated is not true, which is the point I was trying to make.

And all those papers came out when I was a kid. People have started reinventing and republishing some of my stuff too, which I think is not a bad thing in general--there has to be some antidote to the "originality is everything" craze, or we'll never be able to teach future generations anything.

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal```
• posted

Yep. My MC1530 design utilized a "hard" pole followed by a zero. Worked so well it's still being made... 47 years later ;-)

As I pointed out before, the math in...

demonstrates the bounds.

I see all kinds of things, published as original, that I did eons ago.

That doesn't bother me. What bothers me is the crap that gets printed, for example in "Ideas for Design".

I guess the magazines are hard up for filler ?:-) ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
• posted

).

t

I designed some of those DC accurate filters (the Fluke scheme), and used a root solver to perform root locus to find the optimal solution. I will admit that was the last time I ever did root locus.

• posted

I liked those a lot, and used them in several designs--all one-offs, since I was working in a research lab at the time. I also used one of the five-pole universal SC filters to make the baseband phase shifters for SSB mixer to use with a heterodyne interferometer. It had 60 dB sideband rejection on the first try--magic.

Because of all that switching noise, the dynamic range wasn't that great, but with a heterodyne system you don't care as much, since the dynamic range is only the square root of the signal power range. (The detected signal goes like Re{(E_LO) (E_sig*)} instead of |E_sig|**2--it helps a lot.)

Cheers

Phil Hobbs

```--
Dr Philip C D Hobbs
Principal```

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