Relation between slew rate and gain-bandwidth product?

Hi~

Can someone tell me the relation between slew rate and gain-bandwidth product? Though the relation is not direct, I've been thought that the more the slew rate is the higher the gain-bandwidth product is. But I found that the MP103 from Apex(now Cirrus) has 167V/usec slew rate and 250kHz gain- bandwidth product and the MP111 has 130 V/usec and 6MHz gain-bandwidth product. Is there any relationship between these two properties? I've been thought that the high gain-bandwidth product means high speed, so it also means high slew rate. It will be helpful to tell me any kinds of materials on this.

Thanks.

Reply to
Gundal21
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There is not any relation between these two. You might think of GBP as the low (voltage) level frequency response and the slew rate as the high level response. An opamp for low level signals might have a great GBP but crappy slew rate.

George H.

Reply to
George Herold

In general faster op-amps have more of both but they are partly disconnected issues. Slew rate is determined by how fast the op-amp can charge and discharge some internal capacitor before something clips. The "something" usually is the input transistors.

Look up the specs for the LT1028 and the LT1351 to see.

Reply to
MooseFET

Classically, bipolar op amps with no emitter degeneration in the input pair have slew rates of about 0.3V*GBW. (See "The Monolithic Op Amp: A Tutorial Study",

formatting link

Lower input stage transconductance allows higher slew rate, because the op amp runs with more input error voltage (which introduces other problems, e.g. distortion and crosstalk between summing inputs, that you may not want).

Current feedback amps don't suffer from these limitations in the same way, and other modern designs have more complicated tradeoffs, iirc. Still, AN4 is a good place to start.

Cheers

Phil Hobbs

--
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Principal
ElectroOptical Innovations
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Reply to
Phil Hobbs

Regular opamps use the input error to steer the diff pair current source into later gain stages, so there's a limited amount of drive available. CFB opamps steal the drive from the input signal and start applying current gain instantly, and the bigger the loop error the more they steal. This fooled me once recently, and required a board spin: at higher speeds, the CFB amp non-inverting input was severely loading my signal source.

Fet opamps, without input bias current concerns, have a different tradeoff, but have still tended to have low slew rates.

John

Reply to
John Larkin

Happenstance. Burn some current. See my venerable MC1530 for example.

Indirectly. Lower input stage transconductance allows a smaller pole-splitting capacitance, which, for a given current, slews faster.

There have been designs where there are three input stages in parallel, one normal, two with skewed offsets... large input delta turns on a high slew current.

...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

It isn't entirely coincidental, because the GBW and transconductance both go as the input stage current, so cranking up the bias cranks up both together. I'm talking 741 vintage, here--that app note is from about 1970. National AN_4_.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net
Reply to
Phil Hobbs

Think of it this way: fT is independent of voltage (or it's supposed to be), but slew rate is in direct terms of volts. A 100V swing with a slew rate of 100V/us is the same rise time as a 1V swing slewing at

1V/us. But both amps may have the same fT.

Tim

Reply to
Tim Williams

I'd say "little" relation, not "none".

But certainly little enough that you can't count on one given knowledge of the other.

--
www.wescottdesign.com
Reply to
Tim Wescott

Frequency-bandwidth product is a measure of the behavior of the device when it is acting in its linear region. Slew rate is a measure of the point at which the device's behavior departs from the linear region, and a (partial) measure of how it be behaves afterward.

There are no guarantees, but if you look at power consumption as well you will find that low power devices will almost universally have low slew rates relative to bandwidth while high power devices will mostly have high slew rates relative to bandwidth. Amps designed for video will almost always be fairly high bandwidth, fairly high power, and have pretty darn good slew rates.

But if some criterion is important to you then you should always, _always_ check the data sheet to verify that it is met. Further, you should check with physical samples -- sometimes you'll find that a data sheet parameter was established by calculation, or it is only valid for test conditions which don't match your circuit, or that you just plain misread the data sheet, etc.

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Reply to
Tim Wescott

I'm from further back than that ;-) ...Jim Thompson

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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

The slew rate required is determined by the peak voltage in a circuit and the maximum frequency of that voltage. It is only indirectly related to GBWP. The slew rate, p (ro) = dV/dt, the rate of change of the voltage. The voltage is V = Vp*sinwt and the first derivative, dV/dt = Vp*w*coswt. Vp is the peak voltage. w (omega) = 2*pi*f and the maximum cosine value = 1

Therefore the slew rate, p = dV/dt = Vp*2*pi*f.

As an example take a 10 Volt peak voltage at 1 MHz, the slew rate p =

2*Pi*10*1*10^6 = 62.8*10^6 volt per second or 62.8 volts per microsecond. This says that for an amplifier or other device to provide that voltage at that frequency, it must have a slew rate of at least 62.8 Volts per microsecond.

Slew rate is determined by how fast the available current in an amplifier or other device can charge and discharge the internal capacitances. In most amplifiers, there is a dominant capacitance usually used for compensation that must be charged and discharged by the local current. The rate this happens determines the slew rate: dv/dt. The change in voltage, dv = (1/C)int*i*dt determining the slew rate from the current, i, and the capacitance, C. The break point or "Pole" related to this capacitance determines the GBWP but is only a lose relation to the slew rate.

Reply to
Bob Eld

Thanks for the link to the app note Phil. I had always thought these were unrelated.

'Silly me'

Also great (simple) explanation of what's going on inside a 'classic' opamp.

George H.

Reply to
George Herold

There is *NO* relation between slew rate and gain-bandwidth product. That being said, *IF* the slew rate of an op-amp is high, its GBW will (usually) be high.

Reply to
Robert Baer

"Gundal21"

** Try reading again.

Think you will find 250kHz is the full power bandwidth.

The MP103 is a power amplifier.

..... Phil

Reply to
Phil Allison

Of course there is... it's just not an absolute relationship. Slew rate, in an OpAmp, depends on first stage current. This current generally defines first stage transconductance, which critically influences the size of the pole-splitting capacitor (loop compensation). Capacitor size and input stage current define slew rate. So there!

Maybe ;-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
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Reply to
Jim Thompson

I erred.

Suppose we have an ideal OpAmp, bipolar (so I can easily calculate input stage transconductance ;-), input stage biased by current, Ibias, "pole splitting" capacitor of value, C, then...

GBW = (q*Ibias)/(4*k*T*C)

k = Boltzmann's constant q = Charge on electron T = temperature in ºK

Slew Rate = Ibias/C

This is for a classic two-stage OpAmp, with simple "pole-splitting" compensation.

So GBW and Slew Rate are exactly related.

Add extra stage(s) and/or zeroes, and it'll change, possibly getting more GBW without a consequent increase in Slew Rate.

"Real" OpAmps won't be quite as fast as predicted with this simplistic analysis. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

gain-bandwidth

the

gain-bandwidth

product

What!? And the universe didn't end? Gosh, i guess i won't feel so bad when i make an error. Not that they are frequent any more.

Reply to
JosephKK

Throughout my life I have thrown away more designs than survived, by at least an order of magnitude.

To this day I often offend clients by tossing out everything 2/3 of the way to completion, "... because it just doesn't feel right".

Thus my finished projects _always_ work.

Unfortunately there lurk here "designers" who run something up the flag pole, swear by it, and then defend their errors as being of good design practice... despite definitive criticism :-( ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

gain-bandwidth

the

MP103

gain-bandwidth

product

=20

Hmmm.

Interesting, in my current workplace that describes the little new work done by most electrical engineers in my organization. Of course, none of them has built anything (except course projects) with their own two hands.

Reply to
JosephKK

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