Op Amp Selection

[Sambuchi]

Hello all, I have a question!

When selecting the perfect op amp for a design what do you look for in the datasheet?

Say I want to amplify a 50ns pulse wit a period of 90us
I want to amplify it 60x
I am not restricted with the power supply.

I know I have to look at the gain bandwidth of the product but I am not sure what variable can tell me how wide the widow is...
then there's slew rate.. what is a slew?

I'm new to this analog stuff.. so guide me wise ones!

[MrAl]

Hi there,


Slew rate is an important spec about an op amp. It tells us how fast the output can rise or fall
in response to an input, and can not go any faster than the slew rate spec. The slew rate
is usually given in units of volts per microsecond. Volts per microsecond simply means that
the output voltage can rise (or fall) at that rate but not any faster.

For example, lets say we have an op amp with slew rate that is 1 volt per microsecond (1v/us).
That means that no matter what the input is the output can only change at the rate of 1v
every microsecond and that's the fastest it can change. Now if the inputs and the gain
were set up so that the output SHOULD be 3v and the output right now is 0v, it will take
3us before the output will actually reach this level of 3v. Note that this is a serious
limitation of the op amp and needs to be considered carefully for any kind of input
waveform. Also note that during the slewing time period (in our example from 0v to 3v)
the output wave is a ramp no matter what the input is, unless of course the input changes
during this time in a way which would force the output to stop changing.

It's interesting that no matter what the input goes to (as long as it forces the output to change)
the output will not directly follow the input unless the input ramps up (or down) at a rate less
than the slew rate (divided by the gain).
For another example, lets say that we have an op amp with slew rate 1v/us set up with external
resistors for a gain of 10, and the input is 0v and the output is 0v. Then suddenly the input
jumps up to +1v. What does the output do? Well, with a gain of 10 the output SHOULD be 10v,
but because the slew rate limits how fast the output can change at first the output only increases
a little bit as it starts to ramp upward with the goal of getting to 10 volts (1v in, gain of 10, means
10v will eventually appear at the output). At 0.1us for instance, the output is only at 0.1v, and
at 0.2us the output only got to 0.2v, and at 1us the output only got to 1v even though it intends
to get to the full 10v eventually. The output continues to ramp in this manner, 2v at 2us, 3v at
3us, etc., until finally at 10us it gets to 10 volts! So what this means is that for an op amp
with a slew rate of 1v/us that has an output starting at 0v and heading for 10v it is going to
take a whopping 10us to get to the full 10v. That's what i meant by a serious limitation.

In the above we took a brief look at what happens with signals that try to cause the output to
try to change to levels that were greater than about 1v or so. Another point of great interest
is what happens with input signals that DO NOT try to cause the output to change by very much
(say less than 1v for example).
For the same op amp with slew rate 1v/us, lets say that we have it set up to amplify by a gain of
only 1 and we only change the input by 0.1 volt from 0v to 0.1v. With a gain of 1 when the input
changes from 0v to 0.1v the output would like to eventually go to 0.1v, but again the slew rate
stops it from getting there immediately, but because the target output level is so low to begin with
it doesnt take too much time to get there now, so after 0.1us it gets to the required level of 0.1v.
This happened because the slew rate was 1v/us and we only had to change the output by 0.1v
so it only took 0.1us (100ns). That's how the op amp behaves for what are usually referred to
as small signals. The signal is small so the response SEEMS faster, although it really isnt.
Take that same op amp and set up the circuit so it has a gain of 10, and with an input change of
0.1v the output now has to get to a full 1v, so it's going to take a full microsecond. Still, that's
a lot faster than when it had to get to 10 volts, which took 10us.

The slew rate is usually given in volts per microsecond, but sometimes it is given in volts per second.
It's a little easier to think in terms of volts per microsecond so you can convert if you like. For
example, 100000 volts per second (V/s) is the same as 0.1 volts per microsecond. Just divide by
1000000 (one million).

As a last note, this slew rate can also greatly affect a circuit where the op amp is in the control
loop because the response of the circuit is often calculated without taking the slew rate into
consideration and so the op amp is viewed as ALWAYS having control over it's own output
(linear analysis). This isnt true during the slewing period(s) however and might cause problems
because the output doesnt directly follow the input (requires some nonlinear analysis).

ADDED:
Sine waves can be affected by the slew rate also. If the frequency and amplitude are such
that the rate of rise of voltage at any point on the sine wave is faster than the slew rate, the
output sine wave will be distorted. The amount of distortion depends on how much greater that
rate of rise is compared to the slew rate of the op amp. Since the rate of rise is greatest at
the zero crossing, we often look there in order to determine if the sine wave shape will be altered
as it passes through the op amp.

WHAT ALL THIS MEANS for your application is that if you want to choose an op amp you not only
have to consider the pulse time and the gain, but also the intended output variation that will
be required. For example, does the the output have to change by 1v, 2v, 10v, or 100v.
That output change also affects the op amp selection, or even if you can actually use an op amp
at all.

[Sambuchi]

Wow what a great response Mr Al!
I see what you are saying and now have a better idea of whats going on.
Soo.. how does the slew rate relate,or does it, to the gain bandwidth?

[Robert Reed]

Sam
For the hi-speed pulses you are dealing with, I don't think you will find a suitable voltage feedback type to achieve your goals. The relatively new current feed back Op-Amps are very fast . However they do not have much gain and consume somewhat more power. You say you need a gain of 60. That would be from what level to what level?
Most of the IC manus make these chips, just check their websites. I have always found that when it comes to critical rise times and gain - regular transistor circuitry does a much better job. The trouble with hi-speed and Op-Amps is their are just too many PN junctions involved on the die that contribute to slowdowns.

[MrAl]

Wow what a great response Mr Al!
I see what you are saying and now have a better idea of whats going on.
Soo.. how does the slew rate relate,or does it, to the gain bandwidth?

Hi again,


Well thank you

The gain bandwidth says something about the internal gain, the way it drops off with frequency.
The slew rate will tell you how high a frequency (sine wave) can be passed without distortion
at a given amplitude. The formula would be:

Freq=vpus*1e6/(2*pi*A)
where
Freq is the frequency in Hertz,
vpus is the voltage per us spec for the op amp,
A is the amplitude of the output (peak),
pi is the constant 3.14159265 approximately.

For example, say we have an op amp who's slew rate is 0.5v/us (like the LM358) and the output
has to change by 2 volts peak (sine wave). The maximum frequency we can get away with
would be:
Freq=0.5*1e6/(2*3.14159265*2)
where we made the volts per us equal to 0.5 and the amplitude A equal to 2 volts, so we get:
Freq=39789Hz
which is close to 40kHz.
If we try to go over 40kHz we will start to see distortion. If we try to increase the amplitude of the
output (by increasing either the input voltage peak or the gain of the circuit) we will also see some
distortion because amplitude is in the equation too.


Since you seem to want to be able to amplify a 50ns pulse and that represents a 20MHz signal,
you would have to go to at least three times that to 60MHz but even that probably wouldnt be
good enough depending on the accuracy you need.
For a 1v rise with a 0.5v/us op amp it takes 2us which is 10 percent of 20us, so that would work
if a 10 percent error in rise time is acceptable. The formula would be:
vpus=V/Tr

where
vpus is volts per us required,
V is the rise in voltage needed, and
Tr is the rise time needed.

Now for your app if a 10 percent rise time is acceptable that would mean you need to rise in 5ns which
is 0.005us, and say you only have to rise by 1 volt, then we have:
vpus=1/0.005
which of course means the volts per us spec would have to be 200 volts per us (200v/us).
You'd have to find an op amp that can do that.
If the voltage has to rise up to 5v instead of 1v then you would need 5 times that spec, or 1000v/us.

[Sambuchi]

O-Man... I cant wait to get my hands on some op amps!

What you said makes complete sense to what I am seeing in switcher cad.

I want to thank you for your time on this Rob and Al!

[MrAl]

Hi again,


They make quite a few op amps these days that go 2000v/us but im not sure how high the input
impedance can be made. Many of the applications they show are for 50, 75, and 300 ohm applications.
Not sure what input Z you need.

[Bob Scott]

Wow what a great response Mr Al!
I see what you are saying and now have a better idea of whats going on.
Soo.. how does the slew rate relate,or does it, to the gain bandwidth?

Sambuchi,

Op-amps need to be "compensated" so that circuits are stable when negative feedback is used. This means that at high frequencies, the fed-back signal can't be rotated more than 90 degrees or the feedback becomes positive, making the circuit oscillate instead of amplify. The most compensation is required when the circuit has maximum feedback (zero! gain).

General purpose opamps usually have "zero gain" compensation so that they will be stable at all amplification levels. Compensation is accomplished by adding capacitive circuitry inside the chip. These internal capacitors have the secondary effect of limiting the slew rate. The capacitors charge and discharge during operation. As the frequency increases, the current necessary to drive the capacitor compensation also increases. There is a limit to the amount of current available inside the chip. When that limit is reached, the signal goes into slew rate limiting.

Some op-amps can have the ability to be custom compensated if the required gain is more than 1. The data sheets for these op-amps have diagrams and instructions for the required caps and resistors to provide lead/lag tweeking and stabilize the circuit. This custom compensation gives the circuit higher slewing rates because the capacitors are smaller.

Yes, the power bandwidth is reduced. The gain-bandwidth product goes through zero dB earlier in a compensated circuit. After all, you are hacking off the unstable area of the gain graph by adding capacitance.

As a general rule, the slope of the frequency response curve when the response goes through zero must not exceed 6dB per octave. If it is more steep, then the feedback exceeds 90 degrees and the circuit is not stable.

I highly recommend Walter P. Jung's "Op-Amp Cookbook". It is out of print but available used from big booksellers like Amazon. I was a complete dummy at using op-amps and feedback until I read this book. It's way easier to understand than than I thought.

[Bob Scott]

Conservative design says that you should use an op-amp with a minimum gain more than 20dB over the required circuit gain at the upper frequency limit, so you have at least 20dB (x10) for feedback at all frequencies of interest. For a pulse, I guess that would mean harmonic frequencies.

Finding an op-amp with a gain of more than 70 is not hard.
Finding an op-amp with 100MHz bandwidth is not as hard as it used to be.
Finding an op-amp with a gain of 70 at 100MHz is the hard part.

[Robert Reed]

Good points Bob, but don't leave out the - Power Bandwidth.
Sam has not responded to what level he wants to achieve, so it may be a moot point.

[MrAl]

Hello again,


Yes, Bob, i was wondering where you got the gain of 70 from?

[Bob Scott]

Yes, Bob, i was wondering where you got the gain of 70 from?

[Stand by, correcting error]

Um, his requirements have an amplitude gain of 60X. That's the amplification factor. Conservative design says that the open loop gain of the op-amp should be 20dB higher for good corrective feedback, or 10X higher. That's 70X total open loop gain. When the loop is closed as normal operation, the circuit gain is 60x.

Usually a pulse is square. If you want it to remain squarish coming out of the amplifier, the amp should be capable of amplifying some of the harmonics of the pulse. If you consider that the 50nS pulse is a half cycle of a frequency with a period of 100nS, the frequency is 10MHz, plus harmonics. If you make the amp capable of 100MHz, the pulse should still look squarish coming out, right? Sam might not need an accurate pulse amplitude or fast rise time. Then he could use a slower amp with sloppy skirts. I don't know....

The harmonics are subdued in a square wave, so at least the harmonic frequencies don't need to have a rail-to-rail slew rate. The amplitide of the 3rd harmonic is about 1/3rd the amplitude of the fundamental, the 9th, the 1/9th the amplitude etc..

[Sambuchi]

This is a great thread!

The smallest pulse that I will expect will be a 560nA from a photodiode.

I am looking at this part right now and hope to do some modeling with it soon.
http://www.datasheetcatalog.org/datashe ... 9864_1.pdf

[MrAl]

Hi again,


Bob:
Oh ok, thanks for the info.

Sam:
I wish you had mentioned the photodiode before this...
check out the OPA659, that can do 2500v/us. I think they even give a P.D. circuit in the data sheet.

[Robert Reed]

Bob
I am a little confused by:
"Um, his requirements have an amplitude gain of 60X. That's the amplification factor. Conservative design says that the open loop gain of the op-amp should be 20dB higher for good corrective feedback, or 10X higher. That's 70X total open loop gain. When the loop is closed as normal operation, the circuit gain is 60x."

A voltage gain of 60x is a 37.6 db gain. Open loop feed back safety factor increase of 20 db on top of the required 37.6 db would be 57.6 db. 57.6 db of gain is equal to a gain of 600x. A 60x required and 70x (open loop) would represent a gain increase of approx. 1.3 db.
I just got home from an all day ski trip, so my thinking may be a little groggy.

Sam
looks suspiciously like a current feedback amp - low gain, low level output and a current hog.Although since you changed your requirements tenfold (50 ns- 560 ns) it may work for you.