op amp specs for tunable filters

[atferrari]

I've run across a design of a tunable state variable filter using AD825 op amps.

My idea is to replicate the design for a maximum f0 of around 130 KHz.

What could be a correct (and not costly) replacement of those op amps?

I was thinking to use dual / quad ones but I do not know what actual specs to look for. Sure slew rate, low noise. Something else?

As per what I read GBW product has to be 10, 50 or 100 times the highest f0? How much should it actually be?

 

[MrAl]

Hi there,

Well, the bandwidth has to allow the signal to pass unchanged more or less, but also the slew rate has to be high enough to allow the signal to pass without altering the wave shape. Thus if you are to pass a 10kHz signal with a gain of 10, you need at least 100kHz bandwidth but you also need a slew rate which is greater than the maximum slope of the assumed output wave, which if a sine wave the max slope occurs when the sine goes through zero so you can use that to calculate the required slew rate. Remember it's the max slope of the output, not the input that matters, so if you have a gain of 10 that means the max slope will be 10 times that of the input.
Of course a spice simulation would not be a bad idea either

 

[crutschow]

The required GBW depends somewhat on the filter design, but I would think somewhere between 10 and 50 of the highest fo should be adequate.

If you are concerned about noise than you would want a low noise op amp.

The required amp slew rate is 2*pi*f*Vp where f is the highest frequency and Vp is the maximum peak output voltage.

An LM6134 may work for you.

As MrAl suggested you should simulate the circuit with Spice such as LTspice, which is free from Linear Technology.

 

[atferrari]

How to actually calculate slew rate?

The required amp slew rate is 2*pi*f*Vp where f is the highest frequency and Vp is the maximum peak output voltage.

Working out the value as per the equation you gave (for a 5Vp)

2*pi*f*Vp = 2*3,1416*130000*5 = 4084070,45

It does not seem to be anything like a slew rate I could expect to need at this frequency.

There was a post by you or maybe MrAl explaining exactly that but after searching a lot I could not find it.

Could you, or anyone, help? Gracias.

 

[crutschow]

What slew rate would you expect? The value you calculated of 4.08E6 V/s (4.08 V/µs) is correct.

 

[MrAl]

Working out the value as per the equation you gave (for a 5Vp)

2*pi*f*Vp = 2*3,1416*130000*5 = 4084070,45

It does not seem to be anything like a slew rate I could expect to need at this frequency.

There was a post by you or maybe MrAl explaining exactly that but after searching a lot I could not find it.

Could you, or anyone, help? Gracias.

Hi again,

I often can not find things in the past here either. Not sure why yet.

Anyway, if we start with the assumed wave shape of a sine wave:
v(t)=A*sin(w*t)
where
A is the peak amplitude (this is the amplitude of the output, not the input),
w is the angular frequency 2*pi*f,
t is time,
and we differentiate once with respect to time we get:
dv/dt=w*cos(w*t)*A
and if we differentiate one more time we get
d^2v/dt^2=-w^2*sin(t*w)*A
and realizing that w and A dont affect the normalized max we can set those to 1 and we get:
d^2v/dt^2=-sin(t)
and now setting this to zero we get:
-sin(t)=0
and solving for t we get:
t=0
which means we either have a first derivative max or a min at the point t=0.
Now taking w*cos(w*t)*A and realizing that the first w and the A are always positive so dont affect the gain, we are left with the normalized first derivative:
dv/dt=cos(w*t)
Graphing this for any w with t= say -4 to +4 we find that when t=0 we do indeed get a max dv/dt. Thus, at t=0 we get a max so we can use the equation:
dv/dt=w*cos(w*t)*A
with t=0 to calculate the max. Doing this, we get a nice simple equation:
max dv/dt=w*A
which of course is also equal to:
max dv/dt=2*pi*f*A
Now that we have that and knowing that data sheets like to spec their slew rate in volts per microsecond (V/us), all we have to do is divide by 1e6 and we get:
slewrate=2*pi*f*A/1e6 {V/us}
That's the final formula, and it expresses the slew rate in V/us like most data sheets.

Ok, so now we move on to the actual application, which apparently has a frequency of 130kHz and an amplitude of 5. Inserting those quantities into the formula and we get:
slewrate=2*pi*130000*5/1e6
and simplifying we get:
slewrate=4.084 V/us
Then we would look for an op amp that can do at least 4 V/us or maybe a little more so we dont get any distortion.

So you see you had the right answer all along as Carl suggested! You just needed to compare it with the data sheet and maybe convert the units to the same as those used in the data sheet.

One more simple little example...
We have a signal of 1 volt peak that we want to amplify by 10 to get 10 volts peak output. The wave shape is sine and the frequency is 20kHz. What is the minimum slew rate we can look for in the data sheets?
Starting again with the formula:
slewrate=2*pi*f*A/1e6 {V/us}
and since the output amplitude is 10 volts peak and the frequency is 20000 we insert those:
slewrate=2*pi*20000*10/1e6 {V/us}
and calculating this out we get:
slewrate=1.26 V/us
Note that an op amp like the LM358 would not work for this application because its slew rate is only 0.5 V/us. We would need something a little bit better for this application.
Another thing to notice is that because the frequency f works in equation as a linear factor, a multiplication of the frequency by a factor x also multiples the slew rate by that same factor, so if the frequency was 200kHz instead of 20kHz as above we could simply multiply the calculated slew rate of 1.26 by 10 and that would give us 12.6 V/us, which means we would need a much better op amp if we had to pass a signal of 200kHz in this app.

One additional little note here...
We assumed the output wave shape was a sine wave. If it is any other wave you'd have to calculate the max frequency component that is important and its amplitude and use that frequency instead of the first harmonic frequency, or in other words, the max slope of the wave. You can calculate the slope by taking dv/dt of the input signal and hunt for the max slope by taking d^2v/dt^2 as we did in the above derivation.
Of course a spice simulation might help too just to verify the design at least at this stage in the circuit.

 

[atferrari]

What slew rate would you expect? The value you calculated of 4.08E6 V/s (4.08 V/µs) is correct.

Yes. Shame on me! At the moment I did not realize that I had to think in terms of nsec. Gracias.

 

[atferrari]

Hi again,

I often can not find things in the past here either. Not sure why yet.

Gracias MrAl for bringing it back here!