Hi,
Thank you, the responses are nearly identical to mine so I don’t think there are errors, but I don’t know how to calculate the resistors and the capacitors, so I get a 400khz and 500kHz cutoff frequency’s

What you do then is calculate the frequency response with all component values unspecified and solve for the frequency.
The cutoff for a simple RC network is:
w=1/(R*C)

the cutoff for the low pass section of your circuit when both capacitors are the same value and R8 is the same value as R7 is:
w=sqrt(sqrt(2)-1)/(R*C)
with C being the value of one of the caps, and R being the value of R7 or R8 (that one section of your first drawing).

Do you know how to analyze a circuit for the frequency response mathematically?

I'll show an example with just a passive circuit made from one resistor and one capacitor, the classic low pass RC filter.

Start by analyzing the circuit in the frequency domain. For the simple RC LP filter we get:
Vout/Vin=1/(s*C*R+1)

Now replace 's' with "j*w" and get:
Vout/Vin=1/(j*w*C*R+1)

Now take the norm of that and get:
Vout/Vin=1/sqrt(w^2*C^2*R^2+1)

Now multiply by Vin and get:
Vout=Vin/sqrt(w^2*C^2*R^2+1)

Now knowing it is a low pass filter, set w=0 and get:
Vout=Vin

and if we normalize Vin to 1 we get:
Vout=1

That would be the very low frequency response. To get the -3db point we set the equation above equal to 1/sqrt(2):
Vin/sqrt(w^2*C^2*R^2+1=1/sqrt(2)

and again with Vin=1 we get:
1/sqrt(w^2*C^2*R^2+1=1/sqrt(2)

Now we solve for 'w' and we get two solutions:
w=-1/(R*C)
and
w=1/(R*C)

Since we want a positive value for w, we choose the second one:
w=1/(R*C)

Now we can solve that for C:
C=1/(w*R)

and that would be the required value for angular frequency 'w', and of course w=2*pi*f so we could rewrite this as:
C=1/(2*pi*f*R)

That's for the simple passive RC low pass filter but you could use the same technique with your circuits also.

 

What you do then is calculate the frequency response with all component values unspecified and solve for the frequency.
The cutoff for a simple RC network is:
w=1/(R*C)

the cutoff for the low pass section of your circuit when both capacitors are the same value and R8 is the same value as R7 is:
w=sqrt(sqrt(2)-1)/(R*C)
with C being the value of one of the caps, and R being the value of R7 or R8 (that one section of your first drawing).

Do you know how to analyze a circuit for the frequency response mathematically?

I'll show an example with just a passive circuit made from one resistor and one capacitor, the classic low pass RC filter.

Start by analyzing the circuit in the frequency domain. For the simple RC LP filter we get:
Vout/Vin=1/(s*C*R+1)

Now replace 's' with "j*w" and get:
Vout/Vin=1/(j*w*C*R+1)

Now take the norm of that and get:
Vout/Vin=1/sqrt(w^2*C^2*R^2+1)

Now multiply by Vin and get:
Vout=Vin/sqrt(w^2*C^2*R^2+1)

Now knowing it is a low pass filter, set w=0 and get:
Vout=Vin

and if we normalize Vin to 1 we get:
Vout=1

That would be the very low frequency response. To get the -3db point we set the equation above equal to 1/sqrt(2):
Vin/sqrt(w^2*C^2*R^2+1=1/sqrt(2)

and again with Vin=1 we get:
1/sqrt(w^2*C^2*R^2+1=1/sqrt(2)

Now we solve for 'w' and we get two solutions:
w=-1/(R*C)
and
w=1/(R*C)

Since we want a positive value for w, we choose the second one:
w=1/(R*C)

Now we can solve that for C:
C=1/(w*R)

and that would be the required value for angular frequency 'w', and of course w=2*pi*f so we could rewrite this as:
C=1/(2*pi*f*R)

That's for the simple passive RC low pass filter but you could use the same technique with your circuits also.

Thank you,
I know what formulas to use, but because it isn’t said about the gain, resistance and the ripple factor I don’t know how to, I tried rereading the task several times but I don’t get it I found the central frequency and the bandwidth only.I will be very grateful if someone can explain to me how to finish the second part of the task:
output resistance less than 50 ohm, operating voltage range between 4 and 12 V. Supply voltage +-15 V.
Thanks in advance and thank you for all the help

 

One short comment to the shown simulation results (µA741):
A transient simulation (time domain) will show that the circuit can never work in reality due to the limited slew rate of the opamp.

 

output resistance less than 50 ohm, operating voltage range between 4 and 12 V. Supply voltage +-15 V.
Thanks in advance and thank you for all the help

hi k,
The Output of the 2nd stage 741 OPA will be low.

When you say operating voltage range between 4V and 12 V
Do you mean the range of Voutput swing when using a +/-15V supply to the 741 OPA's?
Check this image for the Output voltage swing at +/-15V, with say a 10K RL load,

E

 

One short comment to the shown simulation results (µA741):
A transient simulation (time domain) will show that the circuit can never work in reality due to the limited slew rate of the opamp.

I mentioned that several times in this thread and other threads, but he may have turned to a purely theoretical analysis, for now. That would mean infinite gain and infinite frequency response. I also mentioned that when it comes time to built this thing, he has to get a better, faster op amp.

 

Thank you,
I know what formulas to use, but because it isn’t said about the gain, resistance and the ripple factor I don’t know how to, I tried rereading the task several times but I don’t get it I found the central frequency and the bandwidth only.I will be very grateful if someone can explain to me how to finish the second part of the task:
output resistance less than 50 ohm, operating voltage range between 4 and 12 V. Supply voltage +-15 V.
Thanks in advance and thank you for all the help

You said you did not know what capacitor values to use that's why I showed you that example.

The output resistance will be less than 50 Ohms with a decent op amp. That's because feedback tends to reduce the physical output resistance. The output current will always be limited to what the op amp can do, but the voltage is only limited by the output voltage range which can be very low with a rail to rail op amp, or just plus or minus 1.5 volts below the power rails with a less functional op amp.

To get an output of 4 to 12 volts you need an op amp that can work up to 12 volts peak on the output (I assume that is peak because if it was RMS it would go over the 15 volt power supply). To get that performance at 500kHz you need a fast op amp, and probably have to figure on at least 1MHz. That means you have to choose an op amp that can do 1MHz at least. That's going to take a higher gain bandwidth and higher slew rate op amp.

Do you know how to calculate the power bandwidth of an op amp? That is critical in deciding which one(s) to choose to be able to work at that frequency.

 

You said you did not know what capacitor values to use that's why I showed you that example.

The output resistance will be less than 50 Ohms with a decent op amp. That's because feedback tends to reduce the physical output resistance. The output current will always be limited to what the op amp can do, but the voltage is only limited by the output voltage range which can be very low with a rail to rail op amp, or just plus or minus 1.5 volts below the power rails with a less functional op amp.

To get an output of 4 to 12 volts you need an op amp that can work up to 12 volts peak on the output (I assume that is peak because if it was RMS it would go over the 15 volt power supply). To get that performance at 500kHz you need a fast op amp, and probably have to figure on at least 1MHz. That means you have to choose an op amp that can do 1MHz at least. That's going to take a higher gain bandwidth and higher slew rate op amp.

Do you know how to calculate the power bandwidth of an op amp? That is critical in deciding which one(s) to choose to be able to work at that frequency.

Thank you for the explanation MrAI!
But I don’t know how to calculate the power bandwidth of an op amp.

 

Thank you for the explanation MrAI!
But I don’t know how to calculate the power bandwidth of an op amp.

Hi,

Oh ok, well you start with the general expression for a sine wave:
v=A*sin(w*t)

then because we want to know the maximum rate of change (the slew rate) we take the first derivative with respect to time 't':
sr=dv/dt=w*A*cos(w*t)

with sr the slew rate in volts per second, and because the maximum rate of change occurs at t=0 (you can prove that yourself) you set t=0 and get:
sr=w*A

Can you calculate the required slew rate now?
Remember we started with A being the peak amplitude, so that has to be set to your maximum output peak voltage because we want the op amp output to be able to follow the output wave properly with the maximum rate of change that might occur on the output, and because the slew rate is proportional to 'A' the peak amplitude, we use the maximum value that can ever attain in the application.


Now you probably only have to look for an op amp that can do that slew rate or better because usually an op amp with a fast slew rate will have a fairly high gain bandwidth also, but you can check this easily by looking for the gain bandwidth on the data sheet and make sure it is higher than your highest frequency.
For this example I am guessing that will be 1MHz, but it does depend highly on the application. Since the slew rate for this op amp will be quite high, I doubt the op amp gain bandwidth will become a factor, but it does not hurt to check of course.

If this does not make sense yet, we can look at a drawing or two.

I should also mention one more thing in regards to your most recent schematic with the 1nf capacitors.
It appears that you may not have selected the components correctly for cutoff frequencies of 400kHz and 500kHz. I am wondering if you used the angular frequency 'w' for one of the sections instead of the regular frequency 'f'. You may want to check both sections again.

 

Hi,

Oh ok, well you start with the general expression for a sine wave:
v=A*sin(w*t)

then because we want to know the maximum rate of change (the slew rate) we take the first derivative with respect to time 't':
sr=dv/dt=w*A*cos(w*t)

with sr the slew rate in volts per second, and because the maximum rate of change occurs at t=0 (you can prove that yourself) you set t=0 and get:
sr=w*A

Can you calculate the required slew rate now?
Remember we started with A being the peak amplitude, so that has to be set to your maximum output peak voltage because we want the op amp output to be able to follow the output wave properly with the maximum rate of change that might occur on the output, and because the slew rate is proportional to 'A' the peak amplitude, we use the maximum value that can ever attain in the application.


Now you probably only have to look for an op amp that can do that slew rate or better because usually an op amp with a fast slew rate will have a fairly high gain bandwidth also, but you can check this easily by looking for the gain bandwidth on the data sheet and make sure it is higher than your highest frequency.
For this example I am guessing that will be 1MHz, but it does depend highly on the application. Since the slew rate for this op amp will be quite high, I doubt the op amp gain bandwidth will become a factor, but it does not hurt to check of course.

If this does not make sense yet, we can look at a drawing or two.

I should also mention one more thing in regards to your most recent schematic with the 1nf capacitors.
It appears that you may not have selected the components correctly for cutoff frequencies of 400kHz and 500kHz. I am wondering if you used the angular frequency 'w' for one of the sections instead of the regular frequency 'f'. You may want to check both sections again.

I calculated the slew rate to be 0.5 V/μs

 

I calculated the slew rate to be 0.5 V/μs

What is your opinion? Is such a slew rate sufficient?
You should be able to find an answer by yourself - if necessary, try to find some papers about this subject.
We cannot do all the work for you.

 

I calculated the slew rate to be 0.5 V/μs

Okay.

So, at 500KHz, how many microseconds is the half of the sine wave where the voltage is rising?

And how many volts can the output change during that time?

That is the maximum peak to peak voltage the output can be.

 

I calculated the slew rate to be 0.5 V/μs

Are you sure about that?
What frequency did you use?
If you are after even 500kHz max the slew rate will come out higher than that, quite a bit higher.

 

Okay.

So, at 500KHz, how many microseconds is the half of the sine wave where the voltage is rising?

And how many volts can the output change during that time?

That is the maximum peak to peak voltage the output can be.

Hi there,

The slew rate is important mainly at the zero crossing as the rest of the wave has a slope that is always less than that. So it's really the maximum slope.
I also think he should aim higher, like 1MHz because this is a filter, not a circuit that runs up to 500kHz and then will never go any higher in frequency.

 

Are you sure about that?
What frequency did you use?
If you are after even 500kHz max the slew rate will come out higher than that, quite a bit higher.

Are you sure about that?
What frequency did you use?
If you are after even 500kHz max the slew rate will come out higher than that, quite a bit higher.

Hi, you’re right I redid the calculations and for 500 kHz is 43.98 V/µs and for 400 kHz is 35.2V/µs

 

Hi, you’re right I redid the calculations and for 500 kHz is 43.98 V/µs and for 400 kHz is 35.2V/µs

Hi,

I do not think that is correct but you are getting much closer now. I could be wrong, so tell me how you had arrived at those two values of 43.98 and 35.2 v/us?

You are beginning to understand this much better now, that's nice to see

 

Hi,

I do not think that is correct but you are getting much closer now. I could be wrong, so tell me how you had arrived at those two values of 43.98 and 35.2 v/us?

You are beginning to understand this much better now, that's nice to see

SR = 2 *pi * 400 * 10^3 * 14

SR ≈ 3.52 * 10^7

SR ≈ 35.2 V/µs

So, the slew rate of the 741 op amp with f_max= 400 kHz and Vpeak = 14 volts is approximately 35.2 V/µs. I did the same with the 500 kHz. I do not know if it is correct, but I am grateful for the useful tips and guidance. Thank you!

 

So, the slew rate of the 741 op amp with f_max= 400 kHz and Vpeak = 14 volts is approximately 35.2 V/µs.

The calculated value is the REQUIRED slew rate (minimum) to satisfy your needs.
That means: You need another opamp which meets this requirement.
Note: The parameter "slew rate" is a property of the opamp!

 

The calculated value is the REQUIRED slew rate (minimum) to satisfy your needs.
That means: You need another opamp which meets this requirement.
Note: The parameter "slew rate" is a property of the opamp!

Thank you I read about it, but in laboratory exercises we used only this type for our simulations.That is why I used the 741 op amp.

 

Hi there,

The slew rate is important mainly at the zero crossing as the rest of the wave has a slope that is always less than that. So it's really the maximum slope.
I also think he should aim higher, like 1MHz because this is a filter, not a circuit that runs up to 500kHz and then will never go any higher in frequency.

My calculation was meant to be a sanity check ti quickly tell if the opamp was fast enough. It is the skew rate that would make a sin wave into a triangle wave.

 

Thank you I read about it, but in laboratory exercises we used only this type for our simulations.That is why I used the 741 op amp.

Hello again,

Sorry to say, but I do not think the 741 op amp will work in simulations either because of the limited slew rate, unless of course you reduce the frequency or the maximum peak AC amplitude. If you do either of those though you could see some good results.