I have a project to using LM318 to have a gain of 20.

input: 50 mv peak Square wave 1.5Mhz.
output: 1V peak Square wave 1.5Mhz
Gain: 10

We are using a single inverting setup.
Lm318 OpAmp.

+6VCC
-6Vcc

R1= 1K
R2 = 20K
cap = 100pf across R2 for miller effect.

at lower frequencies we are getting the output right.
But as frequency increases to 1.5Mhz it seems the Overshoot then creates a Sine wave output.

We have a cap across R2 for the Miller Effect.

Any suggestions to get us in the right direction of our goal?

I think we would have to cascade them, but I am not sure why.


Thanks in advance

 

Can you show us a schematic of your setup? A picture or sketch will work.

 

I have a project to using LM318 to have a gain of 20.

input: 50 mv peak Square wave 1.5Mhz.
output: 1V peak Square wave 1.5Mhz
Gain: 10

So which is it? Again of 20 or a gain of 10?

What is the gain-bandwidth product of the LM318?

What is the bandwidth of a 1.5MHz square wave?

 

Its a Inverting Op amp

Rin = R1
Rf= R2
The capacitor is across R2

 

I have a project to using LM318 to have a gain of 20.

input: 50 mv peak Square wave 1.5Mhz.
output: 1V peak Square wave 1.5Mhz
Gain: 10

We are using a single inverting setup.
Lm318 OpAmp.

+6VCC
-6Vcc

R1= 1K
R2 = 20K
cap = 100pf across R2 for miller effect.

at lower frequencies we are getting the output right.
But as frequency increases to 1.5Mhz it seems the Overshoot then creates a Sine wave output.

We have a cap across R2 for the Miller Effect.

Any suggestions to get us in the right direction of our goal?

I think we would have to cascade them, but I am not sure why.


Thanks in advance

Do you want gain of 10 or 20?

Maybe that explains the overshoot. You were expecting gain of 10, but your resistors are chosen for gain of 20.

 

sorry typo ya gain of 20

 

Which brings us back to the question what the gain-bandwidth product for this opamp is?

 

Small signal bandwidth 15 MHz.
http://www.futurlec.com/Datasheet/Linear/LM318N.pdf

I really wish you would give the OP a chance to answer some of these questions for themselves.

 

Right, so that is the GBW? i'm sorry I am very lost and don't have class mates to really talk to.

 

Right, so that is the GBW?

Yes, the GBW is 15 MHz. What is your understanding of what the GBW tells you?

 

So is the 15MHz the highest frequency the op amp can pass with very low slew rate distortion, without significant attenuation. But that means at 1.5Mhz I should be okay and it should be getting the square wave just fine?

 

Or we have GBW of 15MHZ

SO the -3db = 15MHZ/gain+1 = 15Mhz/21 = 357Khz, which means what exactly?

im assuming that doesn't work (I don't know why tho)


and the Max gain we can get (at 10 times) is :

15Mhz/10x 1.5Mhz = 1

so we can only have a gain of 1 on the first op amp so we need op amps?

does this seem correct.

 

So is the 15MHz the highest frequency the op amp can pass with very low slew rate distortion, without significant attenuation. But that means at 1.5Mhz I should be okay and it should be getting the square wave just fine?

Think about the name. Gain Bandwidth Product. The Product of the Gain and the Bandwidth.

If your circuit has a gain of 20 and you try passing a 1.5 MHz sine wave through it, what is the produce of the gain and the bandwidth you need?

And if you have a 1.5 MHz squarewave instead of a 1.5 MHz sine wave, when is the bandwidth you need to pass a waveform that looks like a reasonable squarewave?

 

Or we have GBW of 15MHZ

SO the -3db = 15MHZ/gain+1 = 15Mhz/21 = 357Khz, which means what exactly?

im assuming that doesn't work (I don't know why tho)


and the Max gain we can get (at 10 times) is :

15Mhz/10x 1.5Mhz = 1

so we can only have a gain of 1 on the first op amp so we need op amps?

does this seem correct.

You seem to be working around some of the right ideas.

I don't know where the +1 in your first equation comes from. But you have the basic idea. If you have a gain bandwidth product of 15 MHz, then if you want a gain of 20 you need to constrain the bandwith to about 750 kHz since 20*750 kHz is 15 MHz.

In order to get a reasonable approximation of a square wave, you need to pass the fifth and preferably the seventh harmonic, although sometimes you can live with just the first and third. If you want the fifth harmonic and you're limited to a bandwidth of 750 kHz. That means that you are going to be limited to square waves in the 150 kHz range. Above that, they are going to look more and more like sine waves because only the fundamental is getting through.

Does that sound familiar?

 

alright I think im getting it.

so if I wanted to do a gain of 20 on one amp. I can only go as high as 750Khz?

but for a square wave I have to times the gain by 5 to get the fifth harmonic in there so 15Mhz/ 5x20 = 150K

So I need multiple stages to get the gain I want to achieve with a 1.5Mhz signal within the 15Mhz GBW.

So I f I have the first op amp at gain of 2, but I want the 3rd harmonic:

15Mhz/2x3 = 2.5Mhz which is enough constraint for us at 1.5Mhz correct?

alright I am getting a bit of a square wave. trying to get picture posted

 

alright I think im getting it.

so if I wanted to do a gain of 20 on one amp. I can only go as high as 750Khz?

but for a square wave I have to times the gain by 5 to get the fifth harmonic in there so 15Mhz/ 5x20 = 150K

So I need multiple stages to get the gain I want to achieve with a 1.5Mhz signal within the 15Mhz GBW.

So I f I have the first op amp at gain of 2, but I want the 3rd harmonic:

15Mhz/2x3 = 2.5Mhz which is enough constraint for us at 1.5Mhz correct?

I am still getting a sine wave ...starts at about 700Khz..

I'm not completely following the last couple lines.

For your first stage if you set it up as a gain of 2 then you should be able to get a bandwidth out of that stage of about 7.5 MHz, which should pass the first five harmonics of a square wave up to about 1.5 MHz. So that is no problem. But notice that this is the most you can set the gain to and get five harmonics. That means that, to get a total gain of 20 and pass five harmonics of a 1.5 MHz squarewave that you are going to need five stages. WIth four stages you should be able to get a gain of 16 and you might get a gain of 20 since the actual gain-bandwidths should be a bit higher than the spec.

If you use two stages, you need each stage to give you a gain of √20 = 4.47 (call it 4.5). At a gain of 4.5 you would have a bandwidth of about 15MHz/4.5 = 3.33, which would be enough to pass the third harmonic.

Note that the overall bandwidth is limited by the stage that has the highest gain.

 

.........
I don't know where the +1 in your first equation comes from.
............

I think, the "+1" is correct - in case the word "gain" is identical to the ratio R2/R1.
In this case, the expression (1+R2/R1) gives the so called "noise gain" which must be used to calculate the -3dB point of the opamp´s open-loop gain
(which appears at the frequency where the loop gain LG is unity).

BTSP: Be aware that for squarewave repetition rate of 1.5*1E6 (1/s) you are approaching the limits set by the maximum slew rate of the opamp. (Please note, that I intentionally did not use the word "frequency" for this squarewave).

 

I think, the "+1" is correct - in case the word "gain" is identical to the ratio R2/R1.
In this case, the expression (1+R2/R1) gives the so called "noise gain" which must be used to calculate the -3dB point of the opamp´s open-loop gain
(which appears at the frequency where the loop gain LG is unity).

Ah, okay. Thanks for the clarification.

 

Okay thanks for the help so far , i'm going to be working in the lab a bit more today see what progress i can make.

i was working with the slew rates yesterday and something wasn't making sense ill take a look at what LvW mentioned.

Im sure ill have some more questions tonight.

Thank you

 

So i am reaching the max slew rate...so the overshoot is too great and just becomes a sine wave?