okay so ive measeured the slew rate and it seems really low.

Delta V = .170V
Delta T = .092ms

DElta V/Delta T = 1.85

The datasheet says typ. 70 V/us

and min is 50 V/us

so it seesm my slew rate is alot lower, and im not sure what this means

 

So slew rate is 2piFV

amplitude is 1V and or F is 1.5Mhz

Needed slew rate is 9.4 needed

says minimum is 50 on sheet?

not sure what this means

 

BTSP,
the slew rate SR must carry a dimension voltage/time (mostly V/µs).
How did you measure the SR of your device?
(By the way: what means "2piFV"?)
The following equation connects the slew rate (that gives the max. capability of a device to slew to a certain amplitude Vmax) and the corresponding upper frequency limit (large signal bandwidth BW) :
SR=BW*2Pi*Vmax.

 

okay so ive measeured the slew rate and it seems really low.

Delta V = .170V
Delta T = .092ms

DElta V/Delta T = 1.85

The datasheet says typ. 70 V/us

and min is 50 V/us

so it seesm my slew rate is alot lower, and im not sure what this means

How did you measure this? Your input signal has to be such that it pushes the opamp to operate at it's maximum slew rate. One way to do this is to run it open loop as a comparator.

The numbers you give result in

ΔV/Δt = 0.170V/0.092ms = 1.85 V/ms = 0.00185 V/μs

Units count!

 

So slew rate is 2piFV

amplitude is 1V and or F is 1.5Mhz

Needed slew rate is 9.4 needed

says minimum is 50 on sheet?

not sure what this means

UNITS!!

Saying that the slew rate is 9.4 is meaningless!

The needed slew rate is 9.4V/μs for a 1.5MHz SINE wave.

But you are trying to work with a SQUARE wave. The third harmonic is 4.5MHz and requires up to 28.3V/μs to support (if its amplitude were 1V, which it isn't) and, if the first and third harmonic are passing through zero going in the same direction their needs add. To support a 1V sine wave for the fifth harmonic, you would need a slew rate of 47V/μs. So you can see that you can quickly become slew-rate limited.

Let's look at it another way. If you have a 1V square wave (so going from -1V to +1V), how long does it take to go from one level to another at 50V/μs? The transition time would be

Δt = 2V/(50V/μs) = 40ns

The half period was

T/2 = (1/F)/2 = (1/MHz)/2 = 333ns

So the transition time takes 12% of the total time, which is a very noticeable departure from ideal behavior, but should still yield something that is largely recognizable as a square wave.

 

BTSP,
the slew rate SR must carry a dimension voltage/time (mostly V/µs).
How did you measure the SR of your device?
(By the way: what means "2piFV"?)
The following equation connects the slew rate (that gives the max. capability of a device to slew to a certain amplitude Vmax) and the corresponding upper frequency limit (large signal bandwidth BW) :
SR=BW*2Pi*Vmax.

2piFV is referring to the maximum slew rate required by a sinewave of frequency F and amplitude V.

v(t) = Vsin(ωt)

dv(t)/dt = d(Vsin(ωt))/dt = ωV cos(ωt)

This has a max value of ωV = 2∏fV.

 

2piFV is referring to the maximum slew rate required by a sinewave of frequency F and amplitude V.

WBahn thanks for trying to explain.
But - of course - I could imagine (see the last line in my post#23).
My question was more or less a recommendation to the OP to write and explain equations more clearly (including units).
Just to state F=frequency and V=voltage is by far not sufficient to describe the MEANING of the parameter called "slew rate".
I am sure that you are with me in this question.

 

WBahn thanks for trying to explain.
But - of course - I could imagine (see the last line in my post#23).
My question was more or less a recommendation to the OP to write and explain equations more clearly (including units).
Just to state F=frequency and V=voltage is by far not sufficient to describe the MEANING of the parameter called "slew rate".
I am sure that you are with me in this question.

Yep. For some reason I didn't even see your last line. Strange.

Oh, I know what happened. I saw your question, realized that there was a post by the OP that I had missed so I tracked it down, and then single-mindedly responded forgetting that I hadn't finished reading your post.

My apologies. Getting WAY too late in the semester for me to be thinking straight.

 

My apologies. Getting WAY too late in the semester for me to be thinking straight.

No problem - such things happen from time to time.

 

thank you very much for the explanation of slew rate. And i will try from now on to use proper units and clearly state my calculations.

I measured my rise time and voltage using the oscilloscope reading the slope of the square wave.

So i now understand how the slew rate is a limiting factor, and at high frequency of 1.5Mhz it is on the edge of rising fast enough.

I've come to the conclusion that designing for a gain of 20, with 1.5Mhz on the LM318 is impossible...

The GBW is too low to get the Harmonics we need for a square wave.

BW = 15MHZ / (1+2) = 5Mhz
to include the 5th harmonic i would have a constraint of 1Mhz Which is below our wanted 1.5MHZ.

While analyzing the harmonics i have the 5th harmonic, it is low but there. Im assuming this is cause the GBW on the sheet may be a bit higher than what is written.

I have a gain of 2 on the first stage, using compensation to reduce settling time, i have a decent square wave, the next stage of gain 2 i have an output gain of 4 becoming less of a square wave (still have a bit of the 5th). The next stage gain of 2, output of 8, even worse pretty much a sine wave( no more 5th harmonic).

To get the gain of 20 we need , we would have 5 inverting op amps min. but I know there is something that happens as you cascade more and more Op Amps. the Db drop add up and soon enough it the output will become worse.

I am not clear on this principle.

another note i am wondering about, would using non-inverting op amps be more stable to use?

 

I've looked into the bandwidth shrinking factor of cascading identical op amps.

BW of Cascading Op AMPS = (BW of single OP. Amp)*sqrt(2^(1/N)-1)


N (#of op amps) Shrinking Factor
1 1
2 .644
3 .510
4 .435


So if i have:
GBW = 15Mhz
Gain = 2 (in each stage)
BW of op amp = 15Mhz/(1+2) = 5Mhz


If my thinking is right:

BW of op AMP:
1st stage: 5Mhz
2nd stage: 3.22Mhz
3rd stage: 2.55Mhz

If my thinking is correct to keep the 5th harmonic:

BW 1st: 1Mhz
BW 2nd: 644Khz
BW 3rd: 510Khz

Does this seem right?