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Hello All,

@ Andy& Jim:Thanks for your input on tmax in .tran analysis. It was really useful and I plan to have a different thread on this soon.

Now coming to the issue in thread, I have this feeling that frequency is just a frequency for an opamp, Whether it is that of input source or that of any component variation. This motivates me to put this hypothesis that i can plot the Bode due to rapid variation of R2 by just assuming R2 to be constant at its maximum value and sweeping the frequency of input vi (Vi should be changed to AC source).
Here is setup files to show this:



In inv_opm_freq_respns.asc I have setup to plot frequency sweep. I select three frequncies: 10KHz, 200KHz, 750KHz. I noticed that at the first two frequencies output is constant and at around 750KHz it rolls down by 3dB.

Now in setup inv_opm_tran.asc, I do a transient analysis. I stepped R2 with above three frequencies and found that as suspected, output have same amplitude at 10KHz and 200KHz,and lower at 750KHz.

As always, I especially seek views of people like Helmut, Andy, Jim, John , analogspiceman,chris to see if there is any fallacy in my understanding.
Conference date has been extended to I have ample time to go deep down the problem.

Thanks.

Hello John,

You are getting these results because of your very short
rise and fall times and short 'crossover' pulses.

This was indeed the trick to get a "useful" result for THD
in the .FOUR report. I have had used 10us rise and fall time
in my example which I uploaded yesterday.

Best regards,
Helmut


--- In LTspice@..., John Woodgate <jmw@...> wrote:

In message <kkfbb8+osfi@...>, dated Sun, 14 Apr 2013, Echidna
<mchambin@...> writes:

Indeed, I was tricked by the .FOUR function default calculation on 9
harmonics only. I tried with 99 harmonics and got a much different
result.

You are getting these results because of your very short rise and fall
times and short 'crossover' pulses. If you do an FFT from 'View' you can
see that the harmonic spectrum goes to well over 30 MHz. This isn't
realistic for an audio amplifier, if that is what you are working on.

But calculating THD by 'adding' (r.s.s.-wise) the harmonic amplitudes is
seriously error-prone in my experience. It's much better to notch out
the fundamental with a filter (simulated notch filters always work well,
unlike real ones) and measure the r.m.s. value of what is left.
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
They took me to a specialist burns unit - and made me learn 'To a haggis'.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

In message <kkfbb8+osfi@...>, dated Sun, 14 Apr 2013, Echidna <mchambin@...> writes:

Indeed, I was tricked by the .FOUR function default calculation on 9 harmonics only. I tried with 99 harmonics and got a much different result.

You are getting these results because of your very short rise and fall times and short 'crossover' pulses. If you do an FFT from 'View' you can see that the harmonic spectrum goes to well over 30 MHz. This isn't realistic for an audio amplifier, if that is what you are working on.

But calculating THD by 'adding' (r.s.s.-wise) the harmonic amplitudes is seriously error-prone in my experience. It's much better to notch out the fundamental with a filter (simulated notch filters always work well, unlike real ones) and measure the r.m.s. value of what is left.
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
They took me to a specialist burns unit - and made me learn 'To a haggis'.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

In message <kkf8qp+pa6i@...>, dated Sun, 14 Apr 2013, sawreyrw <sawreyrw@...> writes:

Get over it.

It's well-known to be incurable.
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
They took me to a specialist burns unit - and made me learn 'To a haggis'.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

In message <CALBs-TgwdGnyE+6k-DsnYM_d3xZ2OrmE-xvRep2TEYTBQCGQuQ@...>, dated Sun, 14 Apr 2013, Andy <Andrew.Ingraham@...> writes:

It is a pain in the neck having to download it every time (or to fine the copy I downloaded last week), but it does work.

Can you save it to your desktop, to make it easier to find?

If the problem went away for you, I wonder what you did to make it work.

I didn't do anything.

And I wonder why Helmut never sees this problem. The last time around that this was discussed (a few weeks ago?), it seems everyone had the same problem, except for Helmut. Odd.

That and my experience suggest that it's not a Yahoo thing but something on individual computers, probably a Microsoft 'improvement'.

I have found IE8 going unstable twice in the last few months, showing strange and unpredictable behaviour. I have had to use the Tools => Advanced => Reset option, but that wasn't to fix the 'all_files' problem. Some web sites think I am using an earlier version, perhaps IE6, and complain about it, but I am definitely not.
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
They took me to a specialist burns unit - and made me learn 'To a haggis'.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Hi Jim.
Thanks for your comment on the type of material the transistor. I removed a tag MFG=Germanium-type.
Use the my Standard.zip file.
Bordodynov.

12.04.2013, 20:10, "gr8wi9" <boothjg@...>:

On Sat, 13 Apr 2013 15:47:49 -0000, Tim wrote:

I just uploaded a another copy of the JT, because I wnat to
use it as a techng aid to introduce a family member to some
EE concepts and LTspice. He just graduated high school and
has no exposure but a lot of interest in electronics.

So having read the recent thread on the JT, I though it
might be an excellent resource for him as it exposes
transistor and switching theory and a lot of concepts in an
easy to understand project that he can build and use my lab
tools to verify the expected behavior.

In my posted version, derived from
, I tried to added a second
circuit to compare the effects of using a different
transistor than the B549 in the reference article and I
attempted to estimate the Joules being consumed (atfter all
it is a Joule Thief.

If anyone cares to comment on how I might do this better
(for the intended purpose), please do so.

I just posted up an example I wrote some time back. I added a
comment or two to it before posting, though. It's in the Temp
directory and called Joule_Thief_jk1.asc.

It uses the basic energy equation to develop the expected
peak current required. It's ability to calculate the exact
frequency and the exact peak current given a specific
transistor is only approximate, though. The base resistance
calculation is only very approximate. Some tweaks are
included to allow closer approximations.

Note, the inductor coupling constant is given as 1 -- in
reality it never will be. Feel free to adjust the equations
to take a more realistic value into account.

Also, I have an additional paper I wrote on the Joule Thief
that includes details the above considerations plus others,
such as Bmax, in those cases where an air core isn't used.
The Bmax value places additional constraints on the choice of
inductance value vs frequency, to avoid saturation effects
due to volt-seconds (which the Joule thief doesn't not
require and works perfectly well avoiding.) If interested in
the paper, feel free to write and I will send it along in an
email (PDF form.)

Jon

Jim Wagner <wagnejam99@...> wrote:

You would expect one peak to be quite a different from the other, because it is
basically nonlinear. ...

Thus, when driven with a sine, depending on the dR/dt, the waveform COULD
be quite different at the positive and negative peaks.

Indeed they were; VERY different. But that was not what stood out.

I expected all of the positive peaks (at a given frequency, and
perhaps after the first couple of cycles) to look about the same as
each other and to reach the same value as one another. Instead, what
I saw was that the positive peaks never "leveled off" to the same
value. Some were only 0.38V while others reached 0.55V. Zooming in
showed that some peaks were very narrow and others were broader, and
had a distinctly "sampled" appearance. Turning on "Mark Data Points"
(right-click on the waveform plot window) showed that it had something
to do with how closely spaced the data points were: not enough to
reveal the true detail of the peaks.

The negative peaks, by contrast, were smooth and round. But I wasn't
looking at those.

If this doesn't make sense, run the simulation and see what I mean.

I doubt there is much loop gain at all at 1 MHz.

Andy

But I can't seem to see what differentiates the step version from the slope version.

The difference is the shape of the curves. They are both PWL
(piecewise linear). The step one changes in steps, abruptly, followed
by periods where the resistance stays fixed. The slope one changes
like a triangular wave. Plot out the values to see for yourself.

Andy

The similation file of 74HC193 is bellow site.

...

As you probably know already, the 74HC193 and 74LS193 are completely
different technologies. The two chips might be interchangeable, and
the 74HC part might be easier to obtain today, but note that the SPICE
model for one would not be a good choice to use for the other part.

Just beware.

Andy

Hi

The similation file of 74HC193 is bellow site.
----------------------------------------------------------------------------------------------------------------------------------

-----------------------------------------------------------------------------------------------------------------------------------
I think it is a excellent file and it works correctly in LTspice.

Shiggy



2013/4/14 ltspice_ajax <ltspice_ajax@...>

**


I need 74ls193 for LTSpice

[Non-text portions of this message have been removed]

Hi Ron,
Thanks for sharing the inline table approach - very useful too! But I can't seem to see what differentiates the step version from the slope version. Can you elaborate please?
Thanks,
Mike

You would expect one peak to be quite a different from the other, because it is basically nonlinear. Consider a step input going from low resistance to high; the loop gain will be high and the bandwidth low. Step change in the opposite direction is to a higher bandwidth regime with low feedback resistance and low loop gain. So, rise and fall times could be quite different (so long as neither test exceeds the op-amp slew rate).

Thus, when driven with a sine, depending on the dR/dt, the waveform COULD be quite different at the positive and negative peaks.

Jim Wagner
Oregon Research Electronics

MOHAMMAD A MAKTOOMI <amaktoomamu@...> wrote:

Andy, how could you guess so quickly that waveform was missing some
peaks and the step-size be reduced in the setup

< ()?

I ask this because this will help me start thinking the way an experienced
user thinks.

When I looked at the waveforms, I noticed that the positive peaks of
the 1 MHz output were irregular, even when zoomed in. It looked to me
as if it suffered from aliasing, being sampled but not with a high
enough sampling rate. That led me to check for the plotwinsize=0
option, and then to add the max timestep parameter.

Andy

--- In LTspice@..., Andy <Andrew.Ingraham@...> wrote:

Yes, I did have the problem a few weeks ago, and at that time, I think,
everyone (more or less) did. The problem went away after a week or two,
with no explanation, as far as I know.

The problem never went away for me.

Until a month or so ago, when you clicked on it, it did open directly
(at least in most web browsers). Then that suddenly stopped working,
and hasn't worked since. All it lets me do is download it. This is
true in either Chrome or IE. Once downloaded, I can open it in the
same web browser with no problem. It is a pain in the neck having to
download it every time (or to fine the copy I downloaded last week),
but it does work.

If the problem went away for you, I wonder what you did to make it work.

And I wonder why Helmut never sees this problem. The last time around
that this was discussed (a few weeks ago?), it seems everyone had the
same problem, except for Helmut. Odd.

Whatever this is, I think it is some sort of Yahoo!groups problem.
Something about how they serve up the file to you, makes your web
browser think it needs to save it instead of open it.

Andy

Andy,

I have the same problem. I have no idea what the problem is, but I can live with it, because I don't need to use the file very often. I agree it is a Yahoo problem.

Rick

MOHAMMAD A MAKTOOMI <amaktoomamu@...> wrote:

Thank you, Andy for your hints.
But, in '.AC' we need to have an AC voltage or current source. Here, I don't
have any such thing (as I wish to vary the frequency of R2, NOT that of any
source), so how do I proceed?

Yes indeed, and I think that is why you must do this as a .TRAN
analysis. .AC analysis just doesn't work for your case.

Then .STEP the frequency.

With your 741 op-amp model, you do have an interesting situation
because of the severe distortion at higher frequencies (which by the
way only a .TRAN analysis would show, if there was a way to do this as
a .AC analysis). Given that, you need to consider what parameter of
output amplitude you measure: peak-to-peak vs. RMS vs. fundamental
amplitude?

Andy

Yes, I did have the problem a few weeks ago, and at that time, I think,
everyone (more or less) did. The problem went away after a week or two,
with no explanation, as far as I know.

The problem never went away for me.

Until a month or so ago, when you clicked on it, it did open directly
(at least in most web browsers). Then that suddenly stopped working,
and hasn't worked since. All it lets me do is download it. This is
true in either Chrome or IE. Once downloaded, I can open it in the
same web browser with no problem. It is a pain in the neck having to
download it every time (or to fine the copy I downloaded last week),
but it does work.

If the problem went away for you, I wonder what you did to make it work.

And I wonder why Helmut never sees this problem. The last time around
that this was discussed (a few weeks ago?), it seems everyone had the
same problem, except for Helmut. Odd.

Whatever this is, I think it is some sort of Yahoo!groups problem.
Something about how they serve up the file to you, makes your web
browser think it needs to save it instead of open it.

Andy

Helmut, is it possible to represent the information available from transient analysis of () in frequency domain? I mean, just as one can perform a transient of an RC circuit to see its charging-discharging profile and and AC analysis to see a first order low pass behavior, can the problem in thread have these two views?

Andy, how could you guess so quickly that waveform was missing some peaks and the step-size be reduced in the setup ()? I ask this because this will help me start thinking the way an experienced user thinks.

Hi Richard,
?
Aside from the form Helmut gave you, you can set any resistor to as many changes as you need. The format, which is entered in the resistance window in place of a single value is :
???
???? R=table(time,0,res1,time1a,res1,time1b,res2.....,last res)
?
Example : R=table(time,0,10K,3m,10K,3.00001m,100K,6m,100K,6.00001m,1)
?
This starts with the resistance value at 10K until 3mS after start, then steps to 100K until 6mS and then to 1 ohm for the rest of the run. The run duration sets the max time end point. Careful to place comma's between each entry, and no spaces.
?
If you want a slope change instead of a step, for example a changing value of 100 ohms to 10 ohms over a 10mS time and back to 100 ohms over the next 10mS,?the form would be:
?
???? R=table(time,0,100,10m,10,20m,100)
?
Lastly, always check your entries for pairs. There must be a time/value pair for each entry. This listing approach is invaluable for all sorts of applications which require either a slope or step change such as load testing etc.
?
- Cordially - RC
?


________________________________
From: Richard Norman <rnorman3@...>
To: "ltspice@..." <ltspice@...>
Sent: Sunday, April 14, 2013 2:30 AM
Subject: [LTspice] Step a resistor over time

?

Hi All, is it possible to step a resistor over time? Right now if I use the step command like this:

.step param R 100 1000 100
.tran 0 1ms 0 1

I
end up with a bunch of parallel horizontal lines. What I want is a
bunch of points (maybe even connected) over time. So in other words,
over the 1ms, I want the resistance to step from 100 to 1000. Is it
possible?

Thanks!
Richard

[Non-text portions of this message have been removed]




[Non-text portions of this message have been removed]

Hello Mark,

You could use my uploaded THD simulation. The results from
.FOUR agree with my formula. Give it a try.

THD = 100%*2*pi*sqrt(1/3*(D/T)^3)
=================================

Files > Temp > crossover_distortion.asc

Best regards,
Helmut

--- In LTspice@..., "Echidna" <mchambin@...> wrote:

Thanks a lot, you found where I was wrong.

Indeed, I was tricked by the .FOUR function default calculation on 9 harmonics only.
I tried with 99 harmonics and got a much different result.

The calculated THD is very sensitive to the number of harmonics
With D = 10 us ; T = 1ms
9 harmonics THD = 0.25%
49 harmonics THD = 0.56%
99 harmonics THD = 0.66%
199 harmonics THD = 0.69%
999 harmonics THD = 0.71%

I had done your calculation and did find THD proportional to (D/T)^1.5
Then, to check it, I used LTspice simulations giving me wrong results.
Then got confused.

I am confident the (D/T)^1.5 is right.
However I am unable to check it with LTspice simulations. It is so sensitive about the number of harmonics, I am not sure these simulation results make sense.

Best regards,
Marc.


--- In LTspice@..., "qrx3" <fredh@> wrote:

Greetings,

If we make some assumptions then the math becomes easy enough that even I can do it. First, imagine that you are taking a pure sinewave and subtracting a small signal from it to create the waveform you want. The pure sinewave has no distortion, while the small signal looks roughly like a tiny triangle starting at 0V and reaching some maximum voltage before dropping back to 0V, symmetrically at each zero crossing. That maximum voltage is sine(2 pi D / T), for small values of D/T this is about linear.

If we assume that your 'dent' is small enough that it does not appreciably affect the energy at the fundamental frequency (i.e. the distortion is 'small'), then the total distortion is roughly the ratio of the RMS value of small pulse divided by the RMS voltage of the fundamental.

It's simple to show that the RMS value of a triangular pulse is proportional to (D/T)^1.5 as Helmut stated in the other thread, so that should be what you observe in THD also. The reason this disagrees with what you found has to do with the way THD is calculated by the .FOUR function. Notice that the error log shows Fourier components up to 9th order, or 9kHz in your example. Note also that the magnitude of the distortion components is the SAME for each one (since your signal is symmetric there will only be odd components, ignore the even ones as they are noise), it has not started to drop off by the 9th. The calculated THD is based on only those four frequencies, it should be easy to imagine that if you calculated more components the THD would keep going up.

How many do you need? To get an idea, look at the width of your distortion pulse, 10us in your example. It's a good bet that the 'missing' pulse has components up to at least 100kHz. So you may need to go past the 99th harmonic to get anything accurate. Or, remove the fundamental and measure the RMS of what remains, much easier.

Once this is understood it's easy to see why your observation came out too low (not enough coefficients included in the measurement) and also why the observed exponent is too high (as D increases the number of important coefficients will be reduced, so the THD will not increase as fast as you observed, exponent of 1.5 instead of 2).

HTH,
Fred

--- In LTspice@..., "Helmut" <helmutsennewald@> wrote:

Hello Echidna,

Crossover Distortion
--------------------

I calculated the formula with Fourier series and approximation.

sin(x) = x-1/6*x^3

1-x^2 = 1-2*x

THD = 100%*2*pi*sqrt(1/3*(D/T)^3)
=================================

Files > Temp > crossover_distortion.asc

You can check the distortion calculated by .FOUR with
View -> SPICE Error log

Best regards,
Helmut


--- In LTspice@..., "Echidna" <mchambin@> wrote:

Hello.
How to calculate, using maths and definitions, the THD of this signal: A sine with a small dent at the crossovers.

S(t) = 0 for 0 < t < D where D is much smaller than T
S(t) = sin omega*t for D < t < T/2 where omega is 2*pi/T
S(t) = 0 for T/2 < t < T/2 + D
S(t) = sin omega*t for T/2 + D < t < T

Using LTspice simulations I find THD = 25*( D/T )^2
In this simulation I used a 1000 HZ SINE combined with a PULSE with values of D like 0.1u 0.2u 0.5u 1u 2u 5u 10u 20u 50u
These simulations gave me a THD that perfectly fits with 25*( D/T )^2

Here is the issue.
I was enable to prove this result with maths
(instead of simulations ).
I started computing the RMS value of the error signal (the dents at the crossovers)
With sin omega*t = omega*t ( valid for t << T )
I don' t get the 25*( D/T )^2 result, I do get a fonction of D/T, so far so good, but I don't get the right exponant. It seems my approach is wrong.

Is there a signal theory / maths guru who can give the proof that for such a signal S(t) ( with D << T ) we have THD = 25*( D/T )^2