LTspice@groups.io

Re: All_Files.Htm

--- In LTspice@..., "Basier philippe" <basier.philippe@...> wrote:

Hello
How to open all_files.htm ?
Regards
Philippe

Hello Philippe,

You should open it with any Internet browser.

IE, Firefox, Chrome, Opera, ...

Normally it's automatically opened by your standard browser
when you click on it.

Files > Tables of Contents > all_files.htm

Best regards,
Helmut

Re: All_Files.Htm

John Woodgate

#62663

In message <kker0j+82e6@...>, dated Sun, 14 Apr 2013, Basier philippe <basier.philippe@...> writes:

How to open all_files.htm ?Has it gone wrong again? It should open in any browser.

I just tried it and it opens OK. But it is a very large HTML file, so it may take time to download, or be corrupted during download.
-- 
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

All_Files.Htm

#62661

Hello
How to open all_files.htm ?
Regards
Philippe

CD 4566 / Programmnable Timer / Texas Instruments

[email protected]

#62660

Hello,

I need for LT Spice the simulation model of the CD 4566.
Is the model avaiable anywhere?

Best regards,

Arnd

I need 74ls193 for LTSpice

#62659

I need 74ls193 for LTSpice

Re: inverting opamp simulation: rapid component variation

#62658

Hello,

You can only use .STEP to change the values of components in
the .AC simulation.

Best regards,
Helmut

Show quoted text

--- In LTspice@..., "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?

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

MOHAMMAD A MAKTOOMI <amaktoomamu@> wrote:

... Note: In your circuit, you may need to decrease the maximum timestep
down to about 10ns.  (.tran 0 0.1m 0 10n)  The output waveform is
highly distorted at the highest frequency and LTspice was missing the
narrow peak before doing that, making the amplitude look a lot less.

It probably also means the result is unrealistic at that frequency,
but that's another matter.

Andy

Re: Photo Transistor

#62657

--- In LTspice@..., "mase723777" <mimmo_renato@...> wrote:

Dear all,

I am new to LTspice and electricity(my dad is my teacher), I was looking on google how to add a photo transistor. The only useful thing I found was to add a Photo coupler. I tried this in a circuit with a n-Channel mosfet and it was successful. But when I want to change the parameters of the Photo coupler it gives me a notification it is not possible. How can I override this?

Many thx. for your feedback,

Kr,

RenatoHello Renato,

Nearly all symbols in the LTspice library are intentionally
made as specific symbols. This prevents that users corrupt
their database.

One can modify this symbol to allow to change the parameter
Igain for example.

I have made a new symbol PC817.asy for you. Please try my example.

Files > Temp > PhotoTransWithMOSFET.zip

Its best practice to copy this symbol into every folder where
you have a schematic using this new symbol.  

Best regards,
Helmut

Re: Step a resistor over time

#62656

--- In LTspice@..., Richard Norman <rnorman3@...> wrote:

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
Hello Richard,

You can use formulas instead of a fixed value for a resistor.
Therefore replace the value of 1k on your resistor with a
formula. See below.

R=100*(1+time/10m)

This will rise the resistance from 100Ohm to 1kOhm in 90ms.

Best regards,
Helmut

Re: THD of a sine with a small dent at crossover

#62655

Hello Echidna,

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

I calculated the formula with Fourier series and approximation.

sin(x) = x-1/3*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

Show quoted text

--- 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

Re: THD

#62654

Yes, I found 25*(D/T)^2 by trial.
Running simuations that give me THD, I see it fits perfectly with D/T from 0.02 to 0.
Yes, I meant "unable".

Now, I understand better.
The 25*(D/T)^2 is quite normal.
Because I am looking for small D/T the fonction has to behave like.
THD = a*( D/T) + b* (D/T)^2

The only thing that would need a mathematical proof is that there is no D/T term.
I think this is because the THD fonction is and odd fonction of D/T. By odd fonction, I mean f(-x)=-f(x). May be, the proof of this is not too difficult.

Show quoted text

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

In message <kkckve+t28o@...>, dated Sat, 13 Apr 2013, Echidna
<mchambin@...> writes:

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 )^2I think you mean you found 25*(D/T)^2 by trial. This can be misleading.
I once found a power function that matches half a sine wave very
accurately, but outside the range 0 to pi it doesn't.

Here is the issue.
I was enable to prove this result with maths'unable', I think.

(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.The function you do get might match 25*(D/T)^2 over a range of D/T.

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 )^2Did you try using Fourier analysis in you manual calculations?
--
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

Re: THD

John Woodgate

#62652

In message <kkdr4u+gvv5@...>, dated Sun, 14 Apr 2013, Echidna <mchambin@...> writes:

Indeed, I found 25*(D/T)^2 by trial.
Running simulations that gave me THD values for various D values. I have uploaded the .asc file of that simulaton.Thanks.

"enable" is a spelling mistake. I meant "unable".I just wanted to prevent others being confused.

I did try using Fourier analysis in my manual calculations. Calculation of the Fourrier coefficient is extremly difficult,Agreed. I will try using Mathcad, but I can't do it immediately.

this is why I tried a workaround. I am puzzled to see such a complex calculation while the THD = 25*( D/T )^2 result is so simple.It happens sometimes.

This result fits so well with simuulation trials, I doubt it is so by chance.Maybe.

I notice a small point. Your pulse starts at the zero-crossing. To simulate crossover distortion correctly it should start at time t/2 before the zero-crossing and end at time t/2 after it.
-- 
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

Step a resistor over time

#62653

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

Re: THD

#62651

Indeed, I found 25*(D/T)^2 by trial.
Running simulations that gave me THD values for various D values.
I have uploaded the .asc file of that simulaton.

"enable" is a spelling mistake. I meant "unable".

I did try using Fourier analysis in my manual calculations. Calculation of the Fourrier coefficient is extremly difficult,
this is why I tried a workaround.
I am puzzled to see such a complex calculation while the THD = 25*( D/T )^2 result is so simple. This result fits so well with simuulation trials, I doubt it is so by chance.

Show quoted text

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

In message <kkckve+t28o@...>, dated Sat, 13 Apr 2013, Echidna
<mchambin@...> writes:

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 )^2I think you mean you found 25*(D/T)^2 by trial. This can be misleading.
I once found a power function that matches half a sine wave very
accurately, but outside the range 0 to pi it doesn't.

Here is the issue.
I was enable to prove this result with maths'unable', I think.

(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.The function you do get might match 25*(D/T)^2 over a range of D/T.

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 )^2Did you try using Fourier analysis in you manual calculations?
--
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

Re: Photo Transistor

#62650

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

I am new to LTspice and electricity(my dad is my teacher), I was looking on
google how to add a photo transistor. The only useful thing I found was to add
a Photo coupler. I tried this in a circuit with a n-Channel mosfet and it was
successful. But when I want to change the parameters of the Photo coupler
it gives me a notification it is not possible. How can I override this?This is likely a MS-Windows problem, but you haven't given us nearly
enough information to say for sure.

You probably put a file in a directory owned by Windows (perhaps a
subdirectory of C:/Program Files/ where the LTspice program itself
is), and Windows doesn't let you modify it.  Morale: Never put your
schematics, models, symbols, subcircuits, or netlists under the
Program Files folder.

If your interpretation of "it is not possible" differs from mine, then
maybe I am not on the right track.  If so, you will have to give us a
lot more information.  Or, upload your file(s) as instructed on this
group's main page and show us what didn't work.

Regards,
Andy
Dear Andy,

Thx. for your reply, I indeed saved the library files under program files. When I right click on the photo coupler it gives me following message "High density mounting photo coupler This component cannot be edited". The strange thing is that all other components can be edited and they are saved in the same library. I will upload the file on the group main page to make it more clear.

Many thx. for your feedback,

KR,

Renato

Re: inverting opamp simulation: rapid component variation

John Woodgate

#62649

In message <kkdlr2+p199@...>, dated Sun, 14 Apr 2013, MOHAMMAD A MAKTOOMI <amaktoomamu@...> writes:

Actually, a DC input voltage is present. But, I don't know if it will be ground in AC analysis.It will be ignored, which is equivalent.

If that's grounded then how could I have output (ignore offsets at this moment).I don't see a solution if R2 varies at frequencies as high as the op-amp will handle. If R2 varies more slowly (*), you could put in a sine wave signal at a higher frequency to act purely as a 'carrier' which is amplitude modulated by the variation of R2. Your output would then be the peak-detected and filtered carrier.

(*) However rapidly R2 might vary in reality, you can always slow it down for simulation, e.g. using a frequency range of 0.1 Hz to 1000 Hz with a 10 kHz carrier. Obvious, you need to simulate for several tens of seconds in such a case.
-- 
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

Re: THD

John Woodgate

#62648

In message <kkckve+t28o@...>, dated Sat, 13 Apr 2013, Echidna <mchambin@...> writes:

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 )^2I think you mean you found 25*(D/T)^2 by trial. This can be misleading. I once found a power function that matches half a sine wave very accurately, but outside the range 0 to pi it doesn't.

Here is the issue.
I was enable to prove this result with maths'unable', I think.

(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.The function you do get might match 25*(D/T)^2 over a range of D/T.

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 )^2Did you try using Fourier analysis in you manual calculations?
-- 
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

Re: inverting opamp simulation: rapid component variation

#62647

Hello John,

Actually, a DC input voltage is present. But, I don't know if it will be ground in AC analysis.If that's grounded then how could I have output (ignore offsets at this moment).

Show quoted text

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

In message <kkdb2u+pa7l@...>, dated Sun, 14 Apr 2013, MOHAMMAD A
MAKTOOMI <amaktoomamu@...> writes:

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?Varying the value of R2, with no other input signal, doesn't produce any
output, unless it's from DC offset voltage or current, or noise. You can
add offset and/or noise voltage and current generators at the input and
run, but only by using .TRAN. Time does not exist in an .AC or AC sweep
analysis.
--
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

Re: inverting opamp simulation: rapid component variation

John Woodgate

#62646

In message <kkdb2u+pa7l@...>, dated Sun, 14 Apr 2013, MOHAMMAD A MAKTOOMI <amaktoomamu@...> writes:

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?Varying the value of R2, with no other input signal, doesn't produce any output, unless it's from DC offset voltage or current, or noise. You can add offset and/or noise voltage and current generators at the input and run, but only by using .TRAN. Time does not exist in an .AC or AC sweep analysis.
-- 
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

Re: inverting opamp simulation: rapid component variation

#62645

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?

Show quoted text

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

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

... Note: In your circuit, you may need to decrease the maximum timestep
down to about 10ns.  (.tran 0 0.1m 0 10n)  The output waveform is
highly distorted at the highest frequency and LTspice was missing the
narrow peak before doing that, making the amplitude look a lot less.

It probably also means the result is unrealistic at that frequency,
but that's another matter.

Andy

Re: inverting opamp simulation: rapid component variation

#62644

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

... Note: In your circuit, you may need to decrease the maximum timestep
down to about 10ns.  (.tran 0 0.1m 0 10n)  The output waveform is
highly distorted at the highest frequency and LTspice was missing the
narrow peak before doing that, making the amplitude look a lot less.

It probably also means the result is unrealistic at that frequency,
but that's another matter.

Andy

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