Re: Limit() function in waveforms not working
John Woodgate wrote: "I think it's too simple: it isn't obvious what it could mean. Actually, I don't understand the statement in the Help, either: what is the 'intermediate value' of three other than the middle one, which is
utterly trivial - limit(x,y,z) would be y, for ever." That description in the Help pages had me stumped for a long time too. ?The "intermediate value" just means the one that is numerically between the other two ... no matter in what order they appear.
limit(1,2,3) = 2 limit(2,3,1) = 2 Andy
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Re: Limit() function in waveforms not working
John, The intermediate value is the value that is between the other 2.? The limit function determine which values are the minimum and maximum, and then selects (plots) the third one.
Andy, It is very easy to build small test circuits to see how different functions operate.? Try it. Rick
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Re: charge amplifier using OPA129; but, imported model (OPA129) doesn't work correctly.
I just looked at the op-amp's specs, and an input bias current of 15 pA is too large.
I don't know where that is coming from, but it seems to be in the op-amp model somehow.
Andy
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Good to hear Yahoo keeps on improving... I suppose this means new upload is next year. Thank you for the info.
A Happy New Year to you, all Vlad
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Re: charge amplifier using OPA129; but, imported model (OPA129) doesn't work correctly.
Your simulation file is "20131231_test.asc" located in the "Temp" directory. ?The long URL you posted is useless, because of the way Yahoo does things.
Your simulation might be working correctly. ?I ran it, and the results make some sense, depending on the characteristics of the op-amp (which I haven't investigated).
The op-amp model appears to have an input offset voltage of about 0.936 uV.
It also seems to have an input bias current of about 15 pA. ?As a result, when the switch releases, the output voltage starts slewing at a rate of about -15 mV/s.
When the pulse source representing the photodiode turns on and adds another 100 pA, the output starts moving at -115 mV/s, as expected. ?Then it goes back to the -15 mV/s due to bias current.
Take away the 15 pA input bias current, and the output waveform probably looks almost like you expected. ?Cancel the input offset voltage and it's even better.
Andy
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Re: charge amplifier using OPA129; but, imported model (OPA129) doesn't work correctly.
Hello The opamp doesn't like the pure capacitive load of 1nF in series to 1.1nF. Now I added a 10Ohm resistor (which may be in the diode model anyway) to make the opamp happy.? I also added a V-source to compensate the offset voltage of the opamp.
I have uploaded my circuit. Just unzip it into any folder.
All Files > Temp >? http://groups.yahoo.com/neo/groups/LTspice/files/%20Temp/
Best regards, Helmut
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Re: Limit() function in waveforms not working
In message <CALBs-TjThoJ4iiaU4Hs2LN5HUu1FJWkdf+nHvVOBLhJy298egQ@...>, dated Tue, 31 Dec 2013, Andy <Andrew.Ingraham@...> writes: limit(1,2,3) appears to be one of them, though I only used it here to simplify what I wrote. I think it's too simple: it isn't obvious what it could mean. Actually, I don't understand the statement in the Help, either: what is the 'intermediate value' of three other than the middle one, which is utterly trivial - limit(x,y,z) would be y, for ever. -- OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk Nondum ex silvis sumus John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
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Re: Limit() function in waveforms not working
Then I guess I have to ask ...
What known limitations are there in the way limit() may be used, in the waveform viewer?
Apparently there are some circumstances where it is not accepted, whether the operands are constants or waveforms or formulas. ?Do we know what those constraints are?
limit(1,2,3) appears to be one of them, though I only used it here to simplify what I wrote.
Andy
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charge amplifier using OPA129; but, imported model (OPA129) doesn't work correctly.
Happy New?Year!? ? I have?made a small schematic of a photodiode charge
amplifier - photocurrent integrator?(please find the Filelink
below.) The photocurrent from?the photodiode part (pulsed
current source // diode // capacitor // resistor) flows to the feedback capactor 1000pF of OPA129 with a
reset switch. To model OPA129, I have imported a SPICE code of
OPA129?obtained from the TI website. ? As a result of transient simulation, the result (Vout)
did not show?integrated values? of the photocurrent but quite strange values,
unfortunately.? ? I?tried changing the opamp with?AD549 (Analog Device,
imported model by the same way of OPA129), it was?also as strange as the OPA129's although?it
appeared?a?little bit differently?from the OPA129's). ? Again, I tried changed it with?LT's built-in opamps.
They?showed?good results of integration, suprisingly. ? Could you tell me what?was wrong with my
simulation?? Imported models?often make problem? ? Seongchong ? FileLink:
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Re: Limit() function in waveforms not working
In message <CALBs-TgDOGn8n=-=pebpYJBKQsUMYqAVCheKmzaSqWo2ZPO3gA@...>, dated Tue, 31 Dec 2013, Andy <Andrew.Ingraham@...> writes: Richard Sawrey wrote, "One of the values must be a variable, and the
other 2 may be a variable or constant."
So to use it, I need to add a .PARAM statement to convert my formula
(involving one or two waveforms) into a variable, and then I can use it
in a limit() function?
? Aren't waveforms inherently variables - functions of time? Maybe you need to express your waveform differently from the way you've tried. -- OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk Nondum ex silvis sumus John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
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Re: Using HSPICE model directly in LTSPICE
Perhaps the error is in the conversion to PSpice?? How did you do
it?? To use in PSpice you would need 9 separate models to
cover the PMOS size range.
???? ...Jim Thompson
???? Web Site:
<>
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Re: Limit() function in waveforms not working
Andy, My use of the word "variable" may have been inappropriate.? What I meant was "one of the values must be a signal that is generated during a simulation run."? These expressions will work:?limit(V(x),1,-1), limit(V(x),V(y),1). Rick On Tuesday, December 31, 2013 9:12 AM, Andy wrote:
?
Richard Sawrey wrote, "One of the values must be a variable, and the other 2 may be a variable or constant."
So to use it, I need to add a .PARAM statement to convert my formula (involving one or two waveforms) into a variable, and then I can use it in a limit() function?
?
Hmm, I'll have to give that a try.
Odd that you can't use a waveform itself within the limit() function because it is not a variable ?Who would have known. ?Is that true of all the functions, that waveforms need to be converted into variables first before you can act on them in a waveform function? ?Apparently not all, because the Help page shows that the pow() and abs() functions can act on waveforms directly. ?So why is limit() different, and which other ones can't use waveforms as their operands?
Thanks,
Andy
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Re: Limit() function in waveforms not working
Richard Sawrey wrote, "One of the values must be a variable, and the other 2 may be a variable or constant."
So to use it, I need to add a .PARAM statement to convert my formula (involving one or two waveforms) into a variable, and then I can use it in a limit() function?
?
Hmm, I'll have to give that a try.
Odd that you can't use a waveform itself within the limit() function because it is not a variable ?Who would have known. ?Is that true of all the functions, that waveforms need to be converted into variables first before you can act on them in a waveform function? ?Apparently not all, because the Help page shows that the pow() and abs() functions can act on waveforms directly. ?So why is limit() different, and which other ones can't use waveforms as their operands?
Thanks,
Andy
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Questions to LTC1043 model / SPDT switches
Hello,
I'm going to design a voltage-to-frequency converter (and also f-to-V-converter, need linearity < 0,01% at (500¡À200) kHz carrier) using a switched capacitor circuit e.g. described at LTC1043 datasheet.
Problem: The LTC1043 model does not work with external clock to override the internal oscillator. May be other SPDT switches will do the job, but I found no models.
I tryed to use LT's vswitch model to design a SPDT, but this is awful when you need break-before-make characteristics at different frequencies.
Some ideas? Don't say AD650.
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Re: Limit() function in waveforms not working
Andy, One of the values must be a variable, and the other 2 may be a variable or constant. Rick On Monday, December 30, 2013 10:19 PM, Andy wrote:
?
The Help page for "Waveform Arithmetic" show a limit function, limit(x,y,z).
Every time I try to use it in a waveform, it generates an error:
? Error: undefined symbol in: "<>(1,2,3)"
It seems to be saying it doesn't understand the "limit" function. ?Is this a known problem?
Thanks, Andy
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Limit() function in waveforms not working
The Help page for "Waveform Arithmetic" show a limit function, limit(x,y,z).
Every time I try to use it in a waveform, it generates an error:
? Error: undefined symbol in: "<>(1,2,3)"
It seems to be saying it doesn't understand the "limit" function. ?Is this a known problem?
Thanks, Andy
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Re: Using HSPICE model directly in LTSPICE
Jim, Thanks for the reply. Yes,?pspice doesn't support binning.? Part of reasons we converted hspice model to pspice model was because of this. From what I learned from Mike Engelhardt, ltspice does support binning.
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Re: how can i calculate inverter power efficiency by ltspice?
Thanks for kind replies.
I used??method for solving my problems, but it doesn't work..
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Re: how can i calculate inverter power efficiency by ltspice?
Heres yet another trick I use for all my test schematics to measure input current, efficiency, and total internal power dissappation. Since I use this approach universally I automatically add the necessary components and steps as a template for all my schematics, thereby greatly reducing the task.?Time can also be saved by using a standard nomenclature for the parts such as Rin1 for the input test resister, etc. Once you have done this to a schematic the resulting label
formula's can be copied and pasted from schematic to schematic so only slight modifications will be necessary with new circuits. ? 1) Add a network in the input line between the test circuit and DC power source consisting of a 1 microohm (0.0001 ohm) resister in parrallel with a 1000000 uF capacitor. The capacitor is used to suppress excess ripple and improve integration accuracy. ? 2) Make one display panel to show input current by
clicking on the resister. Add "*-1" to the panel label for positive current flow if necessary. ? 3) Make another display window that will show overall efficiency by clicking on the resister for current and then modifying
the label to the calculation form: ? (v(out1)*I(Rload1))/(v(in)*I(Rin)*-1) ? Note: If there is more than one output voltage/current from the power supply circuit under test then add to the numerator of the formula thus: ? ((v(out1)*I(Rload1))+(v(out2)*I(Rload2)).....+(v(outn)*I(Rloadn)))/(v(in)*I(Rin)*-1) ? 4) Make a third display window that will show overall internal total power dissapation by first clicking on the resister to set the display to input current and then changing the label to the calculation form: ? (v(in)*I(Rin)*-1)-(v(out1)*I(Rload1)) ? Note: again if there is more than one output then the power dissapation of each output must be summed as follows: ? (v(in)*I(Rin)*-1)-((v(out1)*I(Rload1))+(v(out2)*I(Rload2))....+(v(outn)*I(Rloadn))) ? Additional note: It may be neccessary to add a "*-1" multiplier if a negative output voltage is to be added into the calculations. The wrong sign will result in subtracting instead of adding to the overall dissappation. If you are unsure run a window with that output seperated to show the calculated output power to make sure it is positive, adding the "*-1" if neccessary to correct that portion of the formula. ? - The "*-1" added to the input current
label and input power portion of the formulas is to correct for current flow as positive. ? - After the analysis run is complete the integrated values of each panel display measurement?can be displayed by using the mouse cursor to select a range of clock pulses from the display panels. I generally try to select at least 40 to 100?clocks for good integration accuracy. The integrated measurement is then displayed in the usual manner by holding down the
key, positioning the mouse cursor over the panel label of interest, and left mouse button clicking. I often enter the data as text in a tabular format right on the schematic so it will be automatically saved for future reference. ? - A final note. Before making the measurement steps use the "*1" trick added to any of the panel labels?to force vertical expansion. This allows input ripple currents and output ripple voltage peak to peak values
to be measured off the panel traces easily using the mouse cursor "box" feature to display the value at the bottom of the screen in the dynamic data line. Either adding or subtracting the "*1" to/from the label will result in forced vertical?expansion without effecting the calculated values. ? Cordially - RC ? ? ? On Monday, December 30, 2013 1:57 PM, "sawreyrw@..."
wrote:
?
Jungyeon, You could just measure the input and output power and compute it yourself.? The help file on steady also tells you how to do it if the steady state criteria is too tight. Rick
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Hello,
I also had a very few weeks ago this inconvenience or problem.
Be patient. Yahoo will delete it after 2 days.
Best regards,
Helmut
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