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Re: MMBTH81 transistor model
Hello Kalias,
Has this model better worked? ** Product: MMBTH81 ** Package: SOT-23 ** PNP RF Transistor *----------------------------------------------------------------- .model MMBTH81 pnp + IS=1.364229E-15???? BF=109?????????? NF=1????????????????? + ISE=1.59087E-16???? NE=2???????????? VAF=100 + ISC=5.8e-9????????? BR=0.48????????? IKR=0.85 + IKF=0.150527??????? RB=46.6????????? RBM=0.159?????? + IRB=1.51356E-6????? RE=0.23????????? RC=1.0???????????? + CJE=2.63E-12??????? VJE=0.80???????? MJE=0.335??????? + FC=0.5????????????? CJC=2.77E-12???? VJC=0.757??????? + MJC=0.158?????????? XTB=1.59999????? EG=0.86 + XTI=3?????????????? TF=97.83e-12???? ITF=69.29 + VTF=10????????????? XTF=599e-6 *----------------------------------------------------------------- ** Creation: Jan.-15-2009? rev: 1.0 ** Fairchild Semiconductor Best regards, Helmut |
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Re: FFT Resolution
Actually I think we are in complete agreement. ?I just misunderstood the intent of what you wrote before. ?My bad.
Regards,
Andy |
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Re: MMBTH81 transistor model
kalias
Thanks Helmut, I took a look there but didn't see this model.? I ended up taking one from the datasheet. ? cheers ? kalias On Mon, Dec 16, 2013 at 1:11 PM, <helmutsennewald@...> wrote:
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Re: FFT Resolution
John Woodgate
In message
<CALBs-ThuY6qLHD4J3RdQsTdd9YP+1YphDpTzvxun4yTU9wM4DQ@...>, dated Mon, 16 Dec 2013, Andy <Andrew.Ingraham@...> writes: John Woodgate wrote:See 'in practice', words you use yourself in the critique below. I put the factor of 2 in and the words 'in practice' for precisely the reasons you have cited again, as you did in your previous message. We are in violent agreement but I fear that a thirst for rigor can result in added confusion. -- 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: MMBTH81 transistor model
Hello,
Please try this model from Fairchild-Semi. Click on "RF Transistors". http://www.fairchildsemi.com/models/modelDetails?modelType=PSPICE#resultDiv ** Product: MMBTH81 ** Package: SOT-23 ** PNP RF Transistor *----------------------------------------------------------------- .model MMBTH81 pnp + IS=1.364229E-15???? BF=109?????????? NF=1????????????????? + ISE=1.59087E-16???? NE=2???????????? VAF=100 + ISC=5.8e-9????????? BR=0.48????????? IKR=0.85 + IKF=0.150527??????? RB=46.6????????? RBM=0.159?????? + IRB=1.51356E-6????? RE=0.23????????? RC=1.0???????????? + CJE=2.63E-12??????? VJE=0.80???????? MJE=0.335??????? + FC=0.5????????????? CJC=2.77E-12???? VJC=0.757??????? + MJC=0.158?????????? XTB=1.59999????? EG=0.86 + XTI=3?????????????? TF=97.83e-12???? ITF=69.29 + VTF=10????????????? XTF=599e-6 *----------------------------------------------------------------- ** Creation: Jan.-15-2009? rev: 1.0 ** Fairchild Semiconductor Best regards, Helmut |
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Re: FFT Resolution
John Woodgate wrote: To get x Hz resolution, you should, in practice, simulate for 2/x Actually, x Hz resolution needs only 1/x seconds. Simulating for 2/x seconds gives you x/2 Hz resolution, which in practice might be useful if you know you have discrete frequency components at x Hz increments. ?Having those in-between components in the FFT plot both makes it easier to see the components, and can be used as a rough gauge of the FFT's accuracy. ?If components which should be very low, are not low, then something is wrong and you need to take a closer look.
The factor of 2 is needed only for the upper frequency limit.
Andy |
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Re: FFT Resolution
John Woodgate
In message
<CALBs-TipQ2tWb71L_jO+d3CysuyxmNHd1cAx+162RVC-OCHd9g@...>, dated Mon, 16 Dec 2013, Andy <Andrew.Ingraham@...> writes: You need at least two time samples at the highest frequency of the FFT.Thanks. I hope that makes it all clear: To get x Hz resolution, you should, in practice, simulate for 2/x seconds. You can simulate for longer to get a clearer spectrum display, say N/x seconds. To get a spectrum up to X Hz, you then need more than 2*N*X/x samples, preferably the next higher power of 2. I do think that, if it's correct, is a bit more lucid than the Help text. -- 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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MMBTH81 transistor model
kalias
I am trying to create a model for the MMBTH81 transistor.? I extracted the following model from the datasheet and put it in a LTspice schematic with a directive. ? .MODEL MMBTH81? PNP(Is=10f Xti=3
+ Eg=1.11 Vaf=100 Bf=133.8 Ise=1.678p + Ne=2.159 Ikf=.1658 Nk=.901 Xtb=1.5 Var=100 Br=1 + Isc=9.519n Nc=3.88 Ikr=5.813 Rc=7.838 Cjc=2.81p + Mjc=.1615 Vjc=.8282 Fc=.5 Cje=2.695p Mje=.3214 Vje=.7026 + Tr=11.32n Tf=97.83p Itf=69.29 Xtf=599u Vtf=10) ? The reason that I have picked this device is that it should provide a better gain bandwidth than a 3906.? However in my simulation I have two identical circuits and it seems as if the performance is the same.? Is there something wrong with this model?
? I compared the model file of this device to that of the 3906 and there appears to be something missing.? There is no Rb = or Re=.? Could this model be flawed or am I doing something incorrect? ?
kalias |
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Re: FFT Resolution
John Woodgate asked: This is all good stuff, but could you please comment in the OP's You need at least two time samples at the highest frequency of the FFT. 1/90MHz * 0.5 = 5.56 ns Helmut's example simulates for 20 ms, which actually gives you 50 Hz resolution, enough to see the "trough" between alternate components that are 100 Hz apart. ?Given that, you need at least 3600000 for the "Number of data point samples in time" to have time samples no greater than 5.56 ns apart. ?Using Helmut's recommended 4194304 (= 2^22) gives you that. ?That simulation's FFT goes up to (1/2) * 1/(20ms/4194304) = 104.858 MHz, which is what I see when I run it.
Using the bare minimum (3600000) probably stops exactly AT 90 MHz and isn't enough to show you the 90 MHz component. ?So always go higher if there is any doubt.
Andy |
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Re: FFT Resolution
John Woodgate
In message
<CALBs-TgcwTBCyW1+=wNz4R7Yk09QHzk1sbNYC1ii8HU7SecrFg@...>, dated Mon, 16 Dec 2013, Andy <Andrew.Ingraham@...> writes: Here are some tips.This is all good stuff, but could you please comment in the OP's question about analysing a 10 MHz square wave with a resolution of 100 Hz? Given that 20 ms is long enough to get 100 Hz resolution, how many data points are required to get up to, say, 90 MHz? -- 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: CD74HC4017E.sym missing pins
Steve wrote:
That depends on the component's SPICE model. Some SPICE models do not require power connections because the model itself implicitly includes power. A more complete SPICE model should have power and (floating) ground pins too. ?(By floating, I mean not already connected to GND = Node 0.)
Adding power and ground pins to an LTspice symbol should be easy. ?Just add them, like you added every other pin to the symbol. ?Are you trying to modify an existing symbol, or creating your own from scratch? ?It would help if you explained what the actual difficulty is, rather than just saying you "can't add them", since, of course you can.
Note that if you have a SPICE model in the form of a subcircuit for your component, the symbol must have exactly the same pins on it as the subcircuit has.
Regards,
Andy |
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Re: FFT Resolution
ronw6wo?wrote:
I think "the value of the window" as used before, refers to the time interval that was simulated and sent to the FFT. ?Not the number of cycles, but the number of seconds of time.
You could take Helmut's example and change V2 from a Sine wave source to a Pulse source.
Andy |
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Re: FFT Resolution
ronw6wo?wrote, " Here are some tips. (1) ?To get finer frequency resolution in the FFT results, you need to use longer simulation times. ?The two are directly related. For 100 Hz frequency resolution, the total time interval of the waveform data that is sent to the FFT must be 1/100Hz = 0.01 seconds = 10 ms. ?But to examine discrete components that are 100 Hz apart, you might want the frequency resolution to be at least twice as good as that (if not more) ... so simulate for 20 ms if not 40 ms or more. ?Then you can see the dips between the components too. ?Otherwise it tends to look like a big blob and it's hard to tell one component from the next.
(2) ?The "Number of data point samples in time" and the total time interval determine the upper frequency limit of the FFT. ?If the FFT doesn't go up high enough in frequency to show the components you want to see, repeat the FFT with more data point samples in time. ?Hopefully you have simulated with fine enough time steps so that the FFT isn't using fake data (see later, about "Tstep" and "plotwinsize").
An FFT works best when the number of data point samples is a power of 2 (hence LTspice starts with a default of 262144 data points), but I believe LTspice's algorithm does not require a power of 2. ?Just the same, I'd go with the power of 2 in case it impairs accuracy not to use it.
When in doubt, click "Reset to Default Values" at the bottom of the FFT dialog window, then use the up/down arrows to try other powers-of-two.
(3) ?LTspice's FFT display has a truly annoying characteristic when you zoom in to see the individual components of the FFT: ?Each component looks like an arch. ?I think that happens because LTspice does a linear interpolation between the points, BEFORE it computes a logarithm to turn it into Decibels for display. ?Hence you end up with those deceptive arch-like bumps. ?(Note: LTspice's developer says this is actually intentional. ?Not a bug. ?Could've fooled me.)
The apparent "width" of those bumps means nothing and does not imply a width of the components present! ?The tip of each bump is the ONLY thing that matters. ?The sides of the bumps are meaningless.
You can highlight the individual FFT components (right-click in the FFT window, then "Mark Data Points"), and then LTspice draws little dots at the actual FFT components. ?The curves between the dots are interpolations, not real data, and should be ignored. ?Be aware that turning on the display of those data point dots can make LTspice VERY slow and unresponsive when there are a very large number of data points on your screen. ?But this is the only way to see what's really there when your FFT's resolution is 100 Hz and you need to see frequency components every 100 Hz.
(4) ?Many things affect the apparent accuracy of the FFT results, i.e., the "noise floor" of the FFT plot, the appearance of extraneous signal components, etc. Most important is to simulate (and send to the FFT) an integral number of cycles. ?If you have multiple known components present, use an integral number of cycles of ALL of them, if you can. ?In Helmut's example, the time interval (20 ms) is a multiple of both 10.0000 MHz and 10.0001 MHz.
Here's why. The time-domain simulation generates a bunch of data points running from t=0 to t=Tstop = Stop Time. ?Think of turning that into a continuous waveform, by repeating that interval indefinitely. ?This is effectively what the output of the FFT represents. ?When you "splice" the waveform at t=0 to the waveform at t=Tstop, is it continuous, or is there a "glitch" at the splice? ?If there's a glitch, the waveform effectively has a modulation applied to it, which shows up in the spectrum as unwanted sidebands around everything and an increased noise floor.
In the FFT dialog box, the "Time range to include" controls the amount of waveform sent to the FFT. ?Normally one would "Use Extent of Simulation Data". ?A case where you might not do that, is when the simulation includes a start-up transient, and you want to include only the steady-state waveform after that transient has died. ?Make sure the time range you include, has the integral number of cycles, even if the total simulation time does not.
Simulating over many cycles seems to work better than simulating fewer cycles. ?I think it allows mathematical errors to average out better, and this reduces the FFT's noise floor and spurious components in the display.
When it is impossible to simulate for an integral number of cycles of all components, that's when one of the Windowing functions is used. ?My understanding is that they force the pre-FFT, time-domain data to fall somewhat gracefully to zero at t=0 and at t=Tstop. ?Thus, a nasty glitch is avoided when you "splice" the ends together. ?Instead, a gentler modulation is applied to your signal, so it still causes sidebands to appear in the output spectrum.
(5) ?It helps to turn off waveform compression in the time-domain data:
.options plotwinsize=0
Also, make Tstep (Maximum Timestep) as small as practical. ?The less interpolation between data points that the FFT needs to do, the better. (6) ?Enabling double precision may also help: .options numdgt=15 Regards,
Andy |
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Re: LTspice Install problem under linux
"Business Kid" wrote, "Is this a 'known issue?' (m$ speak for A BUG)? Where should I report it?"
The way to report bugs (if it continues to be one) is via email to: "LTspice @ " (remove the spaces).
For you other LTspice users: from within LTspice, click on
? Help > About LTspiceIV and it shows you the email address for bug reports and suggestions.
Regards,
Andy |
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Re: LTspice Install problem under linux
Hello, I also had a problem to update LTspice on two WIN7 PCs the last week. When I clicked on Sync Release, it simply did nothing. It behaved like when another LTspice is open. I rebooted the PCs and then the update worked. Please try immediately after reboot. Best regards, Helmut ? |
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LTspice Install problem under linux
Business Kid
I can't install the latest update in linux. It certainly doesn't go in painlessly like the last one did, and may need shoehorning in. That's a pity, because I uninstalled the older version. Is this a 'known issue?' (m$ speak for A BUG)? Where should I report it? |
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