wikihanya wrote:

"... since the RC value is 100ps. it should introduce a delay of 100 ps ..."

That was your mistake. An RC network has a frequency-dependent delay,
not a constant delay.

For the situation you described, the phase delay would be 32.14
degrees at 1 GHz, or 89.3 ps.

My LTspice simulation is in substantial agreement with that.

Regards,
Andy

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An RC filter does not 'delay' signals, but low pass filters them, with a frequency dependent phase delay. An RC value of 100ps says that the step response gets to 1/e of the final value in 100ps.

--
Richard Damon

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Hello 4cwnirjjs53rhghqimgmcb5zflk7qn7mkdjgphlk

We can't see the screen of your computer.

Please upload your example in the section Files\temp of this group.

Regards
PhB

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Brad,

One of the frustrating things for me about this conversation, is that
I don't think you have made clear what you are simulating, and what
your problem or question is.

You referred to a TIP142 test circuit that (I think) is in the group's
Files area. So I found it (two copies, they are identical, both named
"TIP142_test.zip") and downloaded and ran it. So far, so good. It
would have helped me, if you pointed to that file and said "This is
the thing I am simulating." Sorry, but with 100+MB of Files here, and
a wealth of SPICE stuff elsewhere on the Internet, and unknown what's
on your computer, it can be tricky to guess what you mean.

I had some trouble trying to find what you quoted that says, "Left
click to plot Q10 dissipation. V(c10)*Ix(Q10:C)+V(b10)*Ix(Q10:B)". I
guess what you are referring to, is what happens when you
press-and-hold the Alt key, and hover the mouse over the Q10 symbol.
That text appears as a tip in the lower left corner of LTspice.

Yes, LTspice shows you a formula. It shows you the formula that
LTspice will use when you click there -- the formula that calculates
the power dissipation of Q10.

OK, so now click the mouse button.

Voila -- a plot appears in the waveform window. That squiggly plot is
a plot of the quantity V(c10)*Ix(Q10:C)+V(b10)*Ix(Q10:B), which equals
the sum of powers into Q10. In other words, it is the power absorbed
by Q10. It even goes negative at points -- showing when the device
(through stored charge) is delivering energy back to the external
circuit.

Step back for a moment and remember what's happening. LTspice just
did a time-domain simulation. It knows all the currents and all the
voltages, and all the things that depend on currents and voltages, AS
A FUNCTION OF TIME. All those quantities are functions of time, and
they are known moment-by-moment.

So what is it that you are looking for?

I guess you are looking for the AVERAGE power dissipation of
transistor Q10. Am I right?

Power dissipation, like voltage and like current, is a function of
time. Remember that.

So this is where that second thing I referred to previously, comes in
handy. Go to the Help page for "Waveform Viewer -> Data Trace
Selection", and scroll down almost half way, until you reach the
paragraph between the 3rd and 4th figures on that page, which reads as
follows:

"Yet another schematic probing technique is to plot the instantaneous
power dissipation of a component. To do this, hold down the Alt key
and click on the body of the symbol of the component. The
instantaneous power dissipation will be plotted as an expression of
voltages and currents. It will be plotted on its own scale with the
units of Watts. The mouse cursor turns into an icon that looks like a
thermometer when it's pointing at a dissipation that can be plotted.
You can find the average power dissipation by control-clicking the
trace label."

See that last sentence? "You can find the average power dissipation
by control-clicking the trace label."

OK, so press-and-hold the Ctrl key, and click the mouse (left button)
on the label at the top of the waveform window -- yes, the one with
that formula.

A little pop-up window appears, showing, among other things, that the
Average power is 41.034 mW.

That is the average power over this interval of time in the plot
window. It is important to make sure the time interval is meaningful.
If your signals are periodic, you will want to make sure the time
interval corresponds to an exact whole number of cycles. And check
that nothing "funny" happens, say, at the very start of the simulation
(let's say, if the plotted thing shoots up to 50W for a very brief
moment, but never does that again).

I *think* this is the answer to your question?

Later, you wrote:

"Interestingly, I noticed that R2 said "dissipation = 0" but when I
plot it I see about 850 mW for about 50 us at 50% duty cycle."

It took me a minute to find what you are referring to.

LTspice shows a bunch of things in its lower left corner, depending on
what it knows, and what it thinks you want to do. When you hover the
mouse pointer near a net or wire, LTspice shows you the voltage on the
wire, at t=0, and only at t=0. It can only show you the voltage at
t=0 because the voltage is changing as a function of time. Right?

The value at t=0 is useful to many people because it is the "operating
point" solution. In the early days of SPICE, just finding the
operating point alone was a major accomplishment. Once you knew the
operating point, you could tell most everything about how the circuit
worked.

When you hover the mouse over a resistor, it shows you the current at
t=0, and the instantaneous power of that resistor at t=0. And only at
t=0.

When you Alt-left-click on that resistor, the plot window shows the
instantaneous power of R2 not just at t=0, but for all t (time) where
it was simulated. As you can see, that plot does indeed begin at 0mW
at t=0, and later jumps to 850mW.

Now, what LTspice shows you in the lower left corner, does depend on
the component. There is a difference between 2-pin and 3-pin
components because 2-pin components have only one current (the current
into and out of the component) but 3-pin devices have 3 unequal
currents. And there seems to be a difference between transistors with
.MODEL definitions and transistors with .SUBCKT definitions. I guess
this is where Darlingtons show different results than non-Darlingtons.
This is just the way SPICE works. That information may not be
available to SPICE, so LTspice can't show you. But you can get that
information by plotting Power(time).

I hope some of this helps.

Regards,
Andy

Hello Teodor,

Hello Brad,

Not an answer to your question, but there is **no such thing** as 'RMS power'. The product of RMS voltage and RMS current with zero phase difference is 'average power'.

For pedants, one can calculate a quantity by applying the calculus for RMS to instantaneous power, but it has no physical significance.

With best wishes DESIGN IT IN! OOO – Own Opinions Only
<> www.jmwa.demon.co.uk J M Woodgate and Associates Rayleigh England

Sylvae in aeternum manent.

From: LTspice@... [mailto:LTspice@...]
Sent: Thursday, August 4, 2016 3:47 AM
To: LTspice@...
Subject: [LTspice] Re: Sub circuit heat dissipation not showing


Helmut,

I know you said "All the Darlington transistors in my examples correctly plot power" But that's not my issue.



I can see the plot/s but I was thinking I would see "dissipation" in watts as with BJT transistors or other non-sub circuits.

I did download and run the TIP142 sub circuit files. I get the same thing. "Left click to plot Q10 dissipation. V(c10)*Ix(Q10:C)+V(b10)*Ix(Q10:B)". But as I think I said earlier I was expecting "dissipation = XXXX" at the end of the formula.



Interestingly, I noticed that R2 said "dissipation = 0" but when I plot it I see about 850 mW for about 50 us at 50% duty cycle. So wouldn't there be some power dissipation here??



Although the plot (for dissipation) is very useful I guess I was thinking I would see a single value for power dissipation (probably RMS power) for that darlington or even R2.

What I am missing something??

Remember I am a beginner and Thanks,
Brad


---In LTspice@... <mailto:LTspice@...> , <helmutsennewald@... <mailto:helmutsennewald@...> > wrote :
Hello,

The power dissipation of the TIP142 from our Files section will be correctly displayed.
TIP_142_test.zip
ahoo.com/neo/groups/LTspice/files/%20Lib/ <>

Here are a few more examples. Their power will be displayed too.


All the Darlington transistors in my examples correctly plot power.

In therory there are subcircuits possible where LTspice doesn't plot the power due to special combinations of sources internally connected to the pins of a subcircuit.

You should upload your files for a test.



Best regards,
Helmut






I download and ran the TIP142 sub circuit files. I get the same thing. "Left click to plot Q10 dissipation. V(c10)*Ix(Q10:C)+V(b10)*Ix(Q10:B)" I can see the plot but I was thinking I would see "dissipation" in watts as with BJT transistors.



Interestingly, I noticed that R2 said "dissipation = 0" but when I plot it I see about 850 mW for about 50 us at 50% duty cycle. So wouldn't there be some power dissipation here??



Although the plot (for dissipation) is very useful I guess I was thinking I would see a single value for power dissipation (probably RMS power) for that darlington or even R2.

What I am missing something??

Thanks,
Brad


---In LTspice@... <mailto:LTspice@...> , <helmutsennewald@... <mailto:helmutsennewald@...> > wrote :
Hello,

The power dissipation of the TIP142 from our Files section will be correctly displayed.
TIP_142_test.zip


Here are a few more examples. Their power will be displayed too.


All the Darlington transistors in my examples correctly plot power.

In therory there are subcircuits possible where LTspice doesn't plot the power due to special combinations of sources internally connected to the pins of a subcircuit.

You should upload your files for a test.



Best regards,
Helmut

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From the available data we have not seen Br depending temperature change from 25° C to 100° C.

Bordodynov.

05.08.2016, 15:56, "hkafeman@... [LTspice]" <ltspice@...>:

Borodoynov

Thank you that looks like a good typical material to use.

My understanding is then that given that I am operating at only 50Hz with just some high frequency transients, that any core losses will be insignificant.

Then I can model in LTSpice the effects of ambient temperature using e.g.:

Hc = 22 at 25C decreasing by 0.12 % per C
Bs = 490m at 25C decreasing by 0.27 % per C
Br = 100m at 25C - Does this have a temperature coefficient?

Thanks and Regards
Henry Kafeman

Hello Henry Kafeman.
See material N41
BS (25 °C)=490mT
BS (100 °C)=390mT
Hc (25 °C)=22A/m
Hc (100 °C)=20A/m
Br=0.1T

Power
Transformer (low loss)

f<100 kHz

N27 (?i = 2000 ) low cost, hi-power
N41 (?i = 2800 ) current transformer
N51 (?i = 3000 ) loss min. @ 40°C

Bordodynov.

05.08.2016, 14:27, "HKafeman hkafeman@... [LTspice]" <ltspice@...>: