OP:-
Rectifier circuits are normally explained using sine waves.
The moment you start hitting them with square waves with a fairly fast risetime, the facts of life intervene.
Diodes are not perfect, and a reverse biased diode looks like a capacitor .
You can approximate this, in an (unachievable) circuit using perfect components, by imagining a small capacitor connected in parallel with the diode.
The result will be that the diode does not "clip" the high frequency component of the fast rising square wave.
In this respect, it looks as if LTspice is actually doing its job.
That said, there are different types of diode optimised for different applications---with some, the capacitance in the reverse biased direction is minimal, & others it is quite high, because other parameters are regarded as more important in tne suggested applications.
You say you have a DSO &.function generator----- look at real world circuits rather than simulations, think about how circuits should work, why they are a bit different from what you expected, & so on.
Another thing which would be nice would be if you could find some old transformer type ac "wall warts" & using their low voltage secondaries, build a few rectifier circuirs using real Mains frequencies
The explanations in the text you have shown are horribly "clunky".
I would suggest reading the "ARRL Handbook" (especially the older editions) to get a better idea of what your existing text is trying to communicate.
I've done the same simulation in Falstad and Vout is accurate there( attached screenshot ) . Confirms LTspice is culprit.
I've done the same simulation in Falstad and Vout is accurate there( attached screenshot ) . Confirms LTspice is culprit.
I would not trust Falstad for anything but a simplistic analysis.
Consider the following.
When a diode is forward biased, the depletion region shrinks to a very small (relative value). When you instantaneously reverse bias it, those charges take time to move away from the pn junction. In the mean time, the capacitance across the diode is much higher than when steady state reverse bias. I think that the reverse recovery time specification of a diode captures this feature.
So, the glitch has nothing to do with reverse current and all to do with a very non-linear capacitance.
I've done the same simulation in Falstad and Vout is accurate there( attached screenshot ) . Confirms LTspice is culprit.
I would not trust Falstad for anything but a simplistic analysis.
Consider the following.
When a diode is forward biased, the depletion region shrinks to a very small (relative value). When you instantaneously reverse bias it, those charges take time to move away from the pn junction. In the mean time, the capacitance across the diode is much higher than when steady state reverse bias. I think that the reverse recovery time specification of a diode captures this feature.
So, the glitch has nothing to do with reverse current and all to do with a very non-linear capacitance.
There you go! 1n4148 in this test.
having hard time understanding dB plotting .
the left side scale shows 0.001 to 1 on y-axis . Wondering how 6dB / octave or 20dB / decade were converted from dB to range on y-axis from (0.001 to 1 ) .
having hard time understanding dB plotting .
the left side scale shows 0.001 to 1 on y-axis . Wondering how 6dB / octave or 20dB / decade were converted from dB to range on y-axis from (0.001 to 1 ) .
Does this help?
What I am really showing is dBV which is dB relative to 1 volt.
having hard time understanding dB plotting .
the left side scale shows 0.001 to 1 on y-axis . Wondering how 6dB / octave or 20dB / decade were converted from dB to range on y-axis from (0.001 to 1 ) .
Does this help?
What I am really showing is dBV which is dB relative to 1 volt.
Hello Wim. Thank You . It helped 50% in understanding 20 dB per decade . Can you help me understanding 6dB / Octave on this same plot please.
having hard time understanding dB plotting .
the left side scale shows 0.001 to 1 on y-axis . Wondering how 6dB / octave or 20dB / decade were converted from dB to range on y-axis from (0.001 to 1 ) .
Does this help?
What I am really showing is dBV which is dB relative to 1 volt.
Hello Wim. Thank You . It helped 50% in understanding 20 dB per decade . Can you help me understanding 6dB / Octave on this same plot please.
An equation may make more sense.
... irrespective of whether its octave or decade scales , slope will not change.
If you want to study electronics, you need at least high school level math. You don't need to derive formulas with calculus etc., but you have to know how they work of course. I am a mechanical engineer and i am self studying electronics. A book that is not heavy on math and i think is best for starters is Electronic Principals from Malvino and Bates. I am currently at the chapter where they explain JFET's
Interesting question about "how much math, calculus, differentials, etc." is essential to advance in the hobby or profession of electronics.
It reminds me of the question I had in seminary: Should we be required to learn to read and speak Greek and Hebrew, or is it ok to just learn how to use the tools that express the meanings? I much preferred the second path because I suck at languages, and two semesters of that torture was more than I could stomach. I went back to my day job. The tools can be challenging enough, never mind trying to learn how the tools were made.
If you want to teach it, that may be another matter.
I am assuming something , but want to make sure that's right.
For the problem in attached reference " Imaging putting 1uF Cap across 115 Volt (rms) 60 Hz powerline ? what's current flow was the question.
I understand how the current was derived . But i don't understand how based on the result one can say which one is leading what "Current leading voltage or voltage leading current and by what angle "
My assumption is Given "Voltage" was taken as cosine i.e. 90 degrees before time t=0 begins ( please correct me if i am wrong ) , and the resultant current I came out to be sine (wt) i.e. crosses point t=0 . Therefore voltage is leading current . Is that correct ? (Given in text is opposite i.e. " Current leading voltage". Or is my assumption that cost will always start at 90 degrees before t=0 is incorrect ?
I tried to draw phasor diagram of this current and voltage, I was a bit skeptical to about the current on positive x-axis as the result i received was negative -0.6sine(wt). I made two representations , Appreciate if some one can guide to say which one is right one..
C=1F
v(t) = Acos(wt)
A= 115/root(2) rms volts
v(t) = 163cos(wt)
Ic= C*dv/dt
=> 10^-6 * d ( 163cos(wt) )
=> 10^-6 *w*A*-sin(wt)
=> 10^-6 * 6.2*60*163 =
=>-0.06sin(wt)