Activity for a concerned citizen
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #286114 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Duty cycle of buck-boost converter If you're referring to this Wikipedia page then the topology has a negative output. In that case, the problem as you have it stated in the OP, is misleading: it should be Vo=-20. With that, extracting D will give: $$D=\dfrac{V{out}}{V{out}-V{in}}=\dfrac{-20}{-20-24}=0.4545$$ Sure enough, a quic... (more) |
— | over 2 years ago |
| Comment | Post #286103 |
Can your load take the noise from a switched mode solution? If the load is mostly static (e.g. does not vary, or does not have fast transitions) you can oversize the output filter at the cost of bandwidth, to decrease noise. Otherwise, for 36 V max drop witn 0.1 A you'll have 3.6 W wasted, whereas wi... (more) |
— | over 2 years ago |
| Comment | Post #285861 |
@#53110 Well, if it was an artefact then it's on the owner of the video, feel free to ask him about that. I happen agree with the outcome, but that's just me. (more) |
— | almost 3 years ago |
| Comment | Post #285861 |
@#53110 The whole problem was formulated in an ideal manner, so the lamp would glow instantaneously. It no longer matters that the "bulk" of the current would only appear after 1 s due to the transmission line effect. And it will not start glowing with a brief pulse, then wait for the rest, because t... (more) |
— | almost 3 years ago |
| Comment | Post #285861 |
Have you seen this video? https://youtu.be/2Vrhk5OjBP8 (more) |
— | almost 3 years ago |
| Comment | Post #285792 |
@#54288 I think the author may be saying something else, and that is dependent on different details that are not shown (I don't have the book). But, clearly, any transfer function can be plotted by replacing s->jw, since it's basic complex arithmetic. It's probably still related to the integral part ... (more) |
— | almost 3 years ago |
| Edit | Post #285792 | Initial revision | — | almost 3 years ago |
| Answer | — |
A: H(jω) does not exist for unstable systems, but we still use it when designing controllers - contradiction? Please note that the text refers to the integral, or the mathematical evaluation through integration which, indeed, cannot be obtained. But that doesn't mean you can't obtain the Laplace transfer function directly. A hypothetical RLC filter with a negative resistor is very much possible. In fact, it ... (more) |
— | almost 3 years ago |
| Comment | Post #285687 |
Unless I'm reading it wrong, the datasheet seems to specify the output with a nominal 10k/10p (general data table > output > load nom.). They could probably be around those values. [edit] The corner frequency for those values is certainly less than the maximum 45 MHz, though. (more) |
— | almost 3 years ago |
| Comment | Post #284895 |
@#54668 The picture above is made with LTspice. It should be easy to replicate. (more) |
— | about 3 years ago |
| Edit | Post #284895 | Initial revision | — | about 3 years ago |
| Answer | — |
A: Flyback converter design Try searching for "multiple outputs flyback". Basically, you can't control each, individually, instead you make the voltage feedback as the average between the outputs. Here's a quick'n'dirty test to show what I mean, with a simple voltage mode concoction (don't give it too much thought for the value... (more) |
— | about 3 years ago |
| Comment | Post #284540 |
I don't know the term, but it seems that "switching transients" is a more generic term that can encompass "transient power". (more) |
— | about 3 years ago |
| Comment | Post #283926 |
@#54107s A current source cannot exist as opened (no load), in the same way a voltage source cannot exist as shorted out. The first will want to generate the specified current and since the impedance is infinite the voltage is infinite; similarly, a voltage source's internal resistance is zero, thus ... (more) |
— | about 3 years ago |
| Edit | Post #283636 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Type II compensation network for a non-ideal transconductance amplifier You can use various methods for it, one way would be to simply use the ideal transfer function and make it in parallel with \$Ro\$: $$\begin{align} A(s)&=R{th}+\dfrac{1}{sC{th}} \tag{1} \\\\ B(s)&=A(s)||C{thp} \\\\ {}&=\dfrac{1}{\dfrac{1}{R{th}+\dfrac{1}{sC{th}}}+sC{thp}} \\\\ {}&=\dfrac{1}{sC... (more) |
— | over 3 years ago |
| Comment | Post #283265 |
Logisim has an option for the XOR gates in their properties, regarding `Multiple-Input Behaviour`, but it's either the "one true only", or "an even number of true". (more) |
— | over 3 years ago |
| Comment | Post #282635 |
@#54107 After the switch changes state, the only mesh you have is made of C, R, and L, with C precharged. Therefore there is no Vs, only Vc=Vr+VL, because Vc takes the role of Vs. With Vr=0, all that remains is VL. (more) |
— | over 3 years ago |
| Comment | Post #282635 |
@#54107 Draw an imaginary vertical line separating `C1` from `R2+L1`. Now the voltage on the capacitor is mirrored to the inductor and the resistor. Since the resistor is not a reactive element, the whole initial condition applies now to the inductor, only, therefore `D` ends up with `2`, not `-2`. (more) |
— | over 3 years ago |
| Edit | Post #282635 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Critically damped oscillation issue Your derivation is correct, you just missed the sign: \$VL=-VC=2\;\mathrm{V}\$, because the capacitor charges with +2 V, and the discharge accounts for the negative sign on the inductor. The notation of \$VL=-VC\$ may be confusing, so think of it as \$VC=-VL\$, maybe it makes more sense. Verifying... (more) |
— | over 3 years ago |
| Edit | Post #282265 |
Post edited: mathjax |
— | over 3 years ago |
| Edit | Post #282262 |
Post edited: mathjax |
— | over 3 years ago |
| Suggested Edit | Post #282265 |
Suggested edit: mathjax (more) |
helpful | over 3 years ago |
| Suggested Edit | Post #282262 |
Suggested edit: mathjax (more) |
helpful | over 3 years ago |
| Comment | Post #282205 |
@Chupacabras I've updated the answer, maybe it clears up a few things. (more) |
— | over 3 years ago |
| Edit | Post #282205 |
Post edited: additional answer |
— | over 3 years ago |
| Edit | Post #282205 |
Post edited: text corrections |
— | over 3 years ago |
| Edit | Post #282205 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Meaning of some components around voltage reference in SMPS `R22, C22, C23`, together with `Rfbt` (feedback top) and `Rfbb` (feedback bottom) form what is called a type II compensator, using TL431 as an opamp with built-in reference. The feedback network is a bit different than what you'd see -- usually it's `R22` series with `C23`, and `C22` in parallel with... (more) |
— | over 3 years ago |
| Comment | Post #280357 |
If it were me I'd give up the reputation points and only keep the relevant characteristics, such as the ones that already are. Reputation has the bad tendency that people will want to post for the points, rather than for the contents (just look at *.se). Having a counter that says "guru in this domai... (more) |
— | over 3 years ago |
| Comment | Post #280152 |
Late comment, but that's a nice trick. A 2nd order Bessel might do just a tad better, though at the cost of an opamp. (more) |
— | over 3 years ago |
| Comment | Post #280822 |
@PeteW I've also seen it once, during a storm. The wires failed locally, where the event happened, then the humidity and the aged wires (some from the 60's, 70's) acted as a welding electrode: the isolation burned and continued to burn down the wire. It was faster than what I see in OP's video, but I... (more) |
— | almost 4 years ago |
| Comment | Post #279005 |
Did the edit not work? (more) |
— | about 4 years ago |
| Comment | Post #278756 |
@coquelicot I know that the practical part is what matters here, but the theoretical part might still help along the way. (more) |
— | about 4 years ago |
| Comment | Post #278756 |
The MathJax seems off? `\\` doesn't work, and nither does `\begin{align} ... end{align}`. Are there differences between this and the syntax used in ee.se? (more) |
— | about 4 years ago |
| Edit | Post #278756 | Initial revision | — | about 4 years ago |
| Answer | — |
A: Moving average that uses less memory? I'm adding this as a corollary to Olin's post, in case an exact formula for attenuation is needed. That basic IIR is derived from the exponential moving average, and its impulse response is: $$h[n]=\alpha x[n]+(1-\alpha)x[n-1]$$ which translates into this transfer function: $$H(z)=\frac{\... (more) |
— | about 4 years ago |
| Comment | Post #278614 |
@LvW There are people who prefer Boctor or Deliyannis because, at a first glance, you "only" need one opamp per stage. But when you consider the side effects, really, it really pays to add one more for an almost true zero (and not only). (more) |
— | about 4 years ago |
| Comment | Post #278614 |
I've updated my answer. Be sure to not omit the last paragraph. (more) |
— | about 4 years ago |
| Edit | Post #278614 |
Post edited: additional answer per OP's comments |
— | about 4 years ago |
| Comment | Post #278614 |
Before I continue, you do realize that you'll get something close to (maybe a bit worse) the black trace in the last picture? If you look closely, you'll see that the passband is no longer flat(tish), it has a droop of maybe 3 dB, which means measuring a signal at 90 Hz and one at 100 Hz will give tw... (more) |
— | about 4 years ago |
| Edit | Post #278614 | Initial revision | — | about 4 years ago |
| Answer | — |
A: High pass filter design I saw this link by accident and, while Andy's and Olin's answers are spot on, I thought I'd expand a bit on the complications they describe, just so you know what you're missing (or meeting, should you decide to continue). Also, it might get a bit long, but that's why the voting system exists. Fir... (more) |
— | about 4 years ago |
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