Activity for a concerned citizenâ€
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Comment | Post #287366 |
@#53497 Admittedly, seeing this for the first time without any prior knowledge might seem harsh but, if you would see all this user's history here (preferably in chronological order), you will get a better feeling of why Olin replied as he did (and why things evolved to this point). What I mean is, t... (more) |
— | almost 2 years ago |
| Edit | Post #286690 |
Post edited: corrections |
— | about 2 years ago |
| Edit | Post #287232 |
Post edited: additional note |
— | about 2 years ago |
| Edit | Post #287232 |
Post edited: corrections |
— | about 2 years ago |
| Comment | Post #287232 |
@#52987 No, you're right, that sounds better. Bad wording from my part (but good intentions). (more) |
— | about 2 years ago |
| Edit | Post #287232 |
Post edited: additional note |
— | about 2 years ago |
| Edit | Post #287232 |
Post edited: additional note |
— | about 2 years ago |
| Comment | Post #287232 |
@#52987 Oh, forgot to say: k is the attenuation, related to a unity input. Maybe I should have used A, instead. I'll replace it now. (more) |
— | about 2 years ago |
| Comment | Post #287232 |
I searched online and found [this site](https://circuitsgeek.com/tutorials/t-pad-attenuator/) but, while the formulas are much nicer (no derivation) and the values for the resistors come out as they adveertise it, it doesn't seem to work as intended in terms of attenuation (I get ~51.4 mV, as opposed... (more) |
— | about 2 years ago |
| Edit | Post #287232 |
Post edited: typo |
— | about 2 years ago |
| Edit | Post #287232 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Deriving resistor values for a taper pad attenuator > Yeah, that looks like it's going to be messy. I agree with Olin: the problem starts with three variables and three conditions so, no matter how you look at it, you will end up with a system of equations. However, you can take certain shortcuts (using your 1st picture): $$\begin{align} R{... (more) |
— | about 2 years ago |
| Comment | Post #287027 |
The schematic you shared uses a different symbol for the `ADA4530-1` -- the symbol appears disconnected. While I could correct the connections since they are fairly straightforward, this should have been done by you: either provide the custom symbol, or make a mention of it (preferably the former). A... (more) |
— | about 2 years ago |
| Comment | Post #286868 |
@#53586 Thank you for returning the favour. You had a nice "back of the envelope" approach. (more) |
— | about 2 years ago |
| Edit | Post #286868 | Initial revision | — | about 2 years ago |
| Answer | — |
A: Design high -pass filter with 2 points of the bode plot You still haven't said why you need, or even what order, type, etc. Assuming it's a 2nd order, an exact solution involves creating a generic transfer function and then solving a system of equations with imposed conditions (use squared to get rid of radical): $$\begin{align} H(s)&=\dfrac{s^2}{s^2+... (more) |
— | about 2 years ago |
| Comment | Post #286836 |
I don't see any schematics. (more) |
— | over 2 years ago |
| Comment | Post #286784 |
@#53497 You're right, my words on the previous reply are a little bit ambiguous. I've modified the answer to include the distinction between the choice made for analysis, and the real life case. (BTW, `@` here needs typing the first three letters, no spaces, then the autocomplete should come up; fort... (more) |
— | over 2 years ago |
| Edit | Post #286784 |
Post edited: additional note |
— | over 2 years ago |
| Comment | Post #286792 |
Are you doing it on purpose: not using a schematic editor? I think I was not the only one who told you to use one, so by now it looks like you simply "know better", or you are deliberately not listening. For my part, as long as you post doodles, I won't bother reading your questions. Even if you are ... (more) |
— | over 2 years ago |
| Comment | Post #286784 |
@#54107 Yes but, as I said, their value is chosen such that it doesn't influence the overall oscillator -- after all, you don't want to have a polarizing network that, wehn added, changes the frequency of the oscillations, do you? Therefore, *for the purpose of analysis*, you can ignore them since, i... (more) |
— | over 2 years ago |
| Edit | Post #286784 |
Post edited: typo |
— | over 2 years ago |
| Comment | Post #286784 |
@#53497 You're right, that was a typo, I'll correct it. I was thinking in terms of L1=L2, so I'm probably guilty of "thinking with premeditation". About the input resistance: yes, I've omitted them because they're mostly part of the biasing circuit, and have little to do with the oscillation. They do... (more) |
— | over 2 years ago |
| Edit | Post #286784 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Results of analysis of Hartley oscillator dont make sense If you are pursuing engineering then you can't use doodles instead of schematics generated by dedicated (or not) programs. It will take you just the same amount of time, if not less, and the results would be clear for anyone seeing them. The basic schematic of the Hartley oscillator can be simplif... (more) |
— | over 2 years ago |
| Comment | Post #286766 |
I don't have Altium, or OrCAD, but I think what OrCAD does with those "aliases" is it simply makes copies of them behind your back, either with the symbols or with the info, only, and maybe displaying them transparently to the user. It doesn't make sense otherwise, it has to have some separate info f... (more) |
— | over 2 years ago |
| Comment | Post #286765 |
@#54107 And how is that working for you? The whole pronciple of an oscillator is based on the negative resistance. You need to approximate that, too, with another slope, and then the initial slope, too, if you want your oscillations to start. And you'll end up with a PWL *approximation* which will gi... (more) |
— | over 2 years ago |
| Edit | Post #286765 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Modelling tunnel diode relaxation oscillator Unless you know the exact formula for the I-V curve, you will never find out anything analytically. Fair warning: very unlikely you will get such formula, since they are strongly non-linear in nature. The analysis of such oscillators is done based on approximations (see the Esaki diode, for example),... (more) |
— | over 2 years ago |
| Comment | Post #286730 |
It sounds like an [XY problem](https://xyproblem.info/#). What is it for? (more) |
— | over 2 years ago |
| Comment | Post #286723 |
Olin, if you're reading [this](https://electrical.codidact.com/comments/thread/6474#comment-17851), could you please use it as a reference to force this user to explain every future question, especially when questionable content is asked? I'm not saying to ban the person but, given this blatant abuse... (more) |
— | over 2 years ago |
| Comment | Post #286724 |
@#54107 I see. So you are using volunteers' time for your fun. That is good to know. (more) |
— | over 2 years ago |
| Comment | Post #286724 |
At least now will you say what it is for? (more) |
— | over 2 years ago |
| Comment | Post #286721 |
@#54107 I meant about the 10 kHz part because, as Olin says in his answer, that's not a characteristic of a bandpass filter. But you can define it in terms of center frequency and bandwidth (and attenuations in both pass- and stop-band). You haven't answered, though: what purpose will that filter ser... (more) |
— | over 2 years ago |
| Comment | Post #286721 |
Did you mean "center frequency"? If so, even for a 20 dB attenuation at 1000 times the bandwidth in the stopband, I run into numerical issues and can't get such a high order, and thi sis for a Cauer/elliptic filter. You could try a 2nd order, but that would mean a Q of f0/BW=5000. You'll likely need ... (more) |
— | over 2 years ago |
| Comment | Post #286700 |
@#54107 I don't understand. You have several people telling you that your view is not correct and they all bring arguments, yet you insist on your view. Now, I'm just a random nickname, but Olin is well known. If you think he isn't, take some time to read some of his answers here, or on ee.se. The wa... (more) |
— | over 2 years ago |
| Edit | Post #286706 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Complex frequency of a pole This is mostly another view on @Olin's answer. > For different input signals the frequency of the pole will have different values ,the pole as a pole exists for some specific values of L,C s but the frequency of the pole changes. The poles of an LTI system do not change. Since you're giving an ... (more) |
— | over 2 years ago |
| Comment | Post #286700 |
@#54107 "*For different input signals the frequency of the pole will have different values ,the pole as a pole exists for some specific values of L,C s but the frequency of the pole changes.*" -- This is false. For a transfer function of $1/(s+1)$ the pole is fixed, *at all times*, at $-1$. Otherwise... (more) |
— | over 2 years ago |
| Comment | Post #286680 |
@#54107s Also, what do you mean with "*single phase* transmission line"? Are you thinking of a mains application? You should always say what it is you are after, not what you think you need (that's an [XY problem](https://xyproblem.info/#)), because there may be a different way to look at the problem... (more) |
— | over 2 years ago |
| Comment | Post #286687 |
@#54107 Frankly, it sounds quite dishonest for you to ask about the design of the filter, then say it's only for "educational purposes". It's like me, asking for red fruits which taste like avocado because I'm interested in tasing one, only to say after people searched for me that "oh, I was just rea... (more) |
— | over 2 years ago |
| Edit | Post #286690 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Effect of adding stages to a filter It depends on whther the stages are buffered, or not. If they are buffered then it's more simple, because an RC lowpass has the transfer function: $$H1(s)=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}=\dfrac{1}{sRC+1} \tag{1}$$ And adding $N$ stages will mean $H(s)$ will simply be multiplied $N$ times... (more) |
— | over 2 years ago |
| Comment | Post #286680 |
You can still transmit signals, but they will be attenuated (no such thing as a brick wall filter). Have you seen [this page](https://prc68.com/I/Zo.shtml)? (more) |
— | over 2 years ago |
| Edit | Post #286687 | Initial revision | — | over 2 years ago |
| Answer | — |
A: How to design a low-pass filter when certain conditions must be met Your requirements are odd. Can you specify what purpose that filter is for? I'm saying this because, for a -0.1 dB/Hz, a filter will have a continuously increasing attenuation slope. Think of it like this: at DC, it will have (e.g.) a magnitude of $H(0)=1$. Then, at: $$\begin{align} 1\space\text{... (more) |
— | over 2 years ago |
| Edit | Post #286546 | Initial revision | — | over 2 years ago |
| Answer | — |
A: Unexpected impedance spike when paralleling capacitors Around the 19 min mark in the video (approximately, found with hovering the cursor over the timeline) you'll see that different capacitors have different values of both capacitances and parasitics, and their combined response causes those peaks. Not lastly, there are PCB traces that come with both th... (more) |
— | over 2 years ago |
| Comment | Post #286181 |
To improve a bit on the minimum sample time for accuracy, a buffer (simple repeater) could be added, which will eliminate the 4.7k. It will also add cost, which might not be justified, since the timings are already relaxed. (more) |
— | over 2 years ago |
| Comment | Post #286114 |
@#54288 Just so you know, the formulas that you gave imply a *positive* voltage output, but for the classical topology (non-isolated) the first formula you gave was the correct one. The second one is valid for flyback topologies (or for the awkward case of the classical one with negative input, or sh... (more) |
— | over 2 years ago |