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TimWescott

 Discussion on answer by Hilmar: Expli…

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TimWescott
Jan 13, 2024 18:43
"Should that not be f >= fs/2?" -- that's worth a question in itself, if it's not out there already. The answer is -- nope. The proof is that for $x(t) = \sin \pi f_s$ and samples at $t = \frac{n}{f_s}$, $x_n = 0$.
 

 Discussion on question by kile: How t…

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TimWescott
Apr 15, 2023 22:06
Or at least edit your question to point out that you're doing a calculator that needs to operate at human speed. You may also want to mention the clock speed you're using, your goals for logic footprint, and most importantly the speed you actually need for the computation.
 

 Discussion on question by DKNguyen: L…

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TimWescott
Feb 20, 2023 03:02
You could consider flat flex. Terminate for load, then for 'lectricity. This assumes that you can make it fairly wide. Too bad none of it's black...
TimWescott
Feb 20, 2023 03:02
Personally I'd get some 2-conductor wire to take care of the current, then marry it to something load bearing, taking care that there's enough slack in the conductors that they're never strained. There's a rich selection of load-bearing steel cable and braided plastic line in that load range thanks to people who fish.
TimWescott
Feb 20, 2023 03:02
Alternatively, copper wire made for continuous flex, married to insulation that's abrasion resistant may do the trick, but it'll take some shopping.
 

 Discussion on question by Cyberstrato…

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TimWescott
Feb 20, 2023 03:01
So, in short, are you asking how to make a circuit that counts directly in ternary? If so, you may wish to edit your question to put that right up at the top, perhaps with an expression of the action in more normal notation (i.e., it looks like you want to count 00, 01, 02, 10, 11, etc., or the equivalent). As it stands now it's hard to pick out the actual question from the exposition.
 

 Discussion on question by php_nub_qq:…

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TimWescott
Feb 2, 2023 19:33
I'm glad that you seem to be making progress. I would be more inclined to read your schematic if it looked like a schematic. I see a collection of parts and ports; if I wanted to understand it I'd need to print it out and pencil in all the nets.
 

 Discussion on question by TQQQ: How t…

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TimWescott
Jan 30, 2023 23:14
What is the frequency range of the current? Lowest to highest? What do you mean by "with bandwidth of anything higher than 50kHz" -- do you mean you want to know the current of any component with frequency from 50kHz up? Do you mean all the way up to visible light?
 

 Discussion on question by 2b77bee6-54…

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TimWescott
Oct 6, 2022 21:07
Please edit your question with this information. Comments are only meant to assist in making the question itself clear, through edits. Things have gotten excessive already.
TimWescott
Oct 6, 2022 21:07
Could we take a step back here? Could you edit your question to say what make & model of charge controller you have, whether it's designed to work off of solar panels, and, if not, what is keeping you from buying a charge controller that is?
TimWescott
Oct 6, 2022 21:07
If your charge controller is designed to work off of solar panels, it should provide this functionality already. If it isn't, you need a charge controller that's designed to work off of solar panels.
TimWescott
Oct 6, 2022 21:07
Your hypothetical current limiter could be made (it'd just be a DC-DC converter that servoes the output current). Here's how it'd work: the charge controller would "ask" for more current than the "limiter" wants to supply. The limiter would drop its output voltage. This would make the charge controller ask for more current to make the same power. Then all would stop. So. Yay. Current successfully limited. But that's probably not what you want.
TimWescott
Oct 6, 2022 21:07
It is physically impossible for a device to force an arbitrary voltage and current through some two-terminal device. Your panels are capable of delivering 50A at 12V under certain illumination conditions. But the charge controller can just "decide" to take no more than 20A. In that case, the panel voltage will rise, no farther than it's rated open-loop voltage. As long as that voltage is lower than the charge controller's rated maximum input voltage, you're fine.
TimWescott
Oct 6, 2022 21:07
If it's a decent charge controller then it'll be designed not to take more current than it can handle -- it should just let the panel voltage float up as necessary while not overloading itself. If it does overload itself, then you wanted a new one anyway :O .
 

 Discussion on question by across: Che…

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TimWescott
Jun 20, 2022 15:43
@across that's probably with no discharge current. With discharge current, the cell voltage will go below that -- possibly by a significant amount, if you're discharging the cell heavily. If your discharge current in low and constant, or if you can turn it all the way off and give the battery a minute to recover before reading voltage, then cell voltage is sorta-kinda reliable.
TimWescott
Jun 20, 2022 15:43
Look at the LM339 for starters. Like @Winny said, there's hundreds. The LM339 is possibly the most easily available, it should work from either 12V or 24V, and while it's not known if it meets your requirements for precision, it'll probably add less error than using the 12V from a switching regulator as a reference.
TimWescott
Jun 20, 2022 15:43
Because you do not say to what precision you need to determine your voltage cut-off point, any answer has an implicit "good enough" opinion buried in it. Also, as soon as you implement this you'll begin to appreciate the unique joys of trying to monitor battery usage, in that for almost all battery chemistries, as the battery dischages the output voltage sag is affected more by increasing effective resistance rather than decreasing effective voltage. So unless the load is very constant you end up building a "cut power on high load" device, not a "cut power on discharge" device.
 

 Discussion on question by MrYui: How …

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TimWescott
Jan 13, 2022 18:43
If it's not clear from the comments so far, edit your question to at least give us one example of one of these 3D plots from "alot of people", along with a citation of the paper. The paper should say what the meaning of that 3rd axis is, so perhaps a summary of what papers actually say about the meaning of that 3rd axis would be useful.
TimWescott
Jan 13, 2022 18:43
In an $n \times m$ monochromatic image, you can find $n \cdot m$ dimensions. Three times that if it's full color (for humans). Do you want meaningful dimensions?
 

 Discussion on question by Dan Boschen…

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TimWescott
Dec 4, 2021 16:58
Any polynomial root-finding technique is numerically ill conditioned. It's just the nature of the beast. I can't point you to a specific mathematical reason, but I can say that I've heard it numerous times (since before the eigenvalue decomposition technique was "the way") and my personal experience bears that out.
TimWescott
Dec 4, 2021 16:58
Well, iteratively finding all the zeros inside the unit circle is exactly the same thing as factoring the polynomial, or at least that half of it. Hence, it's numerically ill-conditioned.
TimWescott
Dec 4, 2021 16:58
You could just compute a minimum-phase FIR filter directly.
TimWescott
Dec 4, 2021 16:58
What are you going to do with it? Make minimum-phase (i.e., asymmetrical) FIR filters? Or is this just for research?
TimWescott
Dec 4, 2021 16:58
If you were taking any arbitrary FIR filter and finding its zeros, then you'd just be factoring a ginormous polynomial, and that's known to be horribly susceptible to numerical precision. So your goal boils down to finding the roots of a polynomial that's constrained to have roots that occur as pairs of $\left ( a, \frac{1}{a} \right)$. Somehow I don't think that'll be a big help.
 

 Discussion on question by geocheats2:…

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TimWescott
Jun 16, 2021 19:18
Since you're spending money anyway -- hire a local EE consultant to do a design review for you. They'll tell you the problems, and make suggested fixes. Better, because you're paying them, you can make them sign a non-disclosure agreement (and the decent ones keep their customer's secrets anyway).
 

 Discussion on question by Emmmm: Havi…

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TimWescott
Mar 29, 2021 13:35
You're running into a terminology problem. "I do understand that variance do not change over frequency (or time) for colored Gaussian noise." You're missing a subtlety in the meaning of the word "variance". Yes, the variance of the whole signal won't change over time, but "power spectral density" is roughly synonymous with "variance as a function of frequency". So your declaration is, at best, confusing -- either you mean the variance of the whole signal, in which case "over frequency" is meaningless, or you're talking about some proxy for PSD and you need to listen to Marcus.
TimWescott
Mar 29, 2021 13:35
"It cannot be that $y_n$ has higher "variance" at high frequency and lower "variance" at low frequency because I mentioned earlier that variance should not vary over frequency or time." This statement is incorrect, and if you cannot accept that, you cannot be educated on this point. Either you define variance as being a characteristic of the whole process -- and thus that "variance over frequency" is a meaningless phrase that should not be written, or you accept that "variance over frequency" has a meaning -- typically it would mean "variance if I ran the signal through a bandpass filter".
TimWescott
Mar 29, 2021 13:35
"First, we know that their variance are identical to $2 \sigma^2_n$." No, actually. White Gaussian noise, by definition has finite variance for any finite bit of spectrum, but it has the same variance over the entire spectrum. So when you calculate the variance, you find that white Gaussian noise always has infinite variance. A white Gaussian noise process cannot, by definition have the same variance as a bandlimited (i.e., colored) process.
TimWescott
Mar 29, 2021 13:35
The "colored" in "colored noise" means that the spectral density changes with frequency. That's not opinion, that's what it means, by definition.
 

 Discussion on question by niko20: Hel…

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TimWescott
Dec 30, 2020 07:10
Layout matters. The fact that the emitter is the middle lead is a hint that layout matters. Even though the distance from input to output on that circuit is only 1/100th of a wavelength, that's still significant. If wouldn't expect to make a circuit work at 144MHz with dead-bug construction. You may want to try again, after making yourself some glue-down stripline (basically, cut a 1/8" wide piece of your circuit board, and glue it down for "traces"). Consider that everything is a capacitor and an inductor, too.
 

 Discussion on question by user769708:…

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TimWescott
Dec 7, 2020 23:55
I haven't found any good ones. /r/ece is civil and entirely apolitical, if that's your objection to Reddit
TimWescott
Dec 7, 2020 23:55
If you really want to have a discussion about this -- and it looks like you do -- then find an electronics forum. I suggest reddit.com/r/ece, but there are certainly others. Stackexchange is looking for specific questions that can be answered specifically -- not long discussions in the comments.
TimWescott
Dec 7, 2020 23:55
Get two cheap POTS (plain old telephone service) phones, hook up phantom power at the outside end, and bob's yer uncle.
 

 Discussion on question by user268759:…

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TimWescott
Nov 19, 2020 16:07
Keep in mind that a temperature controlled iron has a temperature sensor in or close to the tip. It's not just reducing the power to the tip -- it's reducing it when it needs to, then increasing it when you start melting solder or trying to heat up something big.
TimWescott
Nov 19, 2020 16:07
A triac-based lamp dimmer should be about as good as you can get. Why the reluctance to use it?
 

 Discussion on answer by relayman357: …

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TimWescott
Nov 18, 2020 15:42
In order for $V_2$ to remain at zero, the current into the node has to equal the current out of the node. Since the ideal op-amp doesn't have any current flowing into the input terminals, that means that the current in the feedback resistor has to be equal and opposite the current in the input resistor.
TimWescott
Nov 18, 2020 15:42
The negative sign is there because it has to be to maintain the condition $V_2 = 0$. It doesn't violate conservation of charge in a real circuit because the op-amp is connected to a power source. It doesn't violate conservation of charge in the circuit above because in theory-land, op-amps are magical devices that can supply or sink charge as needed.
 

 Discussion on question by Talha Israr…

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TimWescott
Oct 28, 2020 18:27
Then, as always when doing homework on Stackexchange, edit your question with your work.
TimWescott
Oct 28, 2020 18:27
I suspect, by the way, that the exact reason you're being given this problem is to show you that things like KVL (or KCL -- either would work here) can be used for more than just finding the currents in resistors.
TimWescott
Oct 28, 2020 18:27
Yes. That's what KVL is all about -- finding the unknown currents from the known voltages. You'll just be doing that upside-down by finding the current in R1 as a function of both the voltage and current in V1, then solving for the current and voltage in V1. Have you tried writing the equation for the current in R1 as a function of the voltage in V1? Perhaps if you did this then all would be made clear!
TimWescott
Oct 28, 2020 18:27
By doing KVL and paying attention to your sign conventions.
TimWescott
Oct 28, 2020 18:27
If you knew the voltage at V1, you could calculate the current at the meter -- right? And with KVL, you can calculate the current at the meter as a function of the voltage at V1 -- right? So what's to keep you from using the current at the meter to calculate the voltage at V1?
TimWescott
Oct 28, 2020 18:27
A voltage source certainly can absorb power! Think of a battery being charged.
TimWescott
Oct 28, 2020 18:27
"how would i know the direction of the current" -- by convention, current flowing into the '+' terminal of an ammeter reads as positive.
TimWescott
Oct 28, 2020 18:27
The OP knows the value of the current from the current source, and the value of the current in R1. So with KVL and some algebra, they can know the voltage at V1.
 

 Discussion on question by namezero: O…

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TimWescott
Sep 6, 2020 14:54
Which is why @KevinWhite was quoting the gain of 18 -- he was actually looking at the resistors on the schematic.
TimWescott
Sep 6, 2020 14:54
Then you're not trying to amplify 600mV to 4V -- you're trying to amplify something much smaller to 4V.
TimWescott
Sep 6, 2020 14:54
What is the purpose of R28, R31, R32 and R33? Why aren't you doing a simple gain stage? What's the purpose of C16? It's putting in a pole at 280kHz, so most of your gain is gone at 500kHz. If you just put a straight amplifying stage in there with a gain of 6.7, your circuit should work to your stated desires -- what am I missing?
TimWescott
Sep 6, 2020 14:54
@KevinWhite: 4V / 600mV = \$6.\bar6\$ -- which translates to a bandwidth of 2.25MHz, which translates to a gain drop-off of only 2.5% at 500kHz. So there's more going on.
TimWescott
Sep 6, 2020 14:54
And why an LM318? In the decades since it came out, other, more capable amplifiers have been designed.