So turns out that those cheap Tower Pro MG996 servos can, in fact, operate at up to 8.4V without bursting into flames, which is good to know.
Also turns out that after less than a minute of this the outer casing gets pretty warm. (like 122F+ warm) :v:
Now my dumb ass just needs to figure out a cost-effective solution to stepping down this 8.4V pack to 6-7V while still delivering the needed amps...
[QUOTE=Zero-Point;51457705]So turns out that those cheap Tower Pro MG996 servos can, in fact, operate at up to 8.4V without bursting into flames, which is good to know.
Also turns out that after less than a minute of this the outer casing gets pretty warm. (like 122F+ warm) :v:
Now my dumb ass just needs to figure out a cost-effective solution to stepping down this 8.4V pack to 6-7V while still delivering the needed amps...[/QUOTE]
eBay buck converter modules, if you can live with around 1 amp output per module.
Even though they're rated for 1.5 ampere, the IC gets really damn hot then.
[IMG]http://puu.sh/sB9AQ/34f5abeb3c.jpg[/IMG]
Finally finished repairing my HP 3478A, all thanks to a tip from ddrl. Turns out to have been a single busted op-amp responsible for a reference voltage to the ADC.
Now it's taken its rightful place on the shelf.
Aaaand I still have a second one to repair, which I suspect has a busted ram chip. A chip that is no longer manufactured, and replacements cost more than I wanna spend on a guess.
[QUOTE=Van-man;51457961]eBay buck converter modules, if you can live with around 1 amp output per module.
Even though they're rated for 1.5 ampere, the IC gets really damn hot then.[/QUOTE]
Lost my job a month ago, money's tight. I'm sure I have some switch-mode chips lying around here somewhere, waiting to be cannibalized.
On a related note, since I'm not too terribly experienced with the practice of putting these things together, if I need more current, I can use a beefier MOSFET to drive the output still, right?
Interesting question.
I've got a dock for my toughbook that has antenna ports for GPS, etc. I've got a GPS antenna that needs 3V +/-10% for the LNA, and I want to try and integrate a bias-t into the dock to inject power. Now, I would like to avoid having to find a regulator so I was wondering if my idea might work. The LNA uses at most 20mA, so would I be able to get by with taking the 5V from the USB power and making a voltage divider? And to add to that, would I be able to use the load of the LNA on the antenna as one leg of the divider so that I only need three components for this? (Inductor, capacitor, and resistor)
That'd make it nice and clean, and easy to stuff in the dock. Just wanting to know people's thoughts.
[QUOTE=papkee;51468018]Interesting question.
I've got a dock for my toughbook that has antenna ports for GPS, etc. I've got a GPS antenna that needs 3V +/-10% for the LNA, and I want to try and integrate a bias-t into the dock to inject power. Now, I would like to avoid having to find a regulator so I was wondering if my idea might work. The LNA uses at most 20mA, so would I be able to get by with taking the 5V from the USB power and making a voltage divider? And to add to that, would I be able to use the load of the LNA on the antenna as one leg of the divider so that I only need three components for this? (Inductor, capacitor, and resistor)
That'd make it nice and clean, and easy to stuff in the dock. Just wanting to know people's thoughts.[/QUOTE]
[URL="http://electronics.stackexchange.com/questions/130509/gps-antenna-when-is-an-active-antenna-really-necessary"]I don't think you'll need an active antenna if you're mounting it relatively close[/URL]. That said (Especially since you mentioned its based on the docking station), I wouldn't really recommend voltage dividers for powering anything beyond 1mA as more power will be wasted in the divider than the device to be powered (Plus it varies depending upon load).
I'd go ahead and just buy a simple linear regulator instead.
I already have an active antenna. It was kind of a "I've got all this stuff, how do I make it work" type thing.
I'm almost wondering if the antenna ports on the bottom of the toughbook are already powered. The recommended antenna that Panasonic specced for use with the dock I have is an active antenna. The dock itself is just a straight-through coax run.
Either way I might just build a simple external bias tee with a regulator and usb plug. I'll try the antenna later today without any external circuitry and see what happens.
So I'm going to install a remote-start module for my mom's car as a gift. The key has some RF tag built into it so the ignition won't fire unless the key is in the ignition...
Here's the super high tech solution recommended by a lot of websites and people:
[IMG]http://i.imgur.com/MaENzsk.jpg[/IMG]
You take an extra key and put it in this module so it can read the RF tag when the remote starter triggers. You also have to install another ring around the ignition switch so this box can broadcast the RF info from the key...
At least there's a better way to do it which involves sticking a resistor between a wire.
[QUOTE=No_Excuses;51494352]So I'm going to install a remote-start module for my mom's car as a gift. The key has some RF tag built into it so the ignition won't fire unless it's detected...
Here's the super high tech solution recommended by a lot of websites and people:
[IMG]http://i.imgur.com/MaENzsk.jpg[/IMG]
You take an extra key and put it in this module so it can read the RF tag when the remote starter triggers. You also have to install another ring around the ignition switch so this box can broadcast the RF info from the key...
At least there's a better way to do it which involves sticking a resistor between a wire.[/QUOTE]
I sincerely hope that also involves installing a "normally open" switch for the gearchanger that's only toggled when the car is actually in Neutral.
[QUOTE=No_Excuses;51494352]So I'm going to install a remote-start module for my mom's car as a gift. The key has some RF tag built into it so the ignition won't fire unless the key is in the ignition...
Here's the super high tech solution recommended by a lot of websites and people:
[IMG]http://i.imgur.com/MaENzsk.jpg[/IMG]
You take an extra key and put it in this module so it can read the RF tag when the remote starter triggers. You also have to install another ring around the ignition switch so this box can broadcast the RF info from the key...
At least there's a better way to do it which involves sticking a resistor between a wire.[/QUOTE]
Kinda the right way. By chance what type of car is it and its year? Some do not need to have a key stored in it. Some can clone an existing one(Less danger then leaving a key under the dash of a car..)
Such as this one:
[url]https://www.amazon.ca/iDatalink-Multi-Platform-Doorlock-Bypass-ADS-ALCA/dp/B001983S0Q/ref=pd_sim_504_7?_encoding=UTF8&psc=1&refRID=147RMZRHCXMK3XRBNCW8[/url]
[quote] have a 512k that the ram went bad and recently de soldered the ram. Whenever i try to re solder it it won't stick. I fluxed the board. But it still wont work. [/quote]
[img]https://68kmla.org/forums/uploads/monthly_10_2016/post-6449-0-68217900-1477890656_thumb.jpg[/img]
He says he desoldered the pins but the chip still wouldn't fall out, so he grabbed it with pliers and pulled until it did. :ohno:
To be fair whatever engineer thought it was a wise idea to directly solder DIP chips (especially ones that run pretty hot and can fail fairly easily) without sockets needs to be shot.
Its already a pain in the butt to try and refurbish C64s and the like without them.
[QUOTE=LoneWolf_Recon;51500428]To be fair whatever engineer thought it was a wise idea to directly solder DIP chips (especially ones that run pretty hot and can fail fairly easily) without sockets needs to be shot.
Its already a pain in the butt to try and refurbish C64s and the like without them.[/QUOTE]
Sockets are expensive
[QUOTE=DrDevil;51502502]Sockets are expensive[/QUOTE]
True, but atleast in the case of the C64, if you're going to pay for a then $10 CPU or $10 video chip, atleast pay an extra $1 to socket the things.
Never would I find myself saying, "Thank God SMD parts have been mainstream for ages".
It isn't that hard to remove DIP chips, the only times they are a real pain is when they use very tight holes with a thick PCB, in that case it's usually better to just snip the leads and remove one by one.
Also good sockets with proper machine pins are expensive, those Chinese ones are pretty shit
[QUOTE=LoneWolf_Recon;51503283]True, but atleast in the case of the C64, if you're going to pay for a then $10 CPU or $10 video chip, atleast pay an extra $1 to socket the things.
Never would I find myself saying, "Thank God SMD parts have been mainstream for ages".[/QUOTE]
Yeah, but sockets are less reliable than directly soldering. The contacts can corrode, or the ICs could fall out.
Pffft, who needs sockets when you have heat guns?
Plastic fumes, ahoy! :disgust:
As of today, I have graduated with my B.S in EE. :excited:
I start monday at my new job designing/prototyping LED arrays and drivers.
congrats! i was originally pursuing a computer engineering and electrical engineering double major but i decided to drop EE so i could graduate a year earlier
It pisses me off more when chips are inserted and the legs are not square to the package or the PCB, so there's always that pressure against the holes that makes desoldering them a royal bitch.
[QUOTE=pentium;51506911]It pisses me off more when chips are inserted and the legs are not square to the package or the PCB, so there's always that pressure against the holes that makes desoldering them a royal bitch.[/QUOTE]
Heat guns~...
Heh. Sure enough, found an old power supply that was pulled from a commercial packaged AC unit that has [url=http://www.ti.com/lit/ds/symlink/lm723.pdf]one of these[/url] on it. Should be just what I need if I can build a proper circuit for it and find an appropriate transistor to boost the output current. (it even has a 2N3055 on it as well)
*update* Hot-damn, turns out I have [I]two[/I] of 'em.
Strange though, they have those 2N3055's on them, but the output of the modules is rated at 24VDC @ 0.225mA, when those transistors are rated for 15A (plenty for my needs). Probably current-limited as a fail-safe for the DDC control systems they were gutted from.
Fun fact: The copyright of the module is from 1979, and the boards are dated 1985. :v:
NASA has a bid out for an SBIR proposal that I'm interested in investigating:
[quote]
Technologies for Large-Scale Numerical Simulation:
 Decrease the barriers to entry for prospective supercomputing users.
 Minimize the supercomputer user's total time-to-solution (e.g., time to discover, understand, predict, or
design).
 Increase the achievable scale and complexity of computational analysis, data ingest, and data
communications.
 Reduce the cost of providing a given level of supercomputing performance for NASA applications.
 Enhance the efficiency and effectiveness of NASA's supercomputing operations and services.
[/quote]
I'd like ot investigate meeting this bid by using the Parallela due to its rather interesting Epiphany SoC. its got an interesting architecture (paper [URL="http://www.adapteva.com/docs/epiphany_arch_ref.pdf"]here[/URL]) that I think could render it useful, and the fact that so much of the software/firmware around the parallela is open-source also would help vouch for it.
The 1024-core Epiphany-V is coming within a few months too, and as long as its out by June I could test that as well. I just think that this being a SoC that could be easily given to someone - as something to plug into their primary computer for example, would meet the project goals. You could get experience thinking about problems in a way oriented towards HPC stuff, and since they are rather efficient power-wise they might scale well instead of something like GPU farms. I mean, the Epiphany SoC has a lot in common with a GPU but it seems like it'll be a lot cheaper and a lot more power efficient.
idk, I'm asking here because I feel that leveraging the different hardware/processor design+pipeline would be vital and I'm curious what you all think of that
Alright guys, it's been 6 hours and I am a dumbass. I literally need this question to pass my course and never touch electronics again (I am doing machining??? Is there electronics in that????)
[URL]http://i.imgur.com/RvfWtvi.png[/URL]
Apart from the awful textbook, confusing web-sources, RLC circuits are hard. All I need to know is HOW to find the total impedance. The rest (finding voltages, current across each component,) should be fairly simple.
Please help me. Your my last hope.
[editline]a[/editline]
I should preface this with saying that I didn't do well on the tests, I have 80% in the course but need at least 75% total combined on the tests to pass. She said if I could do this ONE question I would be allowed to pass. We were never taught this, as we skipped those labs because we ran out of time. So basically she told me to "read the textbook" and figure it out.
Just apply the same [URL="https://en.wikipedia.org/wiki/Electrical_impedance#Combining_impedances"]rules of summing resistors in series/parallel[/URL] to find the total impedance. Fortunately you've already found the inductive/capacitive reactance. Thus you'll represent stuff as a complex number.
For example combine XL1, XL2 and R2 which are represented by 0 + j5k, 0 + j10k, 10k + j0 respectively. Apply parallel combinations of resistors for XL1, XL2 and R2 individually for the real and imaginary part which should result in:
( (0 + j5k)^-1 + (0+j10k)^-1 + (10k + j0)^-1 )^-1 = [B]1k + j3k[/B]
Remember that the reactances are only in the imaginary portion and that capacitive reactances are negative thus R1 & XC are 3.3k + j0 & 0 - j1k respectively. Now you can add them all additively:
(3.3k + j0) + (0 - j1k) + (1k + j3k) = [B]4.3k + j2k[/B]
[QUOTE=LoneWolf_Recon;51531114]Just apply the same [URL="https://en.wikipedia.org/wiki/Electrical_impedance#Combining_impedances"]rules of summing resistors in series/parallel[/URL] to find the total impedance. Fortunately you've already found the inductive/capacitive reactance. Thus you'll represent stuff as a complex number.
For example combine XL1, XL2 and R2 which are represented by 0 + j5k, 0 + j10k, 10k + j0 respectively. Apply parallel combinations of resistors for XL1, XL2 and R2 individually for the real and imaginary part which should result in:
( (0 + j5k)^-1 + (0+j10k)^-1 + (10k + j0)^-1 )^-1 = [B]1k + j3k[/B]
Remember that the reactances are only in the imaginary portion and that capacitive reactances are negative thus R1 & XC are 3.3k + j0 & 0 - j1k respectively. Now you can add them all additively:
(3.3k + j0) + (0 - j1k) + (1k + j3k) = [B]4.3k + j2k[/B][/QUOTE]
This helps me start it, thank you very much.
Ugh just finished my first EE class and I can't say I enjoyed it too much.
Of course it's not designed to be an educational class; it's designed to be a weed-out course to get rid of the incoming students who can't cut it. Boy was that final exam fun.
[editline]14th December 2016[/editline]
Did learn some useful things that I hadn't seen before. Phasors and all the AC stuff was actually pretty informative.
[QUOTE=papkee;51531617]Ugh just finished my first EE class and I can't say I enjoyed it too much.
Of course it's not designed to be an educational class; it's designed to be a weed-out course to get rid of the incoming students who can't cut it. Boy was that final exam fun.
[editline]14th December 2016[/editline]
Did learn some useful things that I hadn't seen before. Phasors and all the AC stuff was actually pretty informative.[/QUOTE]
At least you don't have to do an assignment that was never taught to pass. I did all the exact same stuff you did by the sounds. And yes, the tests are horrible.
P.S, how in gods name do I do (10 + j 0) / (4300 + j 2000) - I have absolutely no clue. The zero is screwing me over.
[editline]a[/editline]
since it's zero you can just ignore the imaginary part and it becomes 10/(4300 + j 2000)
after that you can do a few things. you can put it in your calculator (it should be i instead of j). if you need to do it by hand you multiply by the complex conjugate of the denominator, so you would multiply it by (4300 - j 2000)/(4300 - j 2000). once you do that you simplify it by remembering that j^2 is equal to -1. then, you combine like terms
When dividing complex numbers, its leagues easier (IMO) to turn them [URL="http://www3.ncc.edu/faculty/ens/schoenf/elt115/complex.html"]into polar coors/phasors as shown here[/URL].
Thus it should turn out to be:
[B](10<0°) / (4747.36<24.94°)[/B]
[URL="http://www.allaboutcircuits.com/textbook/alternating-current/chpt-2/complex-number-arithmetic/"]Once turned into polar coors/phasors simply divide each magnitude by each other and subtract the angles.[/URL]
Thus it should be:
[B]0.0021<-24.94°[/B]
Convert that back into complex using the first link, thus it should be:
[B]0.0019 - j0.000885[/B]
[QUOTE=LoneWolf_Recon;51531865]When dividing complex numbers, its leagues easier (IMO) to turn them [URL="http://www3.ncc.edu/faculty/ens/schoenf/elt115/complex.html"]into polar coors/phasors as shown here[/URL].
Thus it should turn out to be:
[B](10<0°) / (4747.36<24.94°)[/B]
[URL="http://www.allaboutcircuits.com/textbook/alternating-current/chpt-2/complex-number-arithmetic/"]Once turned into polar coors/phasors simply divide each magnitude by each other and subtract the angles.[/URL]
Thus it should be:
[B]0.0021<-24.94°[/B]
Convert that back into complex using the first link, thus it should be:
[B]0.0019 - j0.000885[/B][/QUOTE]
This one cracked the puzzle, christ almighty this course is over thank you so much if you want a videogame or something let me know
everybody else thank you as well
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Electrical Engineering V3
