Electrical Engineering V3 - Developers

Why does DipTrace not contain a single audio connector component? Oh well I guess its time to learn how to make my own 3.5mm TRS jack.

[QUOTE=Cakebatyr;49444519]Why does DipTrace not contain a single audio connector component? Oh well I guess its time to learn how to make my own 3.5mm TRS jack.[/QUOTE] Oh, I know that feeling. [IMG]http://i11.photobucket.com/albums/a166/ballsandy/00031.jpg[/IMG] Fuck you, Air Canada.

[QUOTE=Cakebatyr;49444519]Why does DipTrace not contain a single audio connector component? Oh well I guess its time to learn how to make my own 3.5mm TRS jack.[/QUOTE] Fuuuuck. Mind sharing it when done? I need to start using diptrace to work on my end of a project and it was going to require a bunch of audio connectors

I'm hovering between just putting two RCA connectors in place of a 3.5mm jack... In other news I discovered Front Panel Express which seems amazing for little projects but also expensive. So that brings me to the question: Metal project boxes, PCB front/rear panels or metal panels?

Guys, I want to make a subwoofer box that includes the woofer and amp, problem is I want to make the box removable and thus need connectors. What are some good connectors for a thick (1cm²) power and ground cable? What are some good connectors for RCAs?

[QUOTE=Nightrazr;49446799] [B]What are some good connectors for a thick (1cm²) power and ground cable? [/B][/QUOTE] [url]http://s21.postimg.org/sgropd5xj/844907378_850.jpg[/url] They're the same used for charging the batteries of electric forklifts, so they can handle both the current, and a beating.

[QUOTE=Van-man;49446844][url]http://s21.postimg.org/sgropd5xj/844907378_850.jpg[/url] They're the same used for charging the batteries of electric forklifts, so they can handle both the current, and a beating.[/QUOTE] :godzing: I saw that one before somewhere but I couldn't ever find it again.

[QUOTE=Nightrazr;49446976]:godzing: I saw that one before somewhere but I couldn't ever find it again.[/QUOTE] As for audio connector, I presume you're only gonna run 2 RCA plugs (maybe with common ground), so 3-pin or 4-pin XLR connector and plug? [url]https://en.wikipedia.org/wiki/XLR_connector[/url] They're fairly robust, and will also make for a quick-connect solution. But what about the inevitable "remote" signal for turning the amplifier on with the radio/head-unit if used in a automotive setting? then you're gonna need a 4-pin (common ground for audio, left, right and remote) or 5-pin XLR connector.

DipTrace is always lacking on some of the more common parts and connectors but a plethora of the uncommon manufacturer specific chips. But I still hold it in better regard than Eagle for not being a clusterfuck. I would assume the 3.5mm jacks would be under Con RCA & Jacks, but I can't seem to find them there.

So if you recall, last year I gutted a power plant for cool stuff. One such cool stuff was this power meter: [t]http://i.imgur.com/T7z3ste.jpg[/t] Now I want to use it to show my room's power usage. However, I'm a bit confused as to how exactly it works. The meter specifies that 72V and 8A will bring it to full scale, which works out close to half of the standard US 120V and 15A. The way it's supposed to be wired is throwing me for a loop. Here's the wiring diagram I stumbled upon [t]https://i.gyazo.com/470b2dc63373b486c0f7471b74dcae32.png[/t] Which shows you need two current transformers, which makes sense. One for voltage, one for current. Then you've got the burden resistors across the terminals to create a voltage from the current for the voltage measurement, but what I don't understand is why the current side also has a burden resistor when it's supposed to be receiving a current as opposed to a voltage. I wondered if the resistors were built in to the meter, and the voltage sense terminals read a constant 564Ω, but the current sense terminals were near 0Ω. So the test was inconclusive. Does anybody here work in the power industry? Anybody have any ideas? I don't want to just throw 72V at the thing, because it's from the 1930's and really can't be replaced.

Guys, I need some help with this simple circuit. I'm trying to make a NOT gate, with a Darlington gate, but it's the LED is not turning off: [t]http://i.imgur.com/UQqFrxa.jpg[/t]

[QUOTE=Gulen;49459607]Guys, I need some help with this simple circuit. I'm trying to make a NOT gate, with a Darlington gate, but it's the LED is not turning off: [t]http://i.imgur.com/UQqFrxa.jpg[/t][/QUOTE] Do you happen to have a schematic of what you built?

Woops, didn't realise it was that hard to see, it's pretty much this: [t]http://i.imgur.com/89ioHyq.jpg[/t]

Try this instead (Only requires one transistor): [IMG]http://www.electronics-tutorials.ws/logic/log47.gif?81223b[/IMG] VCC is your battery, connect your LED from Out to ground.

[QUOTE=Gulen;49460204]Woops, didn't realise it was that hard to see, it's pretty much this: [t]http://i.imgur.com/89ioHyq.jpg[/t][/QUOTE] Um lol there is direct loop *-> battery -> led -> *, so yeah this is why it is always on.

[QUOTE=LoneWolf_Recon;49460484]Try this instead (Only requires one transistor): [IMG]http://www.electronics-tutorials.ws/logic/log47.gif?81223b[/IMG] VCC is your battery, connect your LED from Out to ground.[/QUOTE] Yeah, I tried following that, but I guess I messed up, because it didn't work, which is why I tried that circuit. [editline]5th January 2016[/editline] [QUOTE=Fourier;49460487]Um lol there is direct loop *-> battery -> led -> *, so yeah this is why it is always on.[/QUOTE] How is that different to the circuit Lonewolf posted? Sorry for my dumb questions, I don't usually dabble with this low voltage stuff.

[QUOTE=Gulen;49461293]How is that different to the circuit Lonewolf posted? Sorry for my dumb questions, I don't usually dabble with this low voltage stuff.[/QUOTE] [t]http://i.imgur.com/btFpgCw.png[/t] I highlighted the path the LED current takes, as you can see the transistors would have no effect. Also those transistors are basically a short circuit, the base to emitter junction is essentially a diode, so you have two series diodes with no current limiting at all. The circuit LoneWolf posted has a base resistor to limit the base-emitter current to a sensible value. Had you tried it with a supply that can deliver significant current you'd certainly get a nice cloud of magic smoke.

So, something like this then? [t]http://i.imgur.com/vuB1tjP.jpg[/t]

[QUOTE=Gulen;49462203]So, something like this then? [t]http://i.imgur.com/vuB1tjP.jpg[/t][/QUOTE] Why do you insist on drawing your battery upside down [editline]5th January 2016[/editline] [QUOTE=Yahnich;49462300]why are you connecting your base and your emitter[/QUOTE] It's called biasing, but I don't think that was the intent here.

[QUOTE=Yahnich;49462300]why are you connecting your base and your emitter[/QUOTE] You mean the base and collector? Because they're both going to be connected to the positive voltage rail. [editline]5th January 2016[/editline] [QUOTE=DrDevil;49462497]Why do you insist on drawing your battery upside down[/QUOTE] I'm not used to drawing this sort of schematics, so I just picked an orientation.

[QUOTE=Gulen;49462582] I'm not used to drawing this sort of schematics, so I just picked an orientation.[/QUOTE] Usually you want your positive voltages on top, makes them easier to read.

[QUOTE=Gulen;49462203]So, something like this then? [t]http://i.imgur.com/vuB1tjP.jpg[/t][/QUOTE] For starters as DrDevil said your battery is the wrong way around, positive voltages at top and negative / ground at bottom. You can put your LED in the collector since current will go through R2, through the collector-emitter junction to ground. For example: [img]http://i.imgur.com/HjSblSm.png[/img] Base current is (9V - 0.7V) / 27k = ~307uA Collector current is (9V - 2V) / 680 = ~10.3mA Current gain is 10.3mA / 307uA = 33.5 (most transistors can manage this easily) 0.7V is from the base-emitter diode drop (0.6V to 0.8V typical) 2V is from the LED diode drop, varies on the color and manufacture.

This might be helpful. [media]https://www.youtube.com/watch?v=Te5YYVZiOKs[/media]

[QUOTE=Chryseus;49462622]For starters as DrDevil said your battery is the wrong way around, positive voltages at top and negative / ground at bottom. You can put your LED in the collector since current will go through R2, through the collector-emitter junction to ground. For example: [img]http://i.imgur.com/HjSblSm.png[/img] Base current is (9V - 0.7V) / 27k = ~307uA Collector current is (9V - 2V) / 680 = ~10.3mA Current gain is 10.3mA / 307uA = 33.5 (most transistors can manage this easily) 0.7V is from the base-emitter diode drop (0.6V to 0.8V typical) 2V is from the LED diode drop, varies on the color and manufacture.[/QUOTE] So, in your schematic, the LED and resistor add up to a resistance greater than the 27k resistor and transistor, causing the current to flow through the base and emitter when the transistor is turned on? But wait a minute, what about when the transistor is off? Then no current should flow through emitter-collector, and therefore none through the LED either. Woops, seems I forgot to put a switch in between the battery and base of the transistor.

[QUOTE=Gulen;49463234]So, in your schematic, the LED and resistor add up to a resistance greater than the 27k resistor and transistor, causing the current to flow through the base and emitter when the transistor is turned on? But wait a minute, what about when the transistor is off? Then no current should flow through emitter-collector, and therefore none through the LED either.[/quote] When there is no base current, I.E the 27k is removed there can be no collector current, since the only path to ground is through the collector to emitter the LED is off. (Current in a BJT is always relative to the emitter, IC = ICE, IB = IBE) IC = 0 when IB = 0 When there is base to emitter current a collector to emitter current is allowed to flow, the amount of current that may flow is dictated by two parameters, the amount of base current and the current gain of the transistor (typically referred to as beta or hFE) IC = IB * beta So if you removed the LED and the 680 ohm resistor and connect it directly to the 9V battery, the maximum collector current is given by the above equation, a typical transistor has a beta between 50 and 150, assuming 100 this means: 300uA * 100 = 30mA It's important to note the BJT is a current controlled device so you should avoid thinking of it as a controlled resistance or a plain switch, it's more of a current source which obeys IC = IB * beta Beta is however very unreliable, it varies with applied collector voltage, base voltage, collector current, transistor and temperature which is why a 680 Ohm resistor is used to limit the current to 10mA rather than rely on just the transistor current gain. In choosing the base resistor size I assume a minimum current gain, this is almost always specified in the datasheet for the transistor, for a desired collector current of 10mA and a minimum beta of 40 this works out as: IB = IC / beta 10mA / 40 = 250uA I increased this to 300uA as is good practice which is given by a common 27k resistor, 300uA ensures that the LED is turned on fully at 10mA on no matter how beta changes. The collector and base current are combined at the emitter so: IE = IC + IB

Weird, I've been messing around with an ESP8266 module (ESP-03 to be specific), and use an Arduino Uno with a board on it as programmer. Thing is, I haven't put a voltage divider on the Arduino TX > ESP RX line while the Uno operates on (I assume after reading documents) 5V and the ESP on 3.3V. Somehow it hasn't caused any issues yet, but I can't imagine it doing any good :v:

[QUOTE=scratch (nl);49463784]Weird, I've been messing around with an ESP8266 module (ESP-03 to be specific), and use an Arduino Uno with a board on it as programmer. Thing is, I haven't put a voltage divider on the Arduino TX > ESP RX line while the Uno operates on (I assume after reading documents) 5V and the ESP on 3.3V. Somehow it hasn't caused any issues yet, but I can't imagine it doing any good :v:[/QUOTE] Usually 3.3V devices will state whether their I/O pins (if no their power pins) are 5V tolerant. But according to the [URL="https://www.adafruit.com/images/product-files/2471/0A-ESP8266__Datasheet__EN_v4.3.pdf"]datasheet[/URL], the GPIO (and subsequently UART) is [U]not[/U] 5V tolerant. I suggest you put a voltage divider on the UNO's TX.

It might be working because I'm bypassing the ATmega328P (RST to GND), thus communication being provided by the ATmega16u2. See [URL]https://www.arduino.cc/en/uploads/Main/Arduino_Uno_Rev3-schematic.pdf[/URL] [sp]Or I'm just talking gibberish and should be grateful nothing broke so far[/sp]

[QUOTE=Chryseus;49463570]When there is no base current, I.E the 27k is removed there can be no collector current, since the only path to ground is through the collector to emitter the LED is off. (Current in a BJT is always relative to the emitter, IC = ICE, IB = IBE) IC = 0 when IB = 0 When there is base to emitter current a collector to emitter current is allowed to flow, the amount of current that may flow is dictated by two parameters, the amount of base current and the current gain of the transistor (typically referred to as beta or hFE) IC = IB * beta So if you removed the LED and the 680 ohm resistor and connect it directly to the 9V battery, the maximum collector current is given by the above equation, a typical transistor has a beta between 50 and 150, assuming 100 this means: 300uA * 100 = 30mA It's important to note the BJT is a current controlled device so you should avoid thinking of it as a controlled resistance or a plain switch, it's more of a current source which obeys IC = IB * beta Beta is however very unreliable, it varies with applied collector voltage, base voltage, collector current, transistor and temperature which is why a 680 Ohm resistor is used to limit the current to 10mA rather than rely on just the transistor current gain. In choosing the base resistor size I assume a minimum current gain, this is almost always specified in the datasheet for the transistor, for a desired collector current of 10mA and a minimum beta of 40 this works out as: IB = IC / beta 10mA / 40 = 250uA I increased this to 300uA as is good practice which is given by a common 27k resistor, 300uA ensures that the LED is turned fully at 10mA on no matter how beta changes. The collector and base current are combined at the emitter so: IE = IC + IB[/QUOTE] Thanks for the info, but it doesn't really help me a lot in this instance, since I'm trying to make a NOT-gate. My end goal here is to have two timing circuits, turning a speaker on and off, I've already gotten one timing circuit provided for me, and now I'm working on reseting that.

[QUOTE=scratch (nl);49463897]It might be working because I'm bypassing the ATmega328P (RST to GND), thus communication being provided by the ATmega16u2. See [URL]https://www.arduino.cc/en/uploads/Main/Arduino_Uno_Rev3-schematic.pdf[/URL] [sp]Or I'm just talking gibberish and should be grateful nothing broke so far[/sp][/QUOTE] I think you're right. Part of the reason the mini, micro's, and nanos are all 5v. The 328 uses 5v for logic level as well, and should engage brown-out mode at about 4v. You can change this, however, just know that you'll have to lower the core clock a fair bit. I guess the ESP has an onboard voltage divider or regulator, independent of the atmega328? The 328p has a typical operation range of 1.8v-5v iirc [editline]5th January 2016[/editline] i mean im pretty sure thats sorta what the mega 2560 does it has a regulator nearly solely to protect the bootloader, and there's a bunch of hocus-pocus magic weird shit going on in there that I forget right now in order for everything to work together all friendly and nice [editline]5th January 2016[/editline] simple question, but are you using the right ICSP pins? or are you not using that? I accidentally wired to and from the bootloaders when trying to flash firmware between my pile of zombie 2560's and I felt dumb after realizing my mistake