• Electronics and Embedded Programming V3
    1,545 replies, posted
Does anybody know how i could make the receiver of a wireless doorbell switch a relay? I've tried tracing some paths to get a pulse when the button gets pressed but couldn't find it. I also considered to add a relay instead of the speaker(there is a tune which sends one pulse, length doesn't matter) but the max voltage is around 1V so i am kinda stuck. Anybody know a solution?
What kind of load are you trying to switch? The issue with trying to switch something from audio output is that it's an AC waveform. What I would do is: common-emitter amplifier -> coupling cap -> peak detection (diode and cap) -> common-emitter amplifier [img]http://i.imgur.com/5rGG6.png[/img] Suddenly, anyone who caught this post before the edit knows just how out of practice I am.
[QUOTE=ROBO_DONUT;35624845]What kind of load are you trying to switch? The issue with trying to switch something from audio output is that it's an AC waveform. What I would do is: common-emitter amplifier -> coupling cap -> peak detection (diode and cap) -> common-emitter amplifier [img]http://i.imgur.com/5rGG6.png[/img] Suddenly, anyone who caught this post before the edit knows just how out of practice I am.[/QUOTE] I'd do roughly the same thing except replace the common-emitter amplifier section with an op-amp, the inverting feedback coming from the emitter of the output transistor to cancel out the VBE drop. So Peak Dector > Op-Amp > Output. You also forgot the protection diode across the relay coil to prevent inductive kickback. A capacitor is also probably a good idea to keep the relay on for a short time, that or a 555 monostable timer.
[QUOTE=Chryseus;35625819]You also forgot the protection diode across the relay coil to prevent inductive kickback.[/QUOTE] Ah, true.
I just thought of something. The separate sound chip needs to know that the button is pressed so maybe it receives a pulse telling it to play the sound. I'm at school right now so i can't check but it should be usable.
This came in the mail today: [t]https://lh4.googleusercontent.com/--azCpcH88DE/T5AT-KmaTsI/AAAAAAAAA1s/ZfdISOJ10-I/s960/tube.jpg[/t] Powered the heater and it gives a good glow so it's most likely good.
The sound chip receives a pulse when the button is pressed. Sadly it's only around 2.2V This should be easier as it is not an audio being sent. Anybody know where to go from here? I have heard about how to step up the voltage but that was some time ago and i have never looked into it.
[QUOTE=0lenny0;35635289]The sound chip receives a pulse when the button is pressed. Sadly it's only around 2.2V This should be easier as it is not an audio being sent. Anybody know where to go from here? I have heard about how to step up the voltage but that was some time ago and i have never looked into it.[/QUOTE] One transistor will do the job: [img]http://i.imgur.com/S0LGX.png[/img]
[QUOTE=ROBO_DONUT;35635511]One transistor will do the job: [img]http://i.imgur.com/S0LGX.png[/img][/QUOTE] Sorry for being stupid but could you try to explain the working of this to me? I am trying to convert 2.2V DC to 12V DC (i see the + connected to a transformer)
Imagine the transistor similar to a mechanical switch. When there is zero voltage on the base, the switch is open (that means not connected), and no electricity can flow through it. But when you apply the 2.2V, it will be like a closed switch and allows electricity to flow through.
[QUOTE=0lenny0;35637397]Sorry for being stupid but could you try to explain the working of this to me? I am trying to convert 2.2V DC to 12V DC (i see the + connected to a transformer)[/QUOTE] The coil you can see is not a transformer but the coil of a relay, the diode that is placed across it (in reverse to normal current flow) is to prevent high voltage damaging the transistor when the coil is de-energized. When the 2.2V enters the base of the NPN transistor it turns it on, allowing current to flow from the collector terminal (shown at the top) to the emitter terminal (the one with the arrow). In this way it acts as a simple switch, the 2.2V causing it to close. Once you have a path for current to flow through the coil the relay contact will close, depending on the relay, most have two contacts, one is normally open, the other is normally closed. So for example if you connect a light through the normally open contact it won't do anything, once the 2.2V signal arrives however the coil will energize and cause the contact of the relay to close, so your light will light up. If you want to convert the 2.2V signal to 12V you can either: Use a relay to control another 12V supply, or a transistor. Build a boost converter. If you want to just use a transistor you can ignore the relay. I suggest you read: [url]http://www.allaboutcircuits.com/vol_3/chpt_4/2.html[/url]
Hey guys look another thing like arduino! [url]http://www.kickstarter.com/projects/teague/teagueduino-learn-to-make?ref=category[/url] And guess what it's only $160! :suicide:
I could see that having a more Signal Analysis use, not much use as a embedded system(IMO) with those molex headers. Remember kiddies, its all about the plugs! :v:
It's a plug and play toy for people who don't want to electronics [editline]20th April 2012[/editline] aka ~artists~
fiddling with the beagleboard [IMG]http://dl.dropbox.com/u/3357334/robots/IMG_20120418_000921.jpg[/IMG]
[QUOTE=GamingRobot32;35646363]fiddling with the beagleboard on a mobile robot [b]Massive Image[/b][/QUOTE] Please use the [t] tag instead so people can actually see the image without needing a massive screen.
[QUOTE=Chryseus;35646369]Please use the [t] tag instead so people can actually see the image without needing a massive screen.[/QUOTE] fixed thanks
Doesn't the beagleboard have GPIO pins?
Decided to try and play with a transistor on a simulator before using it in my project. [img]http://dl.dropbox.com/u/64443729/ele/tran.PNG[/img] Simulator didn't throw an error at me but i don't completely trust them. The 50V should enable the transistor which powers the light from the 120V power source. Did i get it right?
Why do you play with a simulator, just try it out with a real transistor. It's so much more fun!
[QUOTE=0lenny0;35649914]Decided to try and play with a transistor on a simulator before using it in my project. [img]http://dl.dropbox.com/u/64443729/ele/tran.PNG[/img] Simulator didn't throw an error at me but i don't completely trust them. The 50V should enable the transistor which powers the light from the 120V power source. Did i get it right?[/QUOTE] [img]http://i.imgur.com/ZUsUt.png[/img] If you were to build this in real life, expect a huge bang and the lights going out (assuming your 120v source is mains).
Also you should use a 9V battery, and never ever touch mains
[QUOTE=DrLuke;35650256]Also you should use a 9V battery, and never ever touch mains[/QUOTE] Especially if you are a beginner.
[QUOTE=0lenny0;35649914]Decided to try and play with a transistor on a simulator before using it in my project. [img]http://dl.dropbox.com/u/64443729/ele/tran.PNG[/img] Simulator didn't throw an error at me but i don't completely trust them. The 50V should enable the transistor which powers the light from the 120V power source. Did i get it right?[/QUOTE] No you got it very very wrong. V1 is practically shorted through the base-emitter junction of the transistor. Q2 is not a power transistor [del]and even if was have fun getting a 100W heatsink.[/del] (Actually the power would depend on the voltage drop so it is possible to do.) There is no ground node so the simulation results are probably bad. Don't play with high voltage until you have a big long beard and a degree in analog wizardry.
I think he was just seeing if he got the gist of using transistors. Answer: Not quite. You'll want to put a bias resistor on the Base lead for one, on top of the other things Chryseus mentioned. (shorting V1, no ground, etc).
[QUOTE=Zero-Point;35650926]bias resistor on the Base lead for one[/QUOTE] You don't want a 'bias resistor' here because he's not building a class-A amplifier. With no bias, the transistor will switch at 0.7V, which is exactly what he wants. Switching at some reasonable threshold. You just need something to limit the current because the base-emitter junction is pretty much just a diode, and will not limit current on its own. @OP: Don't touch mains until you know what you're doing. It's fine to work with low voltage, and if you want to switch a light or something, I'm sure you can find plenty of LED lighting solutions that work at ~12V.
You probably should think about a relay instead: [URL="http://www.sparkfun.com/tutorials/119"]http://www.sparkfun.com/tutorials/119[/URL] But, be responsible considering the inherit danger of the mains.
[QUOTE=ROBO_DONUT;35651182]You don't want a 'bias resistor' here because he's not building a class-A amplifier. With no bias, the transistor will switch at 0.7V, which is exactly what he wants. Switching at some reasonable threshold. You just need something to limit the current because the base-emitter junction is pretty much just a diode, and will not limit current on its own.[/QUOTE] Yeah, that's what I meant, my bad. I haven't dabbled with this stuff for ages.
[QUOTE=Chryseus;35650366]No you got it very very wrong. V1 is practically shorted through the base-emitter junction of the transistor. Q2 is not a power transistor [del]and even if was have fun getting a 100W heatsink.[/del] (Actually the power would depend on the voltage drop so it is possible to do.) There is no ground node so the simulation results are probably bad. Don't play with high voltage until you have a big long beard and a degree in analog wizardry.[/QUOTE] Seems like i forgot to put a resistor on V1, noticed that now. I am also not planning to use mains. I only used it here for an example as it was the first light could find in the list.
You know what's missing from this thread? Robots. [img]http://imgkk.com/i/ajnv.jpg[/img] [code]#include <Servo.h> Servo motorL; Servo motorR; int Speed = 15; int StopL = 90; int StopR = 90; int LeftPin = 10; int RightPin = 9; int SensorPin = A0; int reading = 0; // forwards is l- r+ void setup() { motorL.attach(LeftPin); motorR.attach (RightPin); motorL.write(StopL); motorR.write(StopR); } void loop() { reading = analogRead(SensorPin); if (reading < 15) { motorL.write(StopL); motorR.write(StopR); delay(1000); motorL.write(StopL + 10); motorR.write(StopR + 10); delay(800); motorL.write(StopL); motorR.write(StopR); delay(1000); } else { motorL.write(StopL - 10); motorR.write(StopR + 10); delay(500); } }[/code]
Sorry, you need to Log In to post a reply to this thread.