hey m/-\n make like an article about some cool modded mouses!!! with and without scrollwheel!!!(though if you don't have a scrollwheel you live in the jura!!)
My mouse looks so nice already mod'in isn't required - http://users.ox.ac.uk/~some0814/logi2.jpg Super's done a cool mod on this intelimouse (painted black - very shiny) - ask him nicely and he may tell you how he did it r.
If its a USB mouse you could drag off some voltage and have LED's, could you not? But not if its PS2....
This Finnish modding site is worth of checking out. http://hw.metku.net/ Altought all text is written in finnish, pics tell all there is to know.. At least I hope so.. =) BTW, you CAN get +5V voltage out of PS/2, USB and Serial mouse, so with correct resistor you can put led(s) in it. Even a fan!
Two things I don't understand about this mod... http://hw.metku.net/logomouse/index.html Why is the resister on gnd? I thought it was meant to be cutting down the power before it gets to the LED. If you put it after the resistor isn't it increasing the load pulled through the LED? And, why on earth would you want a fan in your mouse. The way I see it, it's either blowing your hand away from the mouse. Or it's sucking it down, still don't get it?! Still pretty cool someone has big enough gonads to solder a LED to an IC =)
No it doesn't work like that... e.g. in a potential divider if doesn't matter which way round you put the resistors they will still get the same voltage drop across them. Another way of looking at this is that the actual current flows in the opposite direction - from -ve to +ve so resistors before leds should worry you r.
That's a cool site... anyone know of an online whole page english convertor? Would be nice to read the articles r.
Well I don't understand resisters at all then. I thought they put extra load on so that the things behind them end up with less load Ohwell, Im all confused now.. The backwards thing makes sense though, protons don't move, electrons move towards protons, and protons are +ve All makes sence, but, don't understand resistors now Which is all bad, cuz up until yesterday I understood capictors, cept I have them in this circuit basically making it a short circuit. + Cap - Then this guy tells me, caps don't allow electricity through them, so like WTF?! I thought they store it up and start flowing when it reaches a certain point, and now ytou're telling me to put it in a short circuit and it'll never reach it's peak or what?%#@ Does that have anything to do with short circuit not having a load on it? If that's the case, why have the short at all? Really don't get it =) Yours confusedly Ben
LOL ok well I'll try and explain a bit... You can look at wires like water pipes and a resistor like a partial blobk in the pipe - when it's present it slows down the flow of water in the whole pipe, so it doesn't matter where you put the led in the now flow restricted pipe The electron backwards pictures isn't very good either, you're right in saying protons don't move (metals are crystals and atoms are locally isolated) but you can model the lack of an electron as the presence of a positively charged particle. They call this the hole and it does move the same way as most people think. Capacitors. Now they're fun. A capacitor is in it's most simple form two metal plates with a dielectric material between them. When you put a voltage across the plate charge builds up on the plates (one +ve, the other -ve) and current only flows to build this charge up (they us no real electron transport across the gap with the dielectric in it). So you take a capacitor and place a d.c. curent across it. A current appears to flow corresponding to electrons building up on one plate and electrons evactuating the other (holes building up). This can only go on for so long and so eventually no current will flow, the capacitor will be fully charged and look like an infinite resistor. You can now short circuit the capacitor and it will discharge... Cheers, r.
Ouch. Ok, I have to wade in to this. A capacitor looks like an open circuit to steady-state DC. So if you place a voltage across it, it will build up a stored charge until its internal voltage equals the externally imposed voltage. During that time that it's charging, a current flows. However, to alternating current, a capacitor represents a reactive impedance (an opposition to the flow of current) which is a function of frequency. In general, a small cap looks more like a short circuit at higher frequencies, and more like an open circuit at lower frequencies. So TheCheeseMan's statement "So you take a capacitor and place a d.c. curent across it" is a bit flawed--you can put a DC voltage across it, and that causes current to flow that looks a little like ( V / R ) * exp ( -t / ( R * C ) ) an exponential function of time. R is the series resistance in the charging circuit, and V is the DC voltage supplied. Back to the original LED question, The series resistance is limiting current to the LED. It matters not if it's "in front" or "behind" the LED, because all elements in a series circuit carry the same current. The current-vs-voltage relationship in an LED is not a linear relation like a resistor. The actual function is exponential. However, it's a useful model to think of an LED having a "turn-on voltage" and an "operating current" corresponding to the package ratings. If you apply a DC voltage equal to or greater than the turn-on voltage, current will flow equal to the operating current. The model is deficient, though, because it implies that any voltage whatsoever above the turn-on voltage may be safely applied. The diode has a maximum current it can carry without the junction overheating, thus releasing the magic smoke. The series resistance expression you've all seen for LED circuits, R = ( Vs - Vd ) / Id is based on the above LED model, and some assumptions. First, we assume that the voltage drop across the LED is constant, and equal to the turn-on voltage. Then we assign a current in the series loop equal to the diode's operating current, and then we have enough parameters to solve for R, the resistance in series with the LED. In practice, because the model only approximates reality, the voltage dropped by the LED will differ slightly, as will the current, but it's still a quite useful model because you can solve it easily and it works. TheCheeseMan is right about electrons as charge carriers moving in a direction opposite to "conventional" current flow (the way the arrow points), but that doesn't become interesting until you start to talk about semiconductor physics and charge depletion. I could use some Gorgonzola myself....
LOL that's not true, though band theory does come up most in semiconductor physics, electron flow direction is very important for things like hall probes (which admittedly are now made from SHET samples to limit electron motion to 2 dimensions) but it's not necessary. I only mentioned current flow as his 'behind the led' was really in front in a sense at least. As for applying AC currents across capacitors - I'm surprised you didn't leap into complex phase factors, impedence and waveguides... r.
................oh god now youve done it !!!! Rob's on the war path with his physics degree..............I'd duck and cover if I was you !!!
The Physicist tells you how it works, the Engineer actually makes it work, loosley based on what the Physicist told him! Rules and thumbs spring to mind.
That's not true. I wouldn't be seen dead talking to an engineer.... we make everything ourselves, just only one of everything. r.
