ledmicrocontroller
I'm trying to glow an led using a microcontroller and switch. I have programmed it to blink
Every 2 seconds.when voltage comes from PB1 switch gets tuned on and led . Circuit diagram is shown here.
It doesn't work. Please help me to fix it.is this how to connect to atmega 8?
Since you're new to electronics the power supply is a good way to start. You'll need one for everything you want to do.
AndrejaKo and the others already told you what's probably wrong: you installed it the wrong way, which killed it, because while most diodes can stand the 18V reverse, LEDs can't. What did you do? Just try it one way and reverse it if you saw it was wrong? OK, now you know that you can't afford this with LEDs.
Try to make it a habit to use components you've got information on. No design engineer would use a part if he can't get the datasheet for it. Here's the drawing from the datasheet for a random LED:
There are a few differences between the pins. One pin is longer than the other, the LED's rim isn't complete on one side, and if you look into the LED you may see a thin wire going from the chip to one of the pins. The wire is usually the anode. Usually! Don't try it if you're not sure, remember you don't get a second chance. Use whatever the manufacturer indicates as the anode. In this case that's the longer pin. Only rely on that. That's your anode, which goes to the plus.
If you wouldn't have a datasheet there's still a safe way to try it. Place two of your LEDs in anti-parallel: the anode of one LED to the other's cathode, and vice versa. Place a resistor in series and connect to a 9V battery (or your new power supply!). One LED will be correctly polarized and that will light, the other one will be reverse polarized, but the first one's forward voltage (around 2V for a red LED, 3.5V for a white one) will limit the reverse voltage to a safe level.
Another thing mentioned is your resistor. Nine out of ten it's too high, though you should still see the LED light faintly. Again there's the datasheet. You want to know the LED's nominal current, which is often around 20mA for an indicator LED, but also its voltage drop. The same datasheet says 40mcd at 20mA, and the voltage is 2V typical. Then you use Ohm's Law to calculate the series resistor:
\$ R = \dfrac{V}{I} = \dfrac{V_+ - V_{LED}}{I_{LED}} = \dfrac{18V - 2V}{20mA} = 800\Omega\$
if it's a 20mA LED. The closest value you'll find will probably be 820\$\Omega\$. Note that, due to the high input voltage, the power dissipated will be a bit more than 300mW, so a 1/4W resistor won't do. Pick at least a 1/2W resistor.
There's still another possible explanation why it doesn't work. LEDs are sensitive to ESD, ElectroStatic Discharge. Touching the LED when you're statically charged to a few kilovolt will also kill the LED. It's possible that many of the components you'll use in the beginning won't be so sensitive to this, like resistors or capacitors. CMOS is also ESD sensitive. If you don't have ESD protection (probably not yet) then touch a large metal object before you handle the LED.
I don't think ADCON will make a difference in this case seeing PORTD pins you're using are multiplexed with Enhanced PWM Mode and the Streaming Parallel Port (SPP), rather than Analog Input Module, which take priority over port I/O
You'll want to disable the Enhanced PWM that're also multiplexed with PORTD pins. Also you'll probbably want disable SPP, by clearing SPPCON register bit 0 (SPPEN) to 0
So a solution in this case can be to configure CCP1CON to anything but Dual or Quad PWM by setting CCP1CON to 00xx 11xx
Best Answer
Ok, first we need to tackle that circuit of yours.
I think you're probably trying to run before you can walk, since there are a large number of things fundamentally wrong with it - things that you really should know before you start playing with microchips.
The formula for calculating the resistor for an LED is:
\$R_{LED} = \frac{V_{IN} - V_F}{I_F}\$
Where \$V_{IN}\$ is the supply voltage, \$V_F\$ is the forward voltage drop of the LED, and \$I_F\$ is the current you need to flow through the LED.
Taking some typical values of \$V_F = 2.2V\$ and \$I_F = 20mA\$ we have:
\$R_{LED} = \frac{5 - 2.2}{0.02} = 140\Omega\$ - or the nearest easily available resistor above that value (\$150\Omega\$).
The base resistor's value is less critical, and anything in the region of \$470\Omega\$ should be fine.
Decoupling capacitors would typically be \$100nF\$ per power pin pair, and often \$10{\mu}F\$ per group of around 3 chips or so.
simulate this circuit – Schematic created using CircuitLab
Now, once you have all that sorted, and possibly have replaced your LED which may by now not be functioning, we can look at your software. If it's still not working after you have made all the right changes, then please post your full source code so we can see if you have configured the chip correctly.