Question for Microcontroller

[Tuck3rz]

**broken link removed**

I am to design a circuit to light up a LED at the O/P of the microcontroller. What possible component can i add in the circuit to light up the 3V/2A LED.
Output voltage is 12V.

Sorry for stupid questions.

 

[Gayan Soyza]

You need power & you need the source code to turn on a LED in addition to the ones you posted.

Sorry for the stupid answers.

 

[Tuck3rz]

Urm what i meant was to adding components like resistor to complete the circuit but i do not understand what to add.

 

[Pommie]

There is nothing to add, just connect the resistor to the output and the LED to Gnd.

Mike.

 

[Tuck3rz]

Thanks mike but will the voltage damage the LED? Since the output of the MC is 12V and the voltage rating for the LED is 3V/2A.

 

[ericgibbs]

Thanks mike but will the voltage damage the LED? Since the output of the MC is 12V and the voltage rating for the LED is 3V/2A.

hi,
A MOS FET like this type will drive a 2A diode, add a series resistor to the LED.

Note: a 200R with a 3V LED and a 12V supply will not drive a LED with 2Amps.
More like a 4.7R rated at least 25Watts!

 

[Tuck3rz]

hi,
A MOS FET like this type will drive a 2A diode, add a series resistor to the LED.

Note: a 200R with a 3V LED and a 12V supply will not drive a LED with 2Amps.
More like a 4.7R rated at least 25Watts!

Thanks for the info it helped me alot.

Another qns:
Do you mean we need another additional resistr?
Do we place the series resistor after or before the LED?

 

[ericgibbs]

Thanks for the info it helped me alot.

Another qns:
Do you mean we need another additional resistr?
Do we place the series resistor after or before the LED?

hi,
Look at this image.

 

[Tuck3rz]

thanks alot... sorry for the trouble to draw =)

 

[Tuck3rz]

hi,
Look at this image.

hey... what does the MOSFET?
Does it amplify the current? or?

and the output of the microcontroller is? the 12V is an extra?

 

[jakeselectronics]

Tuck3rz,
I will do my best to help you understand....

A MOSFET (metal oxide semiconductor field effect transistor), or any transistor can act like a switch.
Which is what it will be doing here in your circuit.

The "gate" (aka 'base') only requires a minute amount of current to effectively 'switch' the mosfet/transistor on. That is what the 1K resistor is for. To heavily limit current.

This is what your microcontroller will do. When you 'set' the ouput (logic 1), the MOSFET will turn on, allowing current to flow through the Drain to the Source (or if it were a transistor from the collector to the emmiter)

To put it simply,
- Output of Microcontroller goes high,
- MOSFET turns on (closes)
- Current flows from your 12v source through the big resistor, through the LED, through the MOSFET, to ground.

The low resistance resistor (4.7R @ 25W) is there to limit current seen as though the LED and MOSFET dont have much resistance...
It is only a small amount of resistance to allow a large amount of current to flow. That is why it needs to be rated at least 25W, becasue there will be alot of current passing through it.

Hope that makes it abit clearer....

2A LED hey... that one massive LED...
Got a photo? hope you dont mean .2A or even 2mA???

 

[Tuck3rz]

The 2A value is just a example given to us by our lecturer. We are not required to create the circuit out.

Thanks alot too i somehow able to understand

 

[jakeselectronics]

arrr ok... Makes sense.

Im not sure if the 4.7R resistor will limit to exactly 2A.
I dont know if ericgibbs worked it out, and neither did I.
Not sure if you need to work that out properly. The basic circuit is correct. But the values may need tweeking.

Use ohms Law to work out the right value resitor to limit to 2A.

V = I x R
I = V ÷ R
R = V ÷ I

P (in Watts) = I x V

 

[Tuck3rz]

because my lecturer say what if the resistor is fixed at 200 ohms

 

[Tuck3rz]

and now he changed the resistor to 200milli-ohms
Sorry

 

[jakeselectronics]

So..... you want to know what would happen?

Is that a question?

Use ohms law to find current flowing.

I (current in Amps) = V (voltage in volts) ÷ Resistance (resistance in ohms)

so,
I = 12v ÷ .200Ω
= 60AMPS!!!

in other words, if you used a 200milliΩ, you would blow the **** out of your LED.

 

[Pommie]

Yes, the lecturer should have changed the LED current to 20mA instead. It was an unrealistic circuit with a 2A LED and my guess was that it was a typo and should have read 20mA. He's now realised his mistake but chose to reduce the resistor to 200mΩ and you now have the worlds most powerful LED of 100W.

A normal circuit would drive an LED at 2V and 20mA via a 150Ω resistor from a 5V supply. I think what the lecturer intended was something similar but it somehow got confused.

Mike.