how can i reduce current of 1A to less than 50mA

[FYP]

Hi every1

its a simple question actually, i wan to build an IR sensor which works on a power source of 5V, 10mA...the problem is my power source is 5V,1A which might damage the IR TX & RX, so i need to reduce the current.
i tried to connect parallel resistor ( in Crocodile) but the power dissipation keep blowing the resistor and LED up (LED is just an example).
if i changed the parallel resistor power dissipation (size)  from 2W to other larger value...would it help me overcome the problem ?
i know that this question might be little bit under-level question...but sometimes simple questions force u to ask

thanks for helping

 

[audioguru2]

Most circuits draw only as much current as they need.

An LED must have a current-limiting circuit or series resistor (not a parallel resistor).

Attach your IR TX & RX circuits for us to see if they will try to draw more than 50mA.

 

[FYP]

if i put a series resistor...the resistor will be damaged due to the exceeded power dissipation i guess, correct me if im wrong plz.
any way this is the circuit im using:

thank you

 

[Hero999]

if i put a series resistor...the resistor will be damaged due to the exceeded power dissipation

Why would it be damaged?

Use Ohm's law to calculate the current and power dissipation and choose a resistor with a higher power rating.

 

[audioguru2]

The currents in the simple upside-down circuit are much lower than 50mA.
The resistors will not get warm.

Use simple arithmatic to calculate the currents and the power dissipated in each resistor.

 

[Hero999]

Agreed, it isn't very well drawn but it does seem to be correct.

The correct and normal convention is for the positive power supply to be at the top and the negative to be at the bottom.

Where possible real symbols should be used rather than the IC outline.

The symbol for a comparator is show below.

Why are you using the LM324? I'd recommend using a real comparator IC such as the LM393 or LM311.