Simple circuit question, NEED HELP!

[idtkid]

I need to wire up a simple circuit comprised of the following components.

5x Infrared LEDs 1.28v 30mA
1x preexisting IR camera/bluetooth circuit 3v 300mA

My question is can I wire these up to run off of 1 of 3 ac adapter that I have?

7.5v 700mA
5.2v 650mA
5.1v 700mA

If so would someone please describe or a provide as schematic as to how to create this circuit? I would be so appreciative as this is a school project that needs to be completed soon. (I am a product designer creating a functional prototype for my thesis class) Thanks in advance for any help you may have

 

[Boncuk]

Hi idtkid

5 IR-LEDs of Uf=1.28V connected in series will require a total voltage of 6.4V.

Using the 7.5V power supply there will be 1.1V left to "kill". Applying Ohms law the current limiting resistor must be (US-UF)/IF, (Supply voltage-LED forward voltage)/LED forward current))

(7.5V-6.4V)/0.03=36.333Ohm. Next higher standard resistor value is 39Ohm. Using that value the total LED current will decrease to 28.205mA, which is acceptable.

For the IR-camera use the same power supply and make a voltage regulator circuit using an LM317T (adjustable voltage regulator, 1A). The data sheet contains a suggestion for an adjustable power supply. Instead of using a potentiometer you might use a trimpot to obtain a fixed value output voltage of 3V.

Boncuk

 

[idtkid]

So off the leads of the power supply I run a series chain of 5 LEDs started by a 39ohm (wattage?) resistor. I wire the voltage regulator circuit in after that? or off of the same leads? Thanks for your help, forgive me as I am awfully new to this.

 

[Boncuk]

Hi idtkid,

first one advise: please avoid double postings. It will confuse the forum members as well as yourself.

For power rating of the current limiting resistor use this formula:

P(W)=U(V)*I(A). In your case U=1.1V, I=0.0282. Go figure.

Here is a complete circuit for your power supply including a single sided PCB layout.

With P1 in the center position the output voltage should be around 2.57V. At 66% (right hand turn) the voltage should be exactly 3.0V.

Final adjustment: Adjust voltage without load to approx. 3.0V. Connect the camera and recheck output voltage. A small voltage drop might be compensated by turning P1 a bit more clockwise.

Important: This circuit has no reverse polarity protection. Check polarity before connecting - otherwise poof.

Boncuk

 

[idtkid]

thank you sooo very much for the schematic! appreciate it big time! I will let you know of my results!