How can I boost up current (100mA to 3A) for multiple LEDs used??

I've been working on a project to control multiple LEDs..how can i boost the current from 100mA to 3A so that i can have enough current to use multiple LEDs connected in parallel.

You forgot to post your schematic. Then we do not know what provides the 100mA drive and we do not know your power supply voltage.
We also do not know how you are paralleling the LEDs so that they do not burn out.

schematic cct

this is the schematic.. the 100mA current is sink from every drain from the power logic 8bit shift register (TPIC6C595)..i dont know what is the component and the circuit that can boost the current from 100mA to 3A?? i want to boost every output from the TPIC6C595

You show 8 LEDs in parallel, driven from something that limits the total current to 3A.
But your LEDs are not matched and they are different. So some LEDs will draw way more than 3A/8 like maybe 0.5A and maybe burn out and other LEDs will draw nothing and will not light. Therefore the LEDs must be matched or each LED must have its own current-limiting resistor.
Do you have LEDs that can survive 375mA each?

The 100mA outputs of the TPIC6C595 are their maximum allowed continuous current. If you limit their current to 80mA each then a PNP power transistor can drive the LEDs to a total of 800mA. A TIP125 PNP darlington power transistor can do 3A but it will heat with 6W maximum and will need a heatsink.
A logic-level P-channel Mosfet can do it without a heatsink.

If you use a PNP darlington or a P-channel Mosfet then you do not need the power shift register. An ordinary shift register can drive them.

thanks audioguru..i will try it with 30 LEDs in parallel which each LED can have 100mA current..if there is any problem i will ask again..thanks

Each LED, or series string, will need it's own current limiting resistor. What LED are you using? What is its rated operating current and forward voltage? You need to know those numbers to calculate the resistor value.

But I will ask one other question. What is the power source for your system? Is it 5V only, or is there a higher voltage available? If so, running the LEDs in series or series/parallel is generally a better choice. For instance, if the LEDs have a fv of 2V at 100mA, you can put two of them is series with a 5V source, and, since the same current runs through two parts, the efficiency is more than double that of running a single LED But, if you have 12 volts, you can put 5 or 6 in series using the same current. Even though you do need the higher voltage, you will probably waste less power in the limiting resistor.

sorry there is some changes in this project..the LEDs will be on an external board using 5V as power supply..the circuit that i want to design is the stabilizer circuit that can stand 3A current from the supply..can i know what is the suitable component that can sink the current till 3A??

LEDs are all different. Therefore they cannot be connected directly in parallel. A few might use 300mA (then will burn out) and a few will not light.
Each LED or two LEDs in series needs its own current-limiting resistor. Limiting the total current to 3A will not work.

oh i see..thanks for the info

im going to test a dual n-channel mosfet (FDC6401N) for the relevant current consumption and heat dissipation. the datasheet shows the mosfet maximum id is 3A..i only have 5V 4A power supply..how i can decrease the current from 4A to 3A so i can test the mosfet to sink 3A current??

You must LIMIT THE CURRENT FOR EACH LED then a complicated circuit to limit the maximum current (it won't work) is not needed.

You can use a simple buffer amplifier based on a small high gain transistor (such as, BC547/8/9 2N2222 etc) followed by a power transistor (like 2N3055, TIP3055 etc) - however, as Audioguru has told you, unless you limit the current for each and every LED, you're going to have nothing but smoke and charred silicon, and a whole lot of wasted power heating current limiting resistors.

from this cct, the power supply is 5V,4A..the current that should be drain by the msofet is 3A (max Id=3A)..how can i get the 3A current?? sorry im a newbie.

bopen87 said:

from this cct, the power supply is 5V,4A..the current that should be drain by the msofet is 3A (max Id=3A)..how can i get the 3A current?? sorry im a newbie.

Click to expand...

Yes you are

You need to learn Ohm's Law:

The power supply is rated to deliver up to 4A at 5V.

The current it will actually deliver is determined by the resistance of the circuit connected to it. Your 12V car battery is rated to deliver over a thousand Amps. How much does it deliver to the dome light when you open the door? About 1A.

If you would like to test switching 3A with a suitable NFET, then by Ohms Law, the resistance of the external circuit would have to be equal to the applied voltage (5V in this case) divided by the desired current (3A), or R = E/I = 5/3 = 1.667Ω.

Now, that would be the total resistance connected to the supply; including the resistance of the wires, the On Resistance (Ron) of the NFET when it is turned on, and the actual resistor in the Drain lead of the NFET. If we ignore the first two for now, that means you would need a 1.667Ω power resistor.

Why did I call it a power resistor? The power dissipated in the resistor when the NFET is turned on will be P = IE = 3χ5 = 15Watts, so a little 1/4W or 1/2W wont even come close to handling that much power.... An example of a power resistor that would handle it is here.