Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

How to convert 6V @ 2.5 A to 5v @ 2.0 A

Status
Not open for further replies.

Scarfunkle11

Newbie level 1
Joined
Jul 31, 2015
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
18
I am attempting to program a digital led strip with arduino just getting started on my electronics path in life. I am following this tutorial

https://learn.adafruit.com/digital-led-strip/wiring


I know that part of being a maker is hacking current electronics that you have. I have a AC to 6v 2.5 A dc power adapter right now and don't want to pay for a new one if I don't need to and I would love to learn more by altering or creating a circuit to utilize this power adapter that I do have since it seems to be close in the specs.

The tutorial calls for a 5V 2A power adapter. Would I be safe using this 6v 2.5 A ? If not (and even if so) how would I lower the voltage from 6v to 5v and the current down to 2 A from 2.5?

Thank you so much for any and all help, it is greatly appreciated.
 


    V

    Points: 2
    Helpful Answer Positive Rating
Try placing one or more diodes in series with the output.
 

    V

    Points: 2
    Helpful Answer Positive Rating
Simple series voltage regulator, using NPN transistor and zener diode. The load voltage stays close to 5V.

It is especially useful if the supply is between 6 & 7 V.



Watt ratings must be adequate. Some power is wasted in the bias circuit. For this reason this type of regulator may become unworkable if you exceed a certain range of supply and load situations.
 

    V

    Points: 2
    Helpful Answer Positive Rating
Hi,

if the current is constant and known, then a simple rsistor could work.

Referring to your values (the 2.5A of the 6V supply is not important as long as it is same or bigger as the load current)
R = (6V - 5V) / 2A = 1V / 2A = 0.5 Ohms. Power dissipation is 2W. Please use a 5W rated resistor.

A higher ohmic resistor makes the voltage drop bigger. So it is more safe not to destroy something.

Klaus
 

    V

    Points: 2
    Helpful Answer Positive Rating
  1. Unless you know that your supply is regulated , expect up to 40% variation from full load up to no LED load.
  2. So choice of solution is best with LDO for efficiency and regulation
  3. NPN with zener is ok but not 1% as shown as they also have 7% tolerance typical at a max current and maybe 5% for a 1W type with less dissipation.
  4. A resistor solution is possible but least favorable.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top