Continue to Site

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

Status
Not open for further replies.

Scarfunkle11

Newbie level 1
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

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
Try placing one or more diodes in series with the output.

V
Points: 2
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
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
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.