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What is this things use? Answered

I researched about it and, it told me what it was, but what do I use it for? does it limit current?

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Jack A Lopez
Jack A Lopez

11 months ago

I think it is a DC-to-DC converter, specifically a buck converter, built around the LM2596 IC (integrated circuit).

https://www.alldatasheet.com/view.jsp?Searchword=L...

It's purpose is to convert DC power, at voltage Vin, to DC power at a lower voltage Vout. That is: Vout < Vin. Also the one in your picture is adjustable. The little screw on that pot can be turned, to adjust the set point for Vout.

Actually that is sort of what the word "buck" means in this context. It means Vout less than Vin. Sometimes the phrase "step down" is used as a synonym.

https://en.wikipedia.org/wiki/Buck_converter

I suppose the antonym to "buck" is "boost" or "step up." That is the other, sort of commonly seen switching topology.

https://en.wikipedia.org/wiki/Boost_converter

Although those two are not exhaustive. Other kinds of switching converters exist. Buck and Boost are just the two most often seen.

The buck converter is more efficient than the, uh, sort of old school way, a linear voltage regulator IC (the like LM317, or LM78xx series of voltage regulators) to get a lower DC voltage from a higher one.

https://en.wikipedia.org/wiki/Linear_regulator

However, the higher efficiency of the buck converter comes at the cost of higher complexity. A whole board is needed, instead of just one little printed silicon IC. It needs a physical way to store energy, temporarily. Usually a big inductor is used. And it needs a way to alternately move energy into, and out of, the energy storage.

In contrast, the linear regulator simply disposes of voltage as heat. A linear regulator necessarily wastes an amount of power equal to:

Pwaste = I*(Vin-Vout) where I=Iin=Iout

As far as I know there is no linear regulator analog to the boost converter (with Vout>Vin), and the reason why is because it would involve some impossible physics, like negative power dissipation (Pwaste<0), essentially a black brick that gets cold instead of hot, by mysteriously absorbing heat from its surroundings, and turning that heat into useful work.

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Orngrimm
Orngrimm

11 months ago

It is a DC/DC-Converter. In your case, it is a Step-down converter based on the LM2596. See https://www.aliexpress.com/item/32851488715.html

It takes voltage (4.5V to 40V) at the 2 terminals labeled IN+ and IN- and converts them to an adjustable (Via the trimpotentiometer) voltage which has to be lower athan the input-voltage and will fall in the range of 1.23V to 30V at the two terminal labeles OUT+ and OUT-

You use it as example if you have a 12V or 24V battery or Powersupply and want to power your Arduino pro Micro which only tolerates 5V. You convert the 12 or 24V down to 5V.

You may have heard of LDO (Low dropout regulators) or linear regulators which do the same. Whats the difference?
Well... A DC/DC-Converter basically converts power (with some little losses). A LDO just burns the excess power in the form of heat. MUCH more lossy especially if you go from high voltage the very low voltage but with high current.
Quick math with the following scenario: 12V-Battery, 5V load with 1A.
LDO:
Burns away the difference in voltage at the current flowing:
(Uin - Uout) * I = Ploss
(12V - 5V) * 1A = 7W
We use in our load only 5V * 1A = 5W
So: An LDO gives in this scenario 7W loss and 5W usable. or in other word, an efficiancy of 5/7 or 71%

DC/Dc-Converter:
Converts the incoming power (Voltage & Current-Combo of the supply) to the outgoing power (Voltage & current-combo of output.
A normal DC/DC like the LM2596 has about 90% efficiency. In other word, If we pull 5W (5V * 1A) from it, we have to add 10% mire into the converter to make up for the loss.
So 5.5W In, 5W out, 0.5W loss.

Comparison:
LDO: 7W loss
DC/DC: 0.5W loss
The DC/DC-Converter has in this scenario 14x less loss than the LDO.


It WILL limit the current. No DC/DC-Converter has no limit. In this case the limit is somehere in the region of 2A to 3A a bit depending on the Input- and Output-Voltages...
But as long as you pull less than this limit, not more current will flow as you pull from it. It will not "force" the current out.
This is a constant voltage source. It will keep the voltage as long as possible and will lower it if you run into the current limit.