# Adjustable output current (elementary question)

#### dk31

Joined Oct 30, 2014
36
Hi all,
Just trying to figure out things.
I have a buck-boost converter that takes 12V input from a power supply and outputs 1-30V adjustable, and 5Amps maximum. It has two potentiometers: one to adjust output voltage and one for output current.
I know that if I put a load on output, e.g a common light bulb that draws 1Amp at 12Volts, changing output voltage will dim the light.
What are the effects of changing output current? If I adjust the buck-boost for 12V and 0.5 Amps output, will it be damaged if I connect the 12V/1Amp light bulb?
Maybe sounds like a stupid question - I only ask cause I know that if I was to connect a 12V/0.5A battery to the bulb, that would damage the battery...

#### ScottWang

Joined Aug 23, 2012
6,822
It won't damage the bulb, but the voltage will going down, because V=I*R, the R is the same, the I is smaller, so the V will also became small.

#### dk31

Joined Oct 30, 2014
36
It won't damage the bulb, but the voltage will going down, because V=I*R, the R is the same, the I is smaller, so the V will also became small.
So can we way that
-the voltage pot sets a constant voltage output and
-the current pot sets a constant current output by adjusting voltage ?
And using my example above, if I plugged the light bulb that draws 1Amp at 12Volts, setting the pot for output current 0.5Amps would merely adjust the output voltage to 6volts (an imaginary bulb that would have constant resistance)?
Is that the (simplified) full story?

#### ScottWang

Joined Aug 23, 2012
6,822
So can we way that
-the voltage pot sets a constant voltage output and
-the current pot sets a constant current output by adjusting voltage ?
And using my example above, if I plugged the light bulb that draws 1Amp at 12Volts, setting the pot for output current 0.5Amps would merely adjust the output voltage to 6volts (an imaginary bulb that would have constant resistance)?
Is that the (simplified) full story?
The voltage of power is a fixed voltage when you using the pot to adjust at a position, and assuming the current is enough, the current is not a constant, what you adjust is the Imax, it means that the Iout is according to the load drawing.

If you offering 12V/0.5A for 12V/1A bulb then the voltage will drop down to 6V, that is not you adjust the voltage to 6V, as I mentioned that is V=I*R caused that and assuming the R is constant, actually R is not constant, and it will increasing when the bulb getting hotter.

#### #12

Joined Nov 30, 2010
18,076
He said, "yes".

You can consider the current limit like a safety limit. No matter what the load wants, the current limit will just lower the voltage until the current is as low as you set it.

• #### dk31

Joined Oct 30, 2014
36
Thank you both for making it clear - made my head buzz last night. Now I feel wiser (by 1 nano-click)

#### MrChips

Joined Oct 2, 2009
19,273
You cannot have both constant current and constant voltage at the same time.
Current and voltage are interrelated by Ohm's Law, I = V/R

Think of the current and voltage adjustments as limits. One of the two limits will always be in control, whichever one is reached first.

If the current drawn is under the current limit, the voltage limit takes effect, and I = V/R.
If the current reaches the limit, the voltage V = IR.