Low DC amp through small circuit of 14 AWG copper wire?

AI Thread Summary
The discussion centers on issues with a 2 AMP regulated DC power supply not delivering the expected current in a small circuit with a 1-ohm resistor. Participants highlight that the power supply has a maximum output of 2A, which limits the current regardless of calculations suggesting higher values. Concerns about the internal resistance of the power supply and current limiting features are raised, indicating these could be causing the low current readings. The importance of measuring resistance accurately and avoiding invasive methods with multimeters is emphasized to prevent misleading results. Overall, the conversation underscores the need to understand the specifications and limitations of the power supply being used.
Porter22
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I have a 2 AMP regulated DC power supply, with variable voltage 3 - 12 v. If in a small circuit, i.e, small copper wire with 1 ohm resistor. Why am i not getting at max Amps?
 
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What does your power supply look like? Doesn't it have its own resistance? Have you considered that?

Aside from that, what size copper wire are you using and what is its resistance?

I suggest you draw out a sketch of your circuit and the values of each resistance along with the voltage setting. What do your calculations show for the current when the voltage is 12 V?
 
Porter22 said:
I have a 2 AMP regulated DC power supply, with variable voltage 3 - 12 v. If in a small circuit, i.e, small copper wire with 1 ohm resistor. Why am i not getting at max Amps?

does your power supply have current limiting? if so, then that is what is happening when you are presenting the PSU with a short circuit

as magoo asked ... what sort of PSU ?
Dave
 
magoo said:
What does your power supply look like? Doesn't it have its own resistance? Have you considered that?
]
IMG_4581.JPG

I attached a pic of the PSU..
I have done some calculations assuming (generously and for simplicity, i think) that the wire has .3ohms. So at 3V, shouldn't i be getting 10 A?

I understand there may be "short-circuit" prevention elements in this PSU. Is this what you're talking about?This could be exactly the problem right? Other than that, I have measured the V across the PSU outputs at each of the Voltage levels. Each one is accurate to .05 V. Therefore, i think, internal resistance of the PSU can be taken out of the equation.
 
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davenn said:
does your power supply have current limiting? if so, then that is what is happening when you are presenting the PSU with a short circuit

as magoo asked ... what sort of PSU ?
Dave
Also what kind of batteries could I use (in short bursts to avoid battery malfunction) to measure high A in discharge?
 
Porter22 said:
the wire has .3ohms. So at 3V, shouldn't i be getting 10 A?
Not with that power supply. It's output is rated at 2A max. Try putting 12V across a 6 Ohm power resistor. How much power will that be? (Be sure to use high enough wattage resistor to not smoke the resistor).
Porter22 said:
Also what kind of batteries could I use (in short bursts to avoid battery malfunction) to measure high A in discharge?
Based on your questions, there are no batteries that you should try to use to obtain high output currents. You don't have the experience yet to be working with high powers. That's how batteries fail and burst or catch fire...

http://techbakbak.com/wp-content/uploads/2016/01/hoverboards-are-setting-homes-on-fire.jpg
hoverboards-are-setting-homes-on-fire.jpg
 
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Porter22 said:
I attached a pic of the PSU..
I have done some calculations assuming (generously and for simplicity, i think) that the wire has .3ohms. So at 3V, shouldn't i be getting 10 A?

As Berkeman said, no it won't supply more current that it's designed to ... in this case 2A max

I understand there may be "short-circuit" prevention elements in this PSU. Is this what you're talking about?

Yes
It says it's a regulated PSU. The overcurrent protection will be either be 1) part of that normal protection in the regulator chip
or 2) will be made up of discrete components

just as an example of a voltage regulator chip ... note the two comments I have circled

ss.JPG
Dave
 
berkeman said:
Not with that power supply. It's output is rated at 2A max. Try putting 12V across a 6 Ohm power resistor. How much power will that be? (Be sure to use high enough wattage resistor to not smoke the resistor).

Based on your questions, there are no batteries that you should try to use to obtain high output currents. You don't have the experience yet to be working with high powers. That's how batteries fail and burst or catch fire...

http://techbakbak.com/wp-content/uploads/2016/01/hoverboards-are-setting-homes-on-fire.jpg
View attachment 118370

I got variable 25 W resistor, and set it to 6.0 ohms. Which should give 2 A of current and 24 W power through the resistor. I'm currently reading 0.28 A at 12 V... Whats going on here?

If I measure the resistance of the entire circuit I'm getting 0.685k Ohms which would give .0175 Amp. which also doesn't really make sense... Everything is in series, including multi meter. Have also tested with analog Voltmeters and Ammeters DC. Is the power supply really that that current limiting that i can't even get close?
 
Porter22 said:
I got variable 25 W resistor, and set it to 6.0 ohms. Which should give 2 A of current and 24 W power through the resistor. I'm currently reading 0.28 A at 12 V... Whats going on here?

If I measure the resistance of the entire circuit I'm getting 0.685k Ohms which would give .0175 Amp. which also doesn't really make sense... Everything is in series, including multi meter. Have also tested with analog Voltmeters and Ammeters DC. Is the power supply really that that current limiting that i can't even get close?
In the series current measurement mode, a DMM has a non-trivial resistance. That is how it makes a small voltage to figure out what the series current is.

Take the series DVM out of the circuit and just measure the resistance of your potentiometer before hooking it in. Then just monitor the voltage across that resistance to figure out the current. That's a much less invasive way to figure out the current.

And is your potentiometer capable of handling 25W? Probably not, so just be ready to disconnect it quickly if it overheats...
 
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berkeman said:
In the series current measurement mode, a DMM has a non-trivial resistance. That is how it makes a small voltage to figure out what the series current is.

Take the series DVM out of the circuit and just measure the resistance of your potentiometer before hooking it in. Then just monitor the voltage across that resistance to figure out the current. That's a much less invasive way to figure out the current.

And is your potentiometer capable of handling 25W? Probably not, so just be ready to disconnect it quickly if it overheats...

That worked, and i checked it with an analog ammeter
And its a ceramic variable resistor that said it is 25 W so i think I am good but ill check it for short bursts.
 
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Something to check for:

Search on term "Foldback Current Limiting"
first on 'net to learn what it is
then in your instruction book to see whether your supply has it.
 
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  • #12
jim hardy said:
"Foldback Current Limiting"
And a Quiz Question for the OP @Porter22 -- why would a linear regulator power supply usually use Foldback Current Limiting, and a switching regulator power supply not? :smile:
 
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