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

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Discussion Overview

The discussion revolves around the performance of a 2 AMP regulated DC power supply in a small circuit using 14 AWG copper wire and a 1 ohm resistor. Participants explore why the expected maximum current is not being achieved, considering factors such as power supply characteristics, wire resistance, and circuit configuration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions why the maximum current is not being reached, given the specifications of the power supply and circuit components.
  • Another participant suggests that the power supply may have its own resistance and current limiting features that could affect the output.
  • Some participants propose that the internal resistance of the power supply could be negligible, while others argue it should be considered in calculations.
  • There is a discussion about the implications of short-circuit protection in the power supply and how it may limit current output.
  • Participants discuss the resistance of the wire and the expected current based on Ohm's law, with calculations leading to differing expectations of current output.
  • One participant mentions measuring the resistance of the entire circuit and obtaining unexpected results, raising questions about the accuracy of measurements and the role of the multimeter in the circuit.
  • Concerns are raised about the safety of using batteries for high current measurements, with warnings about potential hazards.
  • There is a mention of "Foldback Current Limiting" as a feature that may be present in the power supply, prompting further inquiry into its function.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the reasons for the low current readings, including the effects of power supply limitations, circuit resistance, and measurement techniques. The discussion remains unresolved as no consensus is reached on the primary cause of the issue.

Contextual Notes

Participants highlight the complexity of the circuit and the potential impact of various resistances, including those from the power supply and measurement devices. There are also concerns about the safety of high current applications and the capabilities of the components being used.

Who May Find This Useful

This discussion may be useful for individuals interested in electronics, circuit design, and power supply characteristics, particularly those exploring current limiting and resistance in small circuits.

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.
 
Last edited by a moderator:
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...
 
  • #10
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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  • #11
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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