Hi everyone

Complete bovine here.

I need to buy an AC to DC power supply with multiple connectors. I've seen ones where the voltage can be changed from 3 to 12V but the current send to be fixed. Could I potentially damage an appliance if, day the appliance input requires 500mA but the power supply is 1A or will any appliance just draw the correct current so long as I select the correct voltage?
Sorry if it's a daft question!
Many thanks
Justin

 

hi Justin,
Welcome to AAC.
The devices you are powering will only draw the current they need to operate, so a 500mA device will only draw 500mA from the power supply at its correct operating voltage.
E

 

Hello,

Is this "universal" power supply regulated?
If not, the voltage can be higher when unloaded.
To be sure, measure the voltage before connecting.

Bertus

 

What is the load (or loads) you want to power?

 

Hi everyone

Complete bovine here.

I need to buy an AC to DC power supply with multiple connectors. I've seen ones where the voltage can be changed from 3 to 12V but the current send to be fixed. Could I potentially damage an appliance if, day the appliance input requires 500mA but the power supply is 1A or will any appliance just draw the correct current so long as I select the correct voltage?
Sorry if it's a daft question!
Many thanks
Justin

Welcome at AAC!

Electrical circuits are governed by Ohm's Law which states that the current through a resistor is directly proportional to the applied voltage and inversely proportional to the resistance.

In mathematical terms,

I = V / R

Thus, current I and voltage V are not independent.
You cannot have 12V and 1A at the same time unless R = 12Ω.

Hence, the current from the power supply is determined by the supply voltage and the resistance of the load, not by the maximum current rating of the power supply.

 

Hi everyone

Complete bovine here.

I need to buy an AC to DC power supply with multiple connectors. I've seen ones where the voltage can be changed from 3 to 12V but the current send to be fixed. Could I potentially damage an appliance if, day the appliance input requires 500mA but the power supply is 1A or will any appliance just draw the correct current so long as I select the correct voltage?
Sorry if it's a daft question!
Many thanks
Justin

Try thinking of it this way: using water in a pipe as an analogy, the pressure is the voltage and the volume of flow is the current. So if you have a pipe (power supply in this case) that has water at 12V pressure and can maintain that pressure with a 1A flow rate then the thing you connect to it that only allows 0.5A to flow will not cause the pressure to drop in a pipe rated for twice that flow rate. I hope this makes sense to you. You can have a thin or huge pipe independent of the pressure. Only a small current can flow in a thin pipe although it may be at huge pressure.

More specifically, the pressure, in this case 12V is the first critical factor. If the pressure (voltage) is too high the appliance can be damaged. Next is the flow (current). The current the appliance needs to operate must be less than the maximum current the supply is designed to provide. So if the voltage is correct then it doesn't matter if the supply is capable of 1000's of amps as the appliance will only draw (allow to flow) the 500mA that it wants. The supply controls the pressure and the appliance controls the flow. If the appliance allows too much flow then the supply will not be able to maintain the pressure => too much load current and the supply voltage falls.

What is the load (or loads) you want to power?

Why on earth would that be relevant? Surely you can answer the guy without knowing that.

Try thinking of it this way: using water in a pipe as an analogy, the pressure is the voltage and the volume of flow is the current. So if you have a pipe (power supply in this case) that has water at 12V pressure and can maintain that pressure with a 1A flow rate then the thing you connect to it that only allows 0.5A to flow will not cause the pressure to drop in a pipe rated for twice that flow rate. I hope this makes sense to you. You can have a thin or huge pipe independent of the pressure. Only a small current can flow in a thin pipe although it may be at huge pressure.

More specifically, the pressure, in this case 12V is the first critical factor. If the pressure (voltage) is too high the appliance can be damaged. Next is the flow (current). The current the appliance needs to operate must be less than the maximum current the supply is designed to provide. So if the voltage is correct then it doesn't matter if the supply is capable of 1000's of amps as the appliance will only draw (allow to flow) the 500mA that it wants. The supply controls the pressure and the appliance controls the flow. If the appliance allows too much flow then the supply will not be able to maintain the pressure => too much load current and the supply voltage falls.

In short, yes, you are correct!
LOL.

Welcome at AAC!

Electrical circuits are governed by Ohm's Law which states that the current through a resistor is directly proportional to the applied voltage and inversely proportional to the resistance.

In mathematical terms,

I = V / R

Thus, current I and voltage V are not independent.
You cannot have 12V and 1A at the same time unless R = 12Ω.

Hence, the current from the power supply is determined by the supply voltage and the resistance of the load, not by the maximum current rating of the power supply.

Assuming the supply is rated for a current equal to or greater than the load current of course.

 

Why on earth would that be relevant? Surely you can answer the guy without knowing that.

Because he wants a supply that...

the voltage can be changed from 3 to 12V but the current send to be fixed

Obviously he lacks a bit of understanding there, so knowing the loads he wishes to drive may well help formulate an answer.
Does he want a variable 3V to 12V supply, capable of 500mA max, or a 500mA constant current supply capable of varying from 3V to 12V?

 

Assuming the supply is rated for a current equal to or greater than the load current of course.

No. I=V/R irregardless. The supply voltage and load resistance will determine the load current. The load current is not magically determined by the "load current".
If the supply cannot supply the current, its voltage drops so the current is still I=V/R. Then the maths fails as the magic smoke comes out
Or the supply internal resistance adds to the load resistance. This is how a pass transistor in the regulator goes higher resistance to limit the current. The total resistance (internal + external load) increases so ohms law still applies.

 

No. I=V/R irregardless. The supply voltage and load resistance will determine the load current. The load current is not magically determined by the "load current".
If the supply cannot supply the current, its voltage drops so the current is still I=V/R. Then the maths fails as the magic smoke comes out
Or the supply internal resistance adds to the load resistance. This is how a pass transistor in the regulator goes higher resistance to limit the current. The total resistance (internal + external load) increases so ohms law still applies.

Ahh Dendad, all I said was (paraphrased) 'as long as the supply can handle the current the load wants'. That is not incorrect in any way, shape or form.

A power source must be rated to provide the current the load requires. What happens beyond that point is conjecture as to the nature of the regulator or power supply. You have assumed a pass transistor with a hard current limit for some reason. What if it is a switching supply? Or uses fold back current limiting? Or uses a fuse? Or uses nothing at all? Or relies on thermal protection for overloads?

You have also for some reason assumed the load is resistive. If the load is a switched mode PSU then the load has a negative resistance, ie lower volts = more current. I bet that makes your brain explode.

 

Yes, I did not talk about IR losses, magnetic losses, saturating transformers..... Host of other losses.

As to my brain exploding, not actually
I think it is still there. I can't see a mess on the ceiling.
The load current is determined by the volts and and "resistance" and the resistance may be part R and part reactance.

Still, it does not matter what the supply is. Ohms law, or at least a form of it applies. I don't see the load being a switch mode supply any different. The load will not be the switch mode supply only, but the switch mode supply and whatever it is driving. I=V/R still applies.
With a switch mode supply, as you no doubt well know, there is not a 1:1 relationship to the input current and the output current, as it is (mostly) with an analog regulator. The switch mode supply has about the same input and output power (if you ignore the losses) and that is a great thing to save power in high power supplies.
But if you take it all into account, Ohms law still applies. The switch mode supply is not the load, just part of the regulator. The input current going down as the input volts increases applies if the output volts to the real load stays more or less constant.
I=V/R still applies. As the input volts go up, in a switch mode supply, the effective R goes up too. So the current will go down. likewise as the input volts go down, the effective R goes down so the current increases. A switch mode supply is active, not a static device. At any instant, if you could know all the variables, Ohms law still works.


EDIT: As much as I am enjoying this discussion, it is very late here and I need to go to bed so my brain does not explode
Good night.

 

Ahh Dendad, all I said was (paraphrased) 'as long as the supply can handle the current the load wants'. That is not incorrect in any way, shape or form.

A power source must be rated to provide the current the load requires. What happens beyond that point is conjecture as to the nature of the regulator or power supply. You have assumed a pass transistor with a hard current limit for some reason. What if it is a switching supply? Or uses fold back current limiting? Or uses a fuse? Or uses nothing at all? Or relies on thermal protection for overloads?

You have also for some reason assumed the load is resistive. If the load is a switched mode PSU then the load has a negative resistance, ie lower volts = more current. I bet that makes your brain explode.

You have gone off-topic.

We are introducing the concept of Ohm's Law to the TS.

I = V / R

R is a resistor and R is resistive. There is no need to complicate the issue and thereby confuse the TS any further.

Quite simply, one can measure the voltage across the resistive load and the current through the load. One does not need to know the internals of the power supply.

I = V / R

 

You have gone off-topic.

We are introducing the concept of Ohm's Law to the TS.

I = V / R

R is a resistor and R is resistive. There is no need to complicate the issue and thereby confuse the TS any further.

Quite simply, one can measure the voltage across the resistive load and the current through the load. One does not need to know the internals of the power supply.

I = V / R

And you are having a go at me over this?
The OP didn't need Ohms law to explain his problem in the first place and I didn't need the rebuke in error from Dendad who is now so seriously off topic as to make this rebuke from you to me very offensive.

 

Yes, I did not talk about IR losses, magnetic losses, saturating transformers..... Host of other losses.

As to my brain exploding, not actually
I think it is still there. I can't see a mess on the ceiling.
The load current is determined by the volts and and "resistance" and the resistance may be part R and part reactance.

Still, it does not matter what the supply is. Ohms law, or at least a form of it applies. I don't see the load being a switch mode supply any different. The load will not be the switch mode supply only, but the switch mode supply and whatever it is driving. I=V/R still applies.
With a switch mode supply, as you no doubt well know, there is not a 1:1 relationship to the input current and the output current, as it is (mostly) with an analog regulator. The switch mode supply has about the same input and output power (if you ignore the losses) and that is a great thing to save power in high power supplies.
But if you take it all into account, Ohms law still applies. The switch mode supply is not the load, just part of the regulator. The input current going down as the input volts increases applies if the output volts to the real load stays more or less constant.
I=V/R still applies. As the input volts go up, in a switch mode supply, the effective R goes up too. So the current will go down. likewise as the input volts go down, the effective R goes down so the current increases. A switch mode supply is active, not a static device. At any instant, if you could know all the variables, Ohms law still works.


EDIT: As much as I am enjoying this discussion, it is very late here and I need to go to bed so my brain does not explode
Good night.

That you feel it necessary to attempt to explain this to a degree qualified electrical engineer is astonishing and insulting.

 

Ohm's Law still applies and this is what the TS needs to learn.

I = V / R

If the power supply cannot deliver the current demanded by the load resistor R, the voltage will decrease.

I = V / R still holds.

 

Ohm's Law still applies and this is what the TS needs to learn.

I = V / R

If the power supply cannot deliver the current demanded by the load resistor R, the voltage will decrease.

I = V / R still holds.

And I am still not the one off topic here but you continue to rebuke me. Why is that?

 

@Pedant Engineer
Maybe you are being overly pedantic?

 

This http://www.minute-man.com/acatalog/Multi_Voltage_Economical_Power_Supply.html is not a bad universal adapter.

It's a linear power supply, so it's low ripple and it's great for powering say an AM radio.

To keep wasted power dissipation down, it switches in different secondary voltages,

Your question about current basically comes down to how the device protects itself. I've run into one device that looks like it depends on the current limiting in the adapter.

There are a number of different kind of adapters. The class of unregulated, regulated and switchmode. 5 V supplies are probably regulated. 12 V can be either.

With any adapter, you have to watch out for polarity. The Radio Shack adapt-a-plug is doesn;t secure well.

Every adapter I label with a p-touch labeler.

#1. The adapter itself
#2. The ends get labels like
5.5/2.1
12 VDC 1A C+

and

e.g.
Belkin
<model>

==

I paced these https://www.ebay.com/itm/DC-Convert...2A-Voltage-Voltmeter-Ammeter-Red/261244844928 DC DC buck converters with ammeter and voltmeter into a Hammond translucent enclosure.

Pick the right color, so you can see the LEDs.

You can basically dial the voltage and the current limit.

The output has to be at least 3V less than the input.

The 5.5/2.1 and 5.5/2.5 plugs are common, so I put those jacks on the case. You can easily flip the polarity by moving it on the binding post.

I did use an adapt-a-plug on the output, but I also heat shrunk the correct polarity.

It proved to be a nice platform to replace adapters, even odd voltages like 6V.

5V adapters at 100 mA or less can be done with USB.

 

@Pedant Engineer
Maybe you are being overly pedantic?

 

Thanks to everyone. I got my answer and it looks like some of you had a lively debate!
Just noticed all my predictive texting spelling mistakes, sorry about those.
Cheers all

 

Thanks to everyone. I got my answer and it looks like some of you had a lively debate!
Just noticed all my predictive texting spelling mistakes, sorry about those.
Cheers all

And what was the answer?
What did you learn from this episode?