What happens when a power supply is has more volts but less amps than you need?

[SleepyFace-_-]

Hi guys and gals, please forgive the lame-o title, first poster here! I recently picked up a pretty fierce interest in electronics and I've been doing plenty of reading and experimentation but I have a few questions that I'm hoping will land on generous ears.

What follows is a "brain dump" of what I have stored in my head, I would really appreciate any corrections!

So I believe I understand V = IR to a "workable" degree.

What I think I know:
-Value V is equal to the value of amperage in amps * resistance in ohms.
-Voltage is a delta in field across two points, the driver for electron flow
-Amperage is a "simple" count of electrons moving across any point (coloumbs/sec)
-Wattage is a bonafide count of work (joules/sec)
-Electron flow and the energy transmitted by electron flow are two inter-related, but very different things.

-Wattage can be calculated by multiplying volts and amps. Failing any voltage info, one can calculate wattage by replacing V with IR, meaning wattage is also IRI.

-Pure amps describes electron flow in qty, but without voltage, you don't know if its a "fast mover" with a "narrow pipe" or a "slow mover" with a "fat pipe".

Here's the lead up question:

-If a laptop power supply, for example, 19.5 V @ 3.4 Amps is plugged in but not connected to anything, a voltage reading across its leads would give me 19.5v, but there would be no amps. Aka, the delta is there, but there is no current.

If I were to plug in a 19.5v fan to this power supply, it would run. If it were to add a second fan in series, would the second fan run at all? Why or why not?

I would think that if there is no voltage left after the first fan, there would be no "power" to drive it. in my head (0V * ?A = 0 Watts no matter what.) In this case, would both fans share the voltage evenly? If not evenly, then what dictates the voltage split?

If I were to put them in parallel, then both fans would get 19.5 V but would have to share amperage, so they would run, given enough source amps.

Here's the real question:

Lets say I connected a 16.5 volt fan to the same power supply, it draws a total of 4.2A. The power supply provides 19.5 volts @ 3.4 amps. So I have a surplus of voltage and a surfeit of amps. What happens in this case?

From my actual experiments, I've had to put in lots of parallel resistors in series with the fan in order to not overdrive it, but yet, I don't understand the relationship...

I knew I had to drop those 3 extra volts, regardless of the lack of amps. I ended up doing this:

3V = 3.4A x R
R=.88 ohms

But then I thought, 3.4A can't be right, can it?

So then I went with watts to see it from another angle:

Power supply is 19.5V x 3.4A = 66 Watts
Fan is 16.5v x 4.2A = 69 Watts.

From resistance calculation : 3V x 3.4A = 10.2 Watts.

If my fan wants/can handle 69 W and the power source provides only 66, why would I would need to "burn off" about 10 Watts of power via the resistor, even though my fan wants more amperage than can be provided by the power supply?

If component "draw" is met on demand by my power supply, up to its max amperage, what happens when the supply doesn't have enough amperage?

It seems that even though W = VA, volts can't be turned into amps, or vice versa?

It feels like voltage is somehow "more important" than amperage? I have a feeling I'm missing something here. Am I crazy? Help!

Thanks!

 

[Zryn]

If I were to plug in a 19.5v fan to this power supply, it would run. If it were to add a second fan in series, would the second fan run at all? Why or why not?

Neither fan would work. The voltage would be split between them (assuming they have exactly the same characteristics) and neither would run off 9.75V.

Look at [URL]http://en.wikipedia.org/wiki/Voltage_divider.[/URL]

I would think that if there is no voltage left after the first fan, there would be no "power" to drive it. in my head (0V * ?A = 0 Watts no matter what.) In this case, would both fans share the voltage evenly? If not evenly, then what dictates the voltage split?

The fans would share the voltage evenly, as per the link above. The Resistance of an element compared to the Total Resistance of the circuit determines the relative split of voltages.

If I were to put them in parallel, then both fans would get 19.5 V but would have to share amperage, so they would run, given enough source amps.

Correct. However if you don't have enough current for 1, you definitely do not have enough current for 2 (at the correct voltage).

To learn how currents split, check out [URL]http://en.wikipedia.org/wiki/Current_divider

Theory and practice do not always mesh. In theory if you don't have enough current or voltage something won't turn on at all. In practice, things may turn on but not work very well when you get to within some fraction of their operating current and voltage. Someone with practical experience with such fans may be able to comment further. (Close enough is good enough perhaps?)

Generally however, if your component draws more current than the supply can handle, either the supply limits itself with protection circuitry; it is designed to not give more than the maximum because its internal ratings can't handle higher current, or a fuse blows and the supply stops working until you replace it. Either that or it smokes and stops working all together. (So the supply blows up)

When your volts is too high, something similar will happen, since the device has some resistance that will use a current at a certain voltage. Double the voltage and keep the resistance fixed, what happens to the current? (Ohms Law) When things aren't rated for the current you give them, they melt / burn / smoke / explode. (So the device blows up)

It seems that even though W = VA, volts can't be turned into amps, or vice versa?

You do clever things with circuitry which will take an input voltage and current, and then output a different voltage and current whilst keeping the power (their product) the same. But in the end, you can't make a 66W power supply give you 69W for your fan. (And you would even lose some power due to efficiency reasons)

It feels like voltage is somehow "more important" than amperage? I have a feeling I'm missing something here. Am I crazy? Help!

You can't run anything without voltage. You can't run anything without current. You can have voltage without current (terminals of a battery) but you can't have current without a voltage (superconducters?). They're both important, but the exercise usually makes one the focus more so than the other is all.

 

[I_am_learning]

One thing I would like tell you (Zryn has already told, just to emphasize it);
the 16.5 V 4.2 A fan won't draw same 4.2 A when connected to the 19.5 V power supply.
Do this:
if it draws 4.2 A if connected to 16.5 V source, then its resistance = 16.5 / 4.2 = 3.92 ohm.
So, if connected to 19.5 V, it will draw 19.5 / 3.92 = 4.97 A. So, that will certainly harm the power supply, if its not adequately protected. Even if the power supply succeeds in delivering that ampere, it will then harm the fan for it is rated only for 4.2A.

Now, the case of adding series resistance, say you add 0.71 ohm (that value comes by calculation), in series with the fan, it will give 16.5 V across the fan. (I hope you followed the voltage divider link).
At this case, the fan will draw 4.2A. This can be also worked out from the powesupply view point; Current from power supply = 19.5/ (3.92 + 0.71) = 4.2A. Or, current for the fan = 16.5 / 3.92 = 4.2A. The same answer isn't co-incident but because of the carefully chosen, 0.71ohm.

So, this time, although, the condition is good for the fan, its not for the powersupply. You are demanding 4.2A from it, but its rated for 3.4A

If you increase the series resistance to make condition good for the power supply, it won't be good for the fan; it won't get enough voltage, and hence not enough current, and that's why not enough power to run.

So, simply put, there is no simple 'linear' solution to your problem.
cheers.

 

[vk6kro]

A fan doesn't really behave like a resistor and there is a big difference between the current it draws when turning or not turning.

However as long as you don't let it stall, the fan can be used at voltages well below its rated voltage.

In this case, you could put a resistor in series with the fan so that the power supply was not overloaded.

So, the load on the power supply should be no less than R, where R = 19.5 volts / 3.4 amps.
So, R = 5.735 ohms.

Now, the resistance of the fan (as far as we know) is 16.5 volts / 4.2 amps or 3.92 ohms.

So if we put a resistance of 1.815 ohms (5.735 ohms - 3.92 ohms) in series with the fan, then the power supply will not be overloaded. It will deliver 3.4 amps to the fan + resistor combination. ( Current = 19.5 volts / 5.735 ohms = 3.4 amps.)

The fan will now have a reduced voltage across it and so the air flow will not be as great as before. It may run quieter, though.
The new voltage for the fan will be = I * R = 3.4 amps * 3.92 ohms = 13.328 volts.

This is still 80% of the normal supply voltage and so the fan should still operate OK, although with reduced performance.
Depending on the fan, voltages as low as 50% of the rated supply voltage will still result in the fan turning and some air flow being achieved.

The 1.815 ohm resistor is close to the standard value of 1.8 ohms and it would dissipate 20.98 watts. (power = I ^ 2 * R = 3.4 * 3.4 * 1.815 = 20.98 watts ).

 

[SleepyFace-_-]

Thank You Gentlemen! A wealth of knowledge!

Voltage and current splitting was great lead!

I am working through all these, thank you for the insight.

One add-on question:

When calculating resistance values for power supplies with the same voltage but different amperages :

Do I use the component draw or the total amperage?

for example: string of LEDS connected to 19.5v 3.4 A power supply.

leds are a total of 15.4v, 4.2A, calculation was : 15.4/3.4= 4.5 ohms ( I figured since the LEDS want more amps than the supply has, i should assume that the supply would feed only up to the max, therefore resist that)

if I connect the same LEDS to a 19.5v, 6.7A power supply do I use the total draw from the LEDS, or the total amperage from the psu?

I have a hazy idea that the power supply will only supply as much as is drawn by attached components but I don't know if this applies when speccing out resistances. I think the calculations I'm seeing tell me that I should always factor in the total amperage...?

Thanks Again!

 

[vk6kro]

Thank You Gentlemen! A wealth of knowledge!

Voltage and current splitting was great lead!

I am working through all these, thank you for the insight.

One add-on question:

When calculating resistance values for power supplies with the same voltage but different amperages :

Do I use the component draw or the total amperage?

for example: string of LEDS connected to 19.5v 3.4 A power supply.

leds are a total of 15.4v, 4.2A, calculation was : 15.4/3.4= 4.5 ohms ( I figured since the LEDS want more amps than the supply has, i should assume that the supply would feed only up to the max, therefore resist that)

if I connect the same LEDS to a 19.5v, 6.7A power supply do I use the total draw from the LEDS, or the total amperage from the psu?

I have a hazy idea that the power supply will only supply as much as is drawn by attached components but I don't know if this applies when speccing out resistances. I think the calculations I'm seeing tell me that I should always factor in the total amperage...?

Thanks Again!

The load will draw current according to the voltage you supply to it.

So, if the power supply can somehow deliver 15.4 volts, then the current will be 4.2 amps.
Whether this is good for the power supply depends on its design.

You are using a laptop power supply which will be a sophisticated switch mode supply with current limiting. So, if you put too much load on it, it will reduce its voltage so that you can't draw so much current.
It might cut off the output completely or just give a few volts out and the load will only draw a very small current, if any, from this.

Another power supply might blow an internal fuse or overheat something inside and need repairs to get it working again.

So, you have to look at it from the power supply's view. If it is a simple supply, without any protection, you could damage it by trying to get too much current from it.

Generally, you shouldn't be operating a power supply close to its current limit. These limits are fairly arbitrary and may be highly optimistic and you could easily destroy a power supply by doing this.

If you connect your string of LEDs that need 15.4 volts to a power supply delivering 19.5 volts, the LEDs will glow a lot brighter and they may be destroyed, depending on the design.

 

[SleepyFace-_-]

From real world measurements:

19.5V

@3.34 A

16.6v drop over leds in series, 2.8v drop over resistors (forgot value). Total drop 19.4V, close to manuspec.

@6.7 A
18.8V drop over same exact led series, .6 Ohm resistor.

18.8 = .6 x 4.2. This tells me I should use the draw only. Great, math matches rl.

Now: What I really want is 4.2 amps @ 15.4V, which would mean I want a 1 ohm resistor in series with the leds, consuming 17.22 W, plus the 65 ought watts from the leds, leaving me well within the 130 W capacity of the 19.5v @ 6.7A power supply, right?

 

[vk6kro]

Yes, I think you have the technique right.

One thing to watch, though.

The rating of power supplies is sometimes given as a voltage AT a current rather than a maximum current.
So, at no load, the power supply might give 25 volts. At 4 amps, it might give 22 volts and at 6 amps it may give 20 volts.
You have to be careful here because if you only draw 4.2 amps, then the real supply voltage would be more like 21.5 volts.

It is easy to check. Just measure the voltage of the power supply with no load on it. If it is a lot higher than the rated voltage, then you need to check it at the current you intend to draw from it.