Too much current out of flyback converter, how to reduce it ?

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Hi,

 

I'm designing a fan that is supplied through a step-up flyback converter (5V > 12V LM2585 from TI).

 

A USB cable is connected to one USB3 port from my laptop and to an Arduino Nano ; The Arduino Nano gives 5V to the input of the converter ; The output of the converter gives 12V to the fan. The fan needs 12V this is why I'm using the converter.

 

The converter's datasheet gives circuits example. I built this one below. Vin is connected to the 5V from the Nano.

 

I get 12V@0.6A at the output. The problem is my fan only needs 0.3A. I don't want to burn it with to much current so the question is simple, how can I do ?

 

I thought of a current divider with on one branch one resistor whose power rating is something around 5W (12V * 0.3A = 3.6W), and on the other branch the fan getting the remaining 0.3A, the nodes connecting the two branches being the 12V from the converter and GND.

 

 

Can I go for it ?

 

 

This topic has a solution.
Last Edited: Wed. Nov 18, 2020 - 01:57 PM
This reply has been marked as the solution. 
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jowell88 wrote:
I get 12V@0.6A at the output. The problem is my fan only needs 0.3A.

Eh?

 

Do you mean you measure 0.6A when you were expecting only 0.3A ?

 

The fan will only "pull" as much current as it needs.

 

The current rating on a power supply output is the maximum it can give - if the load demands less, the power supply does not "force" more ;you don't have to "absorb" the difference.

 

EDIT

 

Just like if you plug a 2kW heater into a wall outlet, it will draw 2kW; but if, instead, you plugged a 40W lamp into that outlet, the lamp will only draw 40W - you don't have to make up the other 1960W with a dummy load.

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Last Edited: Tue. Nov 17, 2020 - 03:26 PM
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Also, USB3 supplies 5V at up to .9A. Doing the math, that is 4.5W. The fan needs 3.6W. (12V at .3A). The boost will have losses. You might have a fan that runs, but you will be pushing it at 80% efficiency. Boost supplies are nice but not magic. 

Last Edited: Tue. Nov 17, 2020 - 05:13 PM
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That is a boost converter, NOT a flyback

 

Note, also that it is impossible to turn off the output (if that is important to you).

 

 I don't want to burn it with to much current so the question is simple, how can I do

Where did you come up with such a question?...when you plug in your TV at home, aren't you worried that the power company will be sending megawatts into your TV & causing an explosion??!!

 

The fan needs 12V  ​​​​​....not such a good idea due to losses​ & very limited power​---get a 5V fan...Really, usb is NOT the place to be powering a fan (unless you mean a 2amp USB wall adapter).

 

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Tue. Nov 17, 2020 - 05:28 PM
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Please note:

 

(1) YES, the schematic has the words 12V @ 0.6A

 

(2) That current rating is what the converter CAN provide, if "requested" by the load.

 

(3) A 0.3A load will ONLY take 0.3A from the converter output.

 

(4) Reduced load current means reduced input current. There is no "extra current" that has to be thrown away.

 

This is a very common misconception among people who are just learning about electrical principles.

 

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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I don't want to burn it with to much current so the question is simple, how can I do ?

Look up Ohm's law?

John Samperi

Ampertronics Pty. Ltd.

www.ampertronics.com.au

* Electronic Design * Custom Products * Contract Assembly

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ka7ehk wrote:
(1) YES, the schematic has the words 12V @ 0.6A

Err ... no:

 

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Faulty eyesight. Looked like 0.6. 

 

Mea culpa!

 

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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It's why I was puzzled about the 0.6A in #2

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Hopefully his supply only has a few Mv of ripple cheeky (yes, I know what you are going to say & I know what you are thinking about that statement)

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Wed. Nov 18, 2020 - 12:18 PM
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Thanks for your replies I understand now. The fan will only pull the current that it needs, and the current rating is what the converter can provide, if "requested" by the load.

 

0.8A is what says the datasheet, but I get 0.6A. The input voltage of the converter is 5V from the Arduino Nano but it is 4.7V when I measure it, I have a 0.3 loss somewhere. So it is maybe the reason why the converter gives me 0.6A instead of 0.8A, or maybe not...

 

 

 

Last Edited: Wed. Nov 18, 2020 - 01:55 PM
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You're welcome - you do now

 

smiley

 

Please mark the solution (here and in your old threads) - see Tip #5 in my signature, below:

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jowell88 wrote:

0.8A is what says the datasheet, but I get 0.6A. The input voltage of the converter is 5V from the Arduino Nano but it is 4.7V when I measure it, I have a 0.3 loss somewhere. So it is maybe the reason why the converter gives me 0.6A instead of 0.8A, or maybe not...

 

How are you measuring that and does it matter as your fan only needs 0.3A?

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jowell88 wrote:
0.8A is what says the datasheet, but I get 0.6A.

As Brian says - how, exactly, do you "get" that?

  • What load are you using?
  • How are you measuring?

 

input voltage of the converter is 5V from the Arduino Nano

Where, exactly, on the Nano ?

 

(easiest way to answer the above would be with a schematic)

 

but it is 4.7V when I measure it

You mean it drops to 4.7V when you put your device on it? Or is it permanently at 4.7V - even with nothing connected?

 

 

 

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jowell88 wrote:
I have a 0.3 loss somewhere

How are you powering the Nano ?

 

You're doing slightly over a 2:1 voltage step-up - so your input current is going to be over twice the output current.

 

So, at 0.6A out, you'll be drawing over 1.2A from the Nano! surprise

 

As  avrcandies suggested in #4, that's not the kind of current to expect from a USB port ... 

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awneil wrote:
You're doing slightly over a 2:1 voltage step-up - so your input current is going to be over twice the output current.

On the other side... if the fan draws 0.3 on the primary side he will get about 0.6

I suggest the OP goes and buy a couple of books on basic electronics, or he might get into dangerous area very quick, or release a lot of magic smoke in the near future as he has no clue what he actually is doing.

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jowell88 wrote:

The input voltage of the converter is 5V from the Arduino Nano but it is 4.7V when I measure it, I have a 0.3 loss somewhere.

 

Cheap USB leads often have significant resistance and can easily drop the voltage.

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I think the USB spec does allow a voltage drop across the cable.

 

Cheap leads will design down to this spec.

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I'm designing a fan that is supplied through a step-up flyback converter

Why do you need a fan? Is it to cool you or some PCB?  Maybe use a heatsink...requires no energy.

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

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awneil wrote:

 

input voltage of the converter is 5V from the Arduino Nano

Where, exactly, on the Nano ?

 

(easiest way to answer the above would be with a schematic)

 

but it is 4.7V when I measure it

You mean it drops to 4.7V when you put your device on it? Or is it permanently at 4.7V - even with nothing connected?

 

 

VCC from the USB connector which is connected to laptop through a USB cable.

 

4.7V permanently even when nothing connected. Could be the leads as you and Brian Fairchild said.

 

avrcandies wrote:

I'm designing a fan that is supplied through a step-up flyback converter

Why do you need a fan? Is it to cool you or some PCB?  Maybe use a heatsink...requires no energy.

 

To cool me.

 

OK. I will not power the fan with USB

 

 

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So, I have to ask, what is the NCP1117 supposed to be doing?

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jowell88 wrote:
4.7V permanently even when nothing connected. Could be the leads as you and Brian Fairchild said.

If there's no current being drawn, there will be no drop across the cable - again, this is basic Ohm's law.

 

But I think 4.7V is within the USB spec?

 

Is your meter reading correctly?

 

You certainly wouldn't want to be connecting a high-power load to a laptop's USB port! surprise

 

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Last Edited: Mon. Nov 23, 2020 - 05:57 PM
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So, I have to ask, what is the NCP1117 supposed to be doing?

Not much by the given schematic---only drives an led  (IF  the reg gets connected to something! )

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

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Look at D1. Now answer my question... :)

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Look at D1. Now answer my question... :)

What's this strange answer supposed to imply?

 

The regulator does nothing, even if running,  it can't drive any PCB circuitry due to D1, and can't do anything in the first place since it isn't even supplied any power!

At most, if is were supplied power it could light up the LED.

 

An insider joke, perhaps?

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Mon. Nov 23, 2020 - 11:25 PM
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Forgive my lack of writing ability. I are a engineer, not a english major.smiley

It seems to me that the USB is supposed to be powering the design. So, if you look at the USB connector and see that the VCC pin is connected to the anode of D1 it 'sort of' implies it might be meant to drive the 1117 but on the output? If the design is supposed to power the USB connector from the undriven 1117 regulator,, at the least D1 is backwards. My post was just a poorly worded (attempt to be polite) question to have the OP take another look at the design before going to layout. 

My lack of writing skills has been a bane during my working life. Especially during self driven and written performance evaluations. Oh well. I enjoy your posts AVRcandies. Take mine with a bit of a smile. Again, just trying to poke the OP into a design review...

 

someOldGuy

WA6MOK (FT4 is fun!)

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Oh, I see; I thought you were the OP asking some strangeness, like trying to insist it was all fine with D1.  A 1100 apologies & a bag of ICE swag.

 

I really wonder what the OP is hoping for.  Maybe some Uno's under the tree, to save us all.

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

Last Edited: Tue. Nov 24, 2020 - 12:02 AM
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A diode in the USB 5V supply is a standard feature of my boards to prevent back-feeding the computer from the device. But you do lose c. 350mV to the diode's Vf.

 

(I believe we discussed this before when I recounted the story of accidentally connecting 12V to a device and almost blowing up a $1500 laptop. It was only saved by the $5 USB hub which died first (RIP). The conversation then progressed on to isolated USB interfaces, etc, etc).

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A diode in the USB 5V supply is a standard feature of my boards

That's a start with the biggie...you also need to worry about gnd & the data sigs.  If gnd is broken ...it will go looking for your PC to supply gnd (which can be bad if it is a 5 amp gnd current rather than a 5 ma gnd current)...your motherboard might become unhappy...so you pretty quickly get to isolating all wires.

I had someone deliver a proto to work on but the main power switch opened gnd instead of V+ .  If my scope happened to be hooked up, the circuit would keep working (or come on when I hooked up my probe!).  Luckily, the current was not very high, maybe 100 mA. 

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!