Will a buck converter set to 5V have a hernia if limited to 4.6V?

[Peter Tryndoch]

The charge/discharge board on my friends bike light has died.
And since the batteries are a bit old as well, I thought of putting a USB plug on the light and running it off a single cell USB powerbank ($AU 9.95 at Officeworks).
Powering the light of my bench supply I noticed that it clipped at 4.6V.
So will the buck converter in the powerbank have a hernia and cook itself or will it just saturate at 4.6V and be happy?
Cheers

[james_s]

Usually if the input of a buck converter sags, the output will also sag and nothing blows up. Or do you mean the light draws so much current that your bench supply can't deliver 5V into it? Does the light have an internal driver or are you trying to power the LED directly?

[Peter Tryndoch]

No the bench supply did not reach it's capacity of 4A. I have attached some pictures of the led controller.
B+ & B- got to the battery, and the other 2 leads to the led.
The led is housed on a star type board (Cree style) with no additional components.
The bottom chip is a mosfet driver and the top is the controller for on/off/dim/blink.

When I adjusted the power supply voltage the brightness of the led changed so there is no on board voltage regulation.
I'm guessing that the input voltage is being clamped by a semiconductor, the mosfet I think.

[Doctorandus_P]

Those power banks usually have Li-Ion batteries in them and these burn pretty well at 4.6V.
Those power banks usually also have a battery charger circuit in them

[james_s]

The single cell power banks have a boost converter in them to get 5V.

I would replace the LED driver with one that is made to accept higher input voltages, they're cheap and easily available.

[Peter Tryndoch]

I took a slightly different approach and put 2 diodes in series to drop the voltage from 5V to 3.8V and a Male USB-A plug on the end.
I tested on a single cell power bank and drew 900mA.
This way my friend can just go out and buy whatever power bank he wants.
I recommended a single cell, but I think he prefers a 4 cell as that's what the original battery was.
I can see his point of view. He commutes about 15Km and it's no fun running out of light before you get home with traffic about!
Then again I did 17Km on a single cell, but hey! Oh and I recharged at work, and I don't think he does.

[fzabkar]

The bottom chip is a mosfet driver and the top is the controller for on/off/dim/blink.

The top chip in your first photo is the MOSFET (AO4468):
http://aosmd.com/pdfs/datasheet/AO4468.pdf

I would think that the two R100 resistors would be used for current sensing. Normally you would regulate the LED's current, not its voltage.

Can you tell us the markings on the other chip?

[Peter Tryndoch]

Yes I realized my mistake after my last post. You can clearly see the pins commoned and going to the led -ve.
The control chip has it's markings rubbed out.

[Peter Tryndoch]

The control chip does not do any current sensing. During testing the current would vary with the supply voltage applied: 2.6A @ 4.5V and 1.7A @ 4V. (Before I added the diodes)
The circuit was designed for a set voltage of 3.6V (with some variation depending on charge state) so adding regulation was not required. Fixed current regulation is set by the 2 X R100

The led driver circuit is quite simple:
Batt -ve -> 2 X R100 -> Mosfet -> LED -> Batt +ve

The drive chip has no current feedback. It has set outputs depending on what mode you select.
On/Off/Flashing/ and 2 dimmed settings which would have preset frequency and pulse width.
No need for the manufacturer to add additional cost of current sensing.
Albeit - not a huge expense, but still.

What I don't understand is why didn't the 2 X !00 resistors sink the additional voltage.
The mosfet and led appear to have a forward bias voltage of 4.6v, but if I adjusted the power supply above that value (lets say to 5V) then the resistors should have dropped the additional voltage across them. But that did not happen. Why?
I had a similar problem when I connected a peltier fridge in my car. The fridge would try to clamp the voltage to somewhere around 9V (I can't remember the exact value) and of course the car would regulate to around 12V. The lead going to the fridge got very hot indeed. One day I'll do the math and put in a suitable dropper resistor.

[james_s]

Are you sure there isn't a fault with the drive circuit? I would reverse engineer it and draw a schematic, then it will make a lot more sense.

You don't want to go adding dropper resistors to a big load like a Peltier fridge. If it's designed for 12V and it's making the cord get hot dropping several Volts then either it's defective or it's using the wrong cord that is much too thin. I worked on a cooler like that for somebody once, seems like it drew about 8 Amps, it had a relatively heavy cord.

[fzabkar]

What I don't understand is why didn't the 2 X !00 resistors sink the additional voltage.
The mosfet and led appear to have a forward bias voltage of 4.6v, but if I adjusted the power supply above that value (lets say to 5V) then the resistors should have dropped the additional voltage across them. But that did not happen. Why?

Two R100 resistors in parallel corresponds to a resistance of 50 milliohms. Therefore a current of 2A would result in a voltage drop of only 100mV.

ISTM that pin #8 of the controller is sensing the LED current via the IR drop across the two resistors. AFAICS, the controller must be using this input in some way, either for current regulation, or current limiting.

[james_s]

Either it is or it's supposed to be. Instead though from the sounds of it, this thing is behaving like a bare LED with no driver or current limiting at all.

[Peter Tryndoch]

Big mistake on my part:
It turns out my PSU only goes to 2.5A and not 4A as I had previously thought.
That may go someway to explaining some of my tests.
Sorry if it caused any of you unnecessary head scratching.
Anyway, the light seems to be still working with the 2 diodes.
My friend has not told me otherwise. Or maybe he's just given up on my fixes.
Hm.....