PIcaxe and micropuck

JPU

Senior Member
Hi All

Thanks for looking. I have what is probably a simple problem to solve if you know what your doing, which I do not!, so I need help please. I don't want to try and solve it with trial and error as I usually would due to the costs of the micro pucks I use which are blown easily.

The circuit is very simple in that I use a PICAXE chip to switch on a N channel mosfet which in turn sinks a micro puck which then drives two high power LEDs in series high power LED. The problem is that the micro puck must be connected to the LED when powered otherwise they burn out. So if an LED blows or a wire is disconnected then the micropuck burns out at a significant cost. Here is a link to the micropuck documentation and I have the puck setup as in page 4, a buck/boost driver.

I was hoping to overcome this by introducing some sort of feedback circuit to the picaxe from the LED. So if the circuit is broken in anyway then the pixace will immediately switch off the mosfet and thus the micropuck will also be turned off.

I have drawn a simplified version of the circuit and my question is what should I do to safely connect the feedback channel to the picaxe. Do I need a resistor in the feedback channel and or should I have a resistor and connect the leg of the chip to ground?

The circuit is powered by a 4.2V battery and the micro puck boast the voltage to 7V and holds the current to 500mah. The pic only works at 3V-5V.

I appreciate there is probably a very simple answer to this but any help would be appreciated as my electronics knowledge is very basic.

Thanks for your help. I am sorry that the schematic is a sketch but I tried to do it in diptrace but it was harder than expected to find the symbols.

JPU
 

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eggdweather

Senior Member
When the Buck is running the cathode of the LED's is pulled (O on the diagram) below zero volts until the required drive voltage for 2xLED or 3xLED is achieved and so when (highly unlikely) an LED fails it presumably tries to drive the O-point voltage negative even more until it fails. So you would need to devise a detection regime with the PICAXE using an ADC input perhaps that measured the cathode voltage and when it went below a pre-defined level your code would switch off the FET and hence Buck.

The problem is detecting that negative voltage, as the ADC can only really measures down to zero volts and the PICAXE might be damaged by a negative voltage input. You could wire two series resisters from the supply and to the cathode, the junction of which is set to be the potential difference, which you then monitor. For example use 2 x 1K R in series and if the Buck normally drives the cathode to -1.0volts then the potential difference would be 5.2volts and 1/2 of that (potential division by two same value R) = 2.6volts, so adjust your code so that the result of an ADC measurement is let's say 0-1024 = 0 to 4.2volts ~= 630 (derived from 2.6/4.2x1024) and if it goes lower than that switch off the FET and hence Buck.

If readadc < 630 then gosub switch_off_FET
 

AllyCat

Senior Member
Hi,

It would have been helpful if the data sheet included an internal circuit diagram of the puck itself, and you gave an indication of the load (volts / current) that you're driving, then we might be able to help more.

Due to the warning that "LED(s) must always be connected whenever power is applied!" it may be that the "boost" is relying on the load (LEDs) to "catch" the overswing voltage, and removal of this load causes over-voltage at (and destruction of) the switching transistor. In this case a higher voltage zener diode directly across the output terminals could provide short-term protection (but remember that zeners also conduct in their "reverse" direction). Then you might need to detect the fault condition with the PICaxe, to switch off the power before overheating occurs.

Are you sure that you're using the most appropriate configuration from the data sheet, in your application?

Cheers, Alan.
 

tmfkam

Senior Member
My assumption would be that the current monitoring internally to the 'Puck' relies on a volt drop monitored across a resistor in series with the LED(s) to limit the current through a switching device and inductor. Should the LED(s) become open circuit I'm imagining that the switching device is turned on for longer and longer as the circuit tries to raise the (non existent) current and subsequent volt drop across the sensing resistor, until eventually either the switching device or inductor are damaged beyond repair.

You could try placing a zener across the R and O terminals that would conduct should the LED(s) become open circuit, a value of 1V greater than the volts dropped across the LED(s) should suffice. The zener might go short circuit but could save the 'puck'?

Alternately, why not consider making your own LED driver unit? A Micrel MIC2287CYD5 should suffice, it claims to drive up to 6 LEDs at 500mA from a 2.5V - 6V supply?
 

AllyCat

Senior Member
Hi,

It's difficult to determine what is the best solution with so little information available (particularly from the manufacturer). I do have some further observations to make, but don't intend to spend any more time until it's confirmed that the OP is still actually watching and interested in the thread.

However, it's probable that the output is actually a (bidirectional voltage) "square wave", so (without further information) I advise against attaching a zener diode directly across the output terminals or attempting to "measure" a (dc) voltage to control the PICaxe.

Cheers, Alan.
 

tmfkam

Senior Member
Alan, you are right, I am speculating about the topology contained. With it being a boost/buck convertor I'd *assumed* it to be similar to others I've encountered which incorporate an inductor and rectifier arrangement.

The zener idea came from the Micrel datasheet for the device I mentioned, the MIC2287CYD5.

Grab.jpg
 

AllyCat

Senior Member
Hi,

One of the (many) issues I have with the application is that the LED cathode is connected to the Positive battery terminal. Thus IMHO it's not ever a Buck converter, just a conventional Boost, with the battery voltage "subtracted" from the load. In view of the very small size, high current and low voltages, it's probable that it doesn't have a (significant) output reservoir/filter capacitor and maybe not even a diode, since the LEDs can potentially perform this function.

Another issue is that the Data Sheet here specifies "Absolute Maximum" Input and Output ratings of 3 volts and 8 volts respectively and Output Power = 1.5 watts, yet the "Application Note" shows Inputs up to 7 volts, Output up to 10 volts and 3 watts (but not in "Buck/Boost") output power. :confused:

It's not clear what the specified "Output Voltage" is, but I'd guess it's between 'O' and 'G'. However in the OP's application the LEDs' drop can be as high as 7 volts, which added to the battery 4.2 volts makes 11.2. With "normal" operation more than a volt outside even the more optimistic Application Note data, it's difficult to decide what level to set for any "protection" circuit.

IMHO, the OP probably should be using either two series LEDs in the pure Boost configuration, or parallel LEDs in Buck. And the Application Note is VERY specific that a LED (or some substitute load) MUST ALWAYS be connected whenever the module is powered.

Cheers, Alan.
 

eggdweather

Senior Member
Alan, well spotted, I missed the unusual if not erroneous connection of the cathode to the + supply. So for the LED's to illuminate let's say there needs to be about 7v potential difference so point O has to to driven to 11v relatively to the 0v line, and something has to sink the LED forward current, let's say 10mA, it's not the battery, unless rechargeable, so it would leave the buck/Boost device, but how does that work, something for nothing! I'm going to study the data sheet a little more, it's an intriguing challenge.
PS: Wiring diagram is wrong, relative to application note, I suggest you re-wire to how the manufacturers have suggested for either boost or Buck mode, deciding what level of efficiency you need will determine the buck or boost choice.
 
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