Cascading AS1109 led drivers with one current set resistor ?

Go To Last Post
8 posts / 0 new
Author
Message
#1
  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Hello everyone,

I'm wondering if it is possible to cascade a number of AS1109 (DATASHEET HERE) but use only one resistor to set the current on all the drivers.

Quote:
The AS1109 scales up the reference current (IREF) set by external resistor (REXT) to sink a current (IOUT) at each output port. As shown in
Figure 3 on page 7 the output current in the saturation region is extremely flat so that it is possible to define it as target current (IOUT TARGET).
IOUT TARGET can be calculated by:
VREXT = 1.253V
(EQ 1)
IREF = VREXT/REXT (if the other end of REXT is connected to ground)
(EQ 2)
IOUT TARGET = IREF*15 = (1.253V/REXT)*15
(EQ 3)
Where:
REXT is the resistance of the external resistor connected to pin REXT.
VREXT is the voltage on pin REXT.

If i just parallel all the current set pins would there be any problems because of small differences of the voltage (VREXT) ?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You could try it, but personally I wouldn't. It's not directly the VREXT that matters but IREF, the current being drawn from the pin. If one of the chips has VREXT a millivolt lower than another, that chip won't be sourcing any current at all and it will have zero brightness. If your intention is to have a common brightness control by varying the single resistor, you might achieve this if you place a separate small resistor on each REXT pin and bring the groundy ends to the common resistor.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

That's what I thought, so a small resistor for each chip and one common potentiometer would do the trick ?

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

I would select the resistor for the maximum LED current, whatever you determine it should be. Say it was 10mA and the resistor value was 1k. Then you could common all those 1k resistors on to a 10k pot and have control from full brightness down. I think you would not need a very high value pot, since it has to pass the total combined current. Maybe 5k or even 1k would give you the control range you need.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Thanks peret, I think I'll do that.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

Why the heck would they give the formula for IOUT in terms of REXT, when the vast majority of people reading the datasheet are will want to calculate the required REXT to acheive a specified IOUT? It really annoys me when datasheets do that--sure it's simple algebra to rearrange the equation to give REXT in terms of IOUT, but that could have been done ONCE by the datasheet author, versus thousands of times by datasheet users. TI's LED driver datasheets are the same way--is there some logic to it that I'm missing?

/rant

Back on topic, I'll second what peret recommended.

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

thavinator wrote:
Why the heck would they give the formula for IOUT in terms of REXT, when the vast majority of people reading the datasheet are will want to calculate the required REXT to acheive a specified IOUT?

I don't know - I've never attempted to calculate it. I use the handy graph included on most data sheets, look up the current on the left and drop down to find the resistor value at the bottom. 8)

  • 1
  • 2
  • 3
  • 4
  • 5
Total votes: 0

You could also PWM the 'groundy' ends using a transistor.