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Driving a group of leds with a buffer?

opolychr

Hello all!

 

I am having 32 input signals that come from an ADUM2400 (isolated buffer) and these signals go as inputs to 32 pins of an FPGA development board.

 

What I would like is to have 32 LEDs in the middle for debuggind reasons but looking at the ADUM 2400 datasheet (without being sure) I saw that the maximum output current per channel is less than 1mA, so not enough. I thought that having 4 8bit buffers would do the job. I found the 74LVC245APW from NXP but I cannot find inside the datasheet if they are able to drive with enough current (10mA per LED) each channel.

 

Does anybody have any idea if this buffer would do the job?

 

Thank you in advance!

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  • in reply to michaelkellett

    Elaborating on Michael's comment, the ADUM2400 data sheet gives three specs for sinking output current:

     

    Vol = 0.0V (typ) 0.1V (max) at Isink = 20 uA

    Vol = 0.04 (typ) 0.1 (max) at Isink = 400 uA

    Vol = 0.2 (typ) 0.4 (max) at Isink = 4 mA

    (Vol = Voltage for an Output Low)

     

    These are the usual characteristics for a CMOS output driver.  You can pull down to almost zero if you're not sinking much current, but if you have to sink current you can only pull down near zero.  Assuming your FPGA board has TTL or LVTTL input levels, they consider 0.8V to be a valid Logic 0.  To get 0.4V noise margin, you want to pull down to 0.4V or lower.  That means you can use the third spec: the ADUM2400 can safely sink 4 mA and still give you Vol = 0.4V, which is the standard TTL or LVTTL Logic 0 output.

     

    A decent modern LED should give enough light with 1 mA of current.  For example, Raspberry Pi LEDs use 1K Ohm resistors with a 3.3V supply, which gives 1mA of current assuming a 2.3V LED diode drop.  Or you could go with 2 mA (470 Ohm pull-up) and get twice as much light.