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What's the best way to drive 3 LEDs (2 mA each) with 1.8 V signals (500 µA max) when I have only the 5 V rail (1.8 V and 3.3 V rails might not yet be available).

(For context: the LEDs are for showing the different steps of the power-up sequence, and to debug those steps if needed; no PWM switching involved).

The simple solution I have for now is, for reach LED, a N-MOS (with pulldown resistor, gate current limiting resistor, and current limiting resistor for the LED):

LED circuit

My main constrain is PCB space. Is there any more compact way to do it?

I was thinking maybe just a 3 channel unidirectional level shifter, but I haven't found one yet that requires just the 5 V supply (I haven't the 1.8 V rail yet).

Or maybe some buffer, but I haven't found one yet accepting 1.8 V logic levels as input while powered with 5 V.

Do you have any idea?

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  • \$\begingroup\$ Turned up blank looking for \$V_{IHmin} 1.2 V@V_{CC}5 V\$. But wouldn't a Darlington with a resistor from "2nd base" to ground do just fine? \$\endgroup\$
    – greybeard
    Commented Mar 3, 2023 at 8:48
  • \$\begingroup\$ Minimal PCB space you say? How about change the blue LED to a red one, and power it directly off the signal? Brightness may vary between boards. \$\endgroup\$ Commented Mar 3, 2023 at 11:23
  • \$\begingroup\$ @greybeard : do you mean to rely on the current gain of the Darlington to get rid of the current limiting resistor (R3)? \$\endgroup\$
    – Sandro
    Commented Mar 3, 2023 at 15:32
  • \$\begingroup\$ @user253751 : for PCB space, it is definitively optimal. I might however be a bit limitted in brigtness if using just 0.5mA. And I'm not very fond of using red for a power good signal (red is often understood as meaning error). But if I don't manage to get everything placed, it's definitively an intersting option. \$\endgroup\$
    – Sandro
    Commented Mar 3, 2023 at 15:34
  • \$\begingroup\$ I was more concerned about an 1.8 V signal not driving \$V_{DS}\$ below 1 V (exactly to leave some voltage for a resistor to set the current). But the same would probably hold if one found the lower \$V_{BE}\$ "ULN2003A-like" array I didn't immediately find, if for \$V_{CEsat}\$ too high rather than \$V_{IHmin}\$. \$\endgroup\$
    – greybeard
    Commented Mar 3, 2023 at 16:25

1 Answer 1

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You could use pre-biased (aka 'digital') transistors such as DTC143TE from Rohm. Three of those and a 4-resistor array for the LEDs.

enter image description here

(in this case R1 = 4.7k and R2 = \$\infty\$, so current from 1.8V will be about 260uA nominal).

There are duals available but with three that wouldn't help so much.

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