Is there a part?

Where? I missed it.

Yup. The green that isn't really a sickly yellow. ;-)

e:

sistor to

t leave

d a

age

ving a 10

if you pick one of the wimpier logic families to drive it won't the output start to act CCish at a led friendly current?

-Lasse

When you get to the page with the switcher at the design review call a restroom break, whistle innocently, and when everyone has left for the loo skip forward one page :-)

te:

esistor to

't leave

nd a

tage

aving a 10

| \ --- | \ | >------| >O------' | / | / |/| | =3D=3D=3D

i(f) @ (Vf =3D 1.7v) =3D nada.

Exactly. :-)

-- Cheers, James Arthur

Yeah, I wasted all my time arguing against using the blue LED instead of just doing it. When the damned thing lit in the existing board, there was no going back. I shuda just done it without bitching.

resistor to

leave

a 10

If I had the 5V available it would have been a walk in the park. Only 3.3V available on that board. The ones that had the battery available I used the NPN & N-FET current source, so the LED output didn't get squirrely until the battery was dead anyway.

Bright, too! Put-your-eye-out-bright! ;-)

How about an open drain device...

te:

y having a 10

ng

d the

the

That's reasonable. Just for fun...

D1 could be an LED, that way you'd have two alternating BLUE LEDs. Marketing + PHB'd love it. :-)

If not, D1 should probably be a dual, else LED won't turn fully off.

-- Cheers, James Arthur

Use a joule thief circuit to squeeze out max battery energy.

ote:

=A0 =A0 =A0 =A0 =A0 =A0 ... Where I had headroom

d at the

nge, or

ed the

just

o going

How about this for a solution?

__________________________ | | + | | / | \ R1 | / | \ | | R2 _/ c +--/\/\--(_) Tr | \ e _|_ | _A_ Zener _|_ | _V_ LEDs | | ______|__________|___________ 0v

Use a low Vce_sat Tr, such as PBSS4130T with 90mV Vce_sat @ 0.1A And give each LED its own very low value R

NT

the

or

Problems:

1. Low voltage zeners suck 2. The problem is that '+' is only 3.3V (3.5 to 4.1 in the raw battery case) 3. You're essentially controlling the LED with a voltage source (or are counting on R2 and beta) 3a.LED current is a function of the Vz, Vbe, and Vf.

Here is what I ended up using:

Vcc | | Vcc V | --- LED \ | /10K | \ | / |-+ | | +--->|-+ | | \| | |----+

e:

sistor to

t leave

d a

age

ving a 10

the gate itself will probably have at least 25ohms too, maybe a good reason use one of the wimpier logic families

-Lasse

I remember John Larkin posted a "LED driver" that was just a resistor feeding a beta-binned transistor. The beta was "tight enough for a LED".

He who lives by the "beta bin" will die by the "beta bin" ;-)

There used to be a CMOS equivalent to the TL431. Look for it. It'll serve you well. ...Jim Thompson

But that drops 600 valuable millivolts, plus tax. You could either prebias the BJT a bit or use an opamp instead. Of course a higher voltage rail wuold be best :-)

According to the datasheet the beta for that transistor at 2mA can be between 380 and 630. If you drive several LEDs that are next to each other and the assembler switches reels that might not look so good.

No tax (560 mV, actually ;-).

Sure (didn't occur to me to throw some extra current at the base but I don't see that buying much). This was the circuit I used when powered directly from the Li-Ion battery (4.1 -> ~3.5V). At the bottom end of the range it behaves exactly as the base-emitter resistor would. Above that the current is pretty much fixed. The current regulation is a lot better than regulating the voltage (down), with a ballast resistor. It works great for the application. For the case where the 3.3V supply is on another board, not so much.

I do poor man's current mode switchers that way, the ones where I don't want to spend the 50c or so for a PWM chip. Or where the client doesn't want to and needs 2nd source on 100% of the BOM stuff. 600mV is too much to leave on the table so I reduce that to 250mV or so via two more resistors.

Two resistors? Splitting Vbe, or some such?