Idea for a simple -1.5V 20mA power supply?

As asked in the header, does anybody have an idea for simply generating a low negative voltage for indicator leds on low output voltages e.g.

Non-red leds (green, yellow, white) need at least 2V or more so a direct connection to low voltage regulators for indicating presence doesn't make sense.

Of course a driving NPN could be used however on many voltage outputs a negative supply would be an alternative.

Using some kind of phase output of buck regulators is one of the possibilities, is there any other simpler idea ?

Thanks for all suggestions, Mike

Doesn't the FPGA have higher power bank voltages?

A simple flyback booster would work.

Am 02.11.2022 um 19:04 schrieb John Larkin:

yes, thats an idea, but how many parts does it cost ?

I'd like to have a voltage that doesn't light a led when the vreg output is 0 or GND.

An ICL7660 with regulated output voltage of 1.5 or 1.23V would be perfect.

These old 7660 are quite pricey ...

Is any higher voltage available to run the LED? FPGA core voltages are usually switched down from something else.

If you do have some voltage to run the LED, all you need is a comparator (or an NPN transistor) to detect that the core voltage is there.

Am 02.11.2022 um 04:08 schrieb snipped-for-privacy@noplace.com:

yeah good point, its an issue to discuss.

The low current leds should have a resistor as well and the current should be about 1mA or below. Modern leds are very bright these days...

And the negative voltage @1.2V or so should not damage an IO pin when current is in the uA range.

Its probably a bad idea because of the high effort to generate a regulated -1.2V. This voltage could be clamped by an IR diode though ?

Thinking about basic design ideas to not get into politics...

Thanks, Mike

Am 02.11.2022 um 21:14 schrieb John Larkin:

yes, some designs use transistors to detect voltage (could also use RET

- resistor equipped transistors) but i like the direct approach.

To drive a low current led with a transistor is like using an i9-13900k for a led blinker :-)

Thanks, Mike

No answer, can't help.

Am 02.11.2022 um 21:54 schrieb John Larkin:

sorry, 3.3V are available...

OK, why not an NPN transistor and two resistors?

onsdag den 2. november 2022 kl. 22.23.01 UTC+1 skrev John Larkin:

not enough Rube Goldberg I guess

OK, live dangerously, an NPN and one resistor.

Or a dual diode and one resistor.

Just barely, one transistor and no resistors.

It's okay if it's an emitter follower.

Or just a LED and a battery. A zillion Chinesium flashlights can't be wrong, surely?

Cheers

Phil Hobbs

How about a resistor in the emitter, to make it a (mostly) controlled current sink?

Yes, that's one way to do it.

Or offend a lot of people and ground the emitter and beta limit. With a beta binned part, like BCX71K or something, that's actually not awful.

Unless Rbb' is fairly huge, the BJT will saturate massively and all the current will come in via the base. A prebiased transistor and a collector resistor would work.

Cheers

Phil Hobbs

Yep, that's a good plan; LED from +3.3V to collector.

Emitter resistance doesn't allow much base bias (i.e. base current) when the transistor ISN"T yet saturated, and... it'll never really saturate if the I*R_emitter + Vbe = 0.9V condition is scaled right. You do want, though, the LED to glow with less than 3.3 -Vce(sat) ~=3V (can't do it with a blue LED on 3.3V)

Very minimalist would be to drive the LEDs from a +2.5V or 3.3V rail and choose highish series resistor so the eye discerns the brightness change between 0V and 0.9V - for comparison include two extra LEDs in the bunch, one at 0V, one at 0.9V to serve as brightness references for the eye?

piglet

Depends how much you can take from the FPGA outputs without annoying them but drive a pair with antiphase square waves AC coupled to a diode doubler using low drop diode won't be too far off LED drive voltage.