Meaning of Vce(sat)?

'Fraid so. It's a Darlington, intended for high voltage outputs.

Cheers

Phil Hobbs

That was what I was afraid of. I should still have enough headroom, but that is wasting a bit of power...

Any suggestion on a replacement part? I need to enable/disable 8 lines with logic level inputs.

It can even be a 1-of-8 decoder I just need it to support up to 1A, and I'd like it to be relatively cheap. My load is going to 24 parallel LEDs (3.3v Vf for 16 of them, 1.8 Vf for 8), each with a constant current sink in line with them.

Cheap, for me, is < $1.25 for 8-in-8-out, or < $2.0 for 3-in-8-out (since I'd be able to replace two chips in that case).

Thanks, Daniel.

Hello Daniel

How about using power MOSFETs?:

e.g. DMN6040SSD:

I should add, I'm talking about price for quantities < 20 at the moment, not including shipping.

You'd think that somebody would make a MOSFET version, but AFAIK they don't.

Cheers

Phil Hobbs

Somebody, I think TI, makes an LED driver with eight built-in current sinks and a serial interface. It was mentioned here not long ago.

We like to use the TPIC6595 for driving relays and LEDs. Serial input and eight open-drain mosfets.

It does not mean that Vce will be 1.3V at that current -- you could have

It DOES mean that the lowest that the transistor will pull the voltage down when 350mA is going through it is 1.3V.

"Saturation" in a BJT means that the Vce is as low as it's going to ever get at that collector current. Basically, the collector isn't acting like a pretty good current source any more -- instead, it's acting like a pretty good short circuit.

"Saturation" in a Darlington is equal to a Vbe + Vce-sat of one NPN, thus the odd-looking number. ...Jim Thompson

There are other chips in the same series that have pregain and a single output stage. I seem to have misplaced the number, but just threw a bunch of them away. Great chips.

Not quite what you asked for, but here's a pin-compatible mosfet equivalent to the ULN2003:

Regards, Allan

Maybe what I really need is just 8 Enhancment-Mode N-Channel MOSFETs?

Such as 8 of these?

Check; it is impossible to saturate a darlington.

The main device... but the driver device saturates. ...Jim Thompson

Looking more carefully, I think I might need P-Channel MOSFETS, and then I'd need to invert my logic driver (replace the 74HC238 with a 74HC138). If I used an N-Channel, I don't think I could get the Vgs to be high enough.

I was thinking this part:

Then I would have the source connected to +5v, the drain connected to to my load, and the gate connected to the output of the 1-of-8.

The load is some parallel LEDs (each in series with a constant-current-sink), so the drain voltage could vary between +1.8v up to +5.0v, The current can vary from 10mA to 1200mA. Depending on how many parallel LEDs are enabled.

Does this sound like it would work?

Thanks, Daniel.

On 11/03/14 19.58, Daniel Pitts wrote: ...

You could control the constant-current-sink directly?

E.g. by sorting the gate/source or base/emitter by a small MOSFET (2N7002)?

Double 2N7002:

/Glenn

Ok, looking at that part it appears it would do exactly what you need.

But I've noticed they have only 1 in stock? Sometimes when you see low stock counts like that, it's a sign of stay away... ;)

Jamie

I'm multiplexing 24*8 LEDs.

The "current sink" is 3 8-bit shift-register constant-current-sink (TLC5196) which power between 0 to 24 LEDs in parallel (depending). The MOSFET is a "row enable" to choose between 1 of 8 sets of 24 LEDS. So in total, I'm powering 196 LEDs, just 24 at a time.

Ah, didn't notice that, thanks...

Here's a "DUAL P-CHANNEL ENHANCEMENT MODE FIELD EFFECT TRANSISTOR" in stock.

About .1? Rds(on) @ 4.5v, which is good for me.

The pinouts also look like they'll be convenient for my application. More than the single-fet packages.