DM74LS244

Only to the rest of the circuit. And the programmer. ;-)

Exactly. I don't know why he can't find optocouplers that can handle

For example:

Or is this eight relays simultaneously? Then, what the other guys said is right.

Thanks, Rich

--- The coupler can only pull down, but to answer your question, yes. Your suggestion requires current to be flowing in the LED in order for the relays to be turned off. In other words, the relays default to the ON condition with no current into the LED. In most normal circumstances, one would expect the relays to default to the OFF state and to be turned on when current is forced through the opto's LED.

---

--- Yes, but that's immaterial at the moment.

-- John Fields Professional Circuit Designer

On 11 Jan 2006 08:34:16 -0800 in sci.electronics.design, "john" wrote,

How does a 74ls244 functionality enter in to this at all?

If you want to drive all eight relays at once, put the coils in parallel. 8 x 16 = 128 mA, one medium general purpose transistor will do, e.g. 2n4401

Hi john,

Marte

On Wed, 11 Jan 2006 12:52:40 -0600 in sci.electronics.design, John Fields wrote,

Does it matter whether the coupler pulls up or pulls down? It's isolated, right?

It is more the custom to specify the nominal relay coil voltage and not the current. Checking your datasheet

It is only the standard 24V coil relays that activate at the 16mA currents. You can't drive those with the 5V 74LS244.

On Wed, 11 Jan 2006 17:45:59 -0600 in sci.electronics.design, John Fields wrote,

Oh, nonsense.

Is this a contest? ;-)

Cheers! Rich

On 12/01/2006 the venerable Rich Grise etched in runes:

Don't know, but it's OK I'll fix it.

--- Thinking about your circuit some more, with a forced beta of ten in the NPN, the degradation of the opto's current transfer ratio because of the addition of the emitter resistor may cause it to not work.

Try this:

where the opto's emitter is raised a fixed about 0.7V above ground, or this:

where the LED drive and the CTR of the opto is all that's limiting the current into the NPN.

Unfortunately, all of these approaches allow the base to float when the transistor's cut off, so for simplicity and reliability, I still think the PNP approach is best.

-- John Fields Professional Circuit Designer

I prefer the pulldown R to ensure a good clean turn-off. I guess with a PNP, that'd be a pullup, wouldn't it? ;-) (which I'd still use) Plus, I'd arrange them like this:

That just "feels" like it gives a more "decisive" pulldown. Of course, for PNP, I'd put the opto on the bottom of the string it's on here. And the load, of course, I'd move to the collector, which would be on the - - oh, the hell with it:

Both of these are "High == On"; making it a "Low == On" is left as an exercise for the reader. [Gaw! I LOVE saying that!!!!] ;-)

Uh, yeah, that's kind what "isolator" means. ;-)

Cheers! Rich

On Fri, 13 Jan 2006 09:30:27 -0600 in sci.electronics.design, John Fields wrote,

Nonsense. The output transistor of the optoisolator is isolated. It is in series with the base resistor, and passes some amount of current. It DOESN'T MATTER which one of those two series elements are higher or lower.

More nonsense. It is fine with the base floating, or add your pulldown R if you prefer. But the NPN and PNP versions work exactly the SAME WAY except for polarity. The optoisolator absolutely doesn't care whether it is pulling the NPN base up or the PNP base down and I cannot guess why you ever said it did. Either way you may have an R pulling the other way if you feel you must change something, or you may omit it.

On Fri, 13 Jan 2006 22:37:43 GMT in sci.electronics.design, Rich Grise wrote,

I am more interested in a good turn-on, so I don't care for unnecessary opposing resistance. ;-)