[clarified] Trying to understand a datasheet (TTP223-BA6)

Hi,

I've got some TTP223-BA6 soldered to some self-designed PCBs. My question is about the datasheet and the IC, respectively.
How much current can I get through this little IC before it get's destroyed? Mostly I find something like "absolute maximum ratings". In this case, there is no "absolute maximum rating" für sink/source current but voltage. Maybe I lack the knowledge to derive it from the specs?
On page 3 there is just

Output Port Sink Current IOL VDD=3V, VOL=0.6V - 8 - mA
Output Port Source Current IOH VDD=3V, VOH=2.4V - -4 - mA

.
And do I understand correctly, that output port source current means "pin is high"?

Would be great if someone could help me improve my datasheet reading skills.

Thanks & best

Can't go wrong keeping it under the recommended levels.

You are correct. IOH means Current, output high and IOL means Current, output low. Current flowing into the device (sunk) is marked positive, and current flowing out (sourced) is marked negative.

There probably should be an Absolute Maximum rating for the output currents, but that datasheet gives out a faint whiff of Chinglish in a few places so I don't know how high the standards would be.

So this in effect says:-
When you are sourcing current typically when you get to 4mA the output voltage has dropped from 3.3V to 2.4V.
and

When you are sinking current typically when you get to 8mA the output voltage has risen from 0V to 0.6V.

I would not expect an Absolute Maximum rating as this is just a logic output designed to drive CMOS logic inputs.

I posted and then deleted it because I explained source/sink in general, not per the datasheet.

You said:

And do I understand correctly, that output port source current means "pin is high"?

Yes, that's what it means. That is what the term IOH is for, IO HIGH. So it stands to reason IOL is a low output.

IOH is spec'ed as VDD=3V and VOH=2.4V means you can pull -4 ma out of the pin, maximum. This would be current pulled through the device, out the pin, into your load. This would be what you'd get if you attached a resistor from the output pin to gnd.

The IOL spec is higher at 8 ma which means it can absorb more than it can push. Think of a resistor with one end to the output pin, the other end attached to VDD.

Not that you'd do this, but you could calculate the maximum values for your resistors this way:

Sourcing: IOH=-4ma, VOH=2.4. R=E/I so 2.4/.004 = 600 ohms

Sinking: IOL=8ma when VDD=3V, VOL=0.6V. VDD-VOL=2.4V/.008 = 300 ohms

Does that help?

Thank you all for your help. I really appreciate it.

Your explanations about the VOH of 2.4V were also interesting. I had been wondering what that voltage drop means.

I think I'm good to go and I learned another memorable lesson about the fuzziness of datasheets.
In fact I was also irritated by my practical trial and error: I used a TTP226 (8-touch pads) ready made board to drive a fan directly by an output pin which worked like a charm. But when I replaced the TTP226 by TTP223 the fan did not move.
Surprisingly, as I checked just now, both datasheets state the same 8/4 mA current.

Thanks & best

I used a TTP226 (8-touch pads) ready made board to drive a fan directly by an output pin which worked like a charm.

Are you sure their was no buffer on the ready made board?

Remember a data sheet is the minim performance that is guaranteed, any individual device may outperform the data sheet but the data sheet is what you must design to.

Grumpy_Mike:
Are you sure their was no buffer on the ready made board?

Actually, no, I'm not sure :wink:
I'm neither sure that I really got a TTP226 because the IC does not provide any markings. I've something that looks identical to that:


Despite the IC I don't see anything fancy that could explain the better performance.

Thanks again & best.

I agree I can't see anything that would suggest a buffer either.