What is the maximum current that each pin can source?
samuell
Posts: 554
Hi,
What is the maximum current that each pin can source? And conversely, what is the maximum current each one can sink? I'm asking this because I've modified the prototyping board to light one LED per output. I've used 82R resistors, which gives roughly 14mA for the green LEDs I'm using. Is that too much?
Kind regards, Samuel Lourenço
What is the maximum current that each pin can source? And conversely, what is the maximum current each one can sink? I'm asking this because I've modified the prototyping board to light one LED per output. I've used 82R resistors, which gives roughly 14mA for the green LEDs I'm using. Is that too much?
Kind regards, Samuel Lourenço
Comments
Kind regards, Samuel Lourenço
We designed things very conservatively regarding power delivery. For every four I/O pins, there is a GND and a VIO bond wire that are each 1mil diameter gold, which have about 1A capacity for their 1mm length. The top metal layer of the chip (M6) is 3um thick and used as the power grid on the die.
Well, those GND and VIO wires are in series, effectively, so only 16 Amps. I think the bond wires have much greater capacity than the GND and VIO pads they connect to.
Yes, the core power connections would make it 32 Amps. We won't have a wiring problem.
Taking the 30mA mentioned already, over a hypothetical 60 pins (getting quite rare) that's 1.80 Amps (tolerable, with that many pins bonded), and at a nominal 10 ohms FETs that is 0.54W added from the IO i2r effects alone
That's more of a factor to consider - it might be enough to bump a design from 2L to 4L for example, depending on target MHz.
Because of this I may stress test a P2 chip if I'm game, at least up to 2A. Isn't there a problem with too much current flowing through the substrate and the resultant voltage drop/rise?
You could, for example drive Blue LEDs to a 5V rail from P2 pins, and not have any additional SMPS load.
The PAD ring has the bonding pads, and the wide metal, so most of the drop is across the mosfet, but you probably can measure some millivolts across the metal on groups of 4 IOs, if you look for them.
You can also load 3 pins, and just CMOS out the 4th, to read the internal GND potential shift, caused by those 3 loaded pins.
Silicon substrate currents are near zero. The metal layers move all the current and the output FETs will exhibit the most drop.
That would be great.
I think P1 is 40 mA per pin and 300 mA total. So, only a few pins can be drawing max current...
I think it all comes down to acceptable heat contribution, not design limits of pins.
At around 20 ohm switch impedance thats around 165mA into a short circuit, or 80 mA into a matched impedance load
When doing DAC testing, there was a few millivolts difference between outputting a digital strong low, to analog dac level 0, that I think reveals the impedance of the gold wires and bus on the pin pads.
A 99.6 ohm resistor from VIO to an output and it sinked to 0.468 (-2.838) volts. VIO = 3.306 which comes to 16.42 ohms sinking.
Those were single separate pin tests.
Wow! Is there ANY other microcontroler that will do anything close?
Yes, taking 60 pins at that 28.5mA, is ~800mW of added power load, which is quite significant.
MOSFET Rds is also positive tempco, so is prone to thermal runaway, needing more care with cooling.
Running all the DACs is likely to be a challenge thermally too.
Chip mentioned before I think the sourcing and sinking impedances should be close near midrange.
I think the 30mA figure Chip gave is probably *really* conservative [img][/img]
With P = U*U/R
Rising R should not run away ?
Pmos = I² * Rds
So, if R increases, the dissipated power across the MOSFET, P, also increases.
Kind regards, Samuel Lourenço
Thanks
Makes sense