Hello,
I want to connect a LED with Uf=1.8V and If=20mA to a CPLD (XC2C512). But from the datasheet I get that the current is limited to I_OH and I_OL = 8mA. Is this right? Means this that I couldn't use this LED with
1.8V? Have I to operate the LED with 8mA and the corresponding voltage (using a resistor in series)?Regards Deniz
just use leds with 2mA (for a small price premium) or use a resistor to limit the curent to Hello,
-- __ / /\/\ Aurelian Lazarut \ \ / System Verification Engineer / / \ Xilinx Ireland \_\/\/ phone: 353 01 4032639 fax: 353 01 4640324
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You can easyly drive a 20mA LED with the Coolrunner-II. The current given in the datasheet is the guarenteed maximum under certain measurement conditions (Rise time, output LOW/HIGH level). For a LED this doesnt matter at all. There is a typical I/O curve in the datasheet, from there you can see that the output resistance is ~25 ohm. So if you draw 20mA, the output voltage will be 0.5V away from GND or VCC (depending if you sink or source the current). Just go ahead, calculate the appropiate current limiting resistor value and drive the LED with 20mA. Only keep in mind that you should not drive a logic inputs with this LED in parallel, since the the output voltages ar not more full specification compliant (but it will work thou)
Regards Falk
Using a LED with a forward voltage similar to that available is likely to give problems. The series resistor normally fitted gives a determination of currrent but does need some voltage itself. The smaller the notional voltage used across the resistor the wider the span of currents that may result due to tolerances. In short using small resistor voltage gives relatively un-predicatable currents. No resistor at all is asking for a LED meltdown assuming it actually lights at all.
Look at the MAX DC voltage specifications, and the Typical I/O curves. Vol is < 0.4V @ 8mA, or 50 ohms (Max) RdsON.
Typical is just over 2x beter than the worst case, and Typical current is ~30mA, so a worst-case port might deliver only 15mA (and still be in-spec) - thus 20mA is pushing it, from a single pin, for volume production.
You can draw a load-line on the I/O curves, to see the LED current. eg assume a 3.3Vcc, and 1.8V Vf on LED, your load line will start at X=1.5V, and will hit the Y axis at 1.5V/Rseries
A Y intercept of 30mA is appx 50 Ohms, and the Vol curve intercepts the 1.5V30mA load line, at appx 22mA A 100 Ohms load line, gives ~12.5mA etc.
-jg
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