ESP8266 Community Forum

- Sat Mar 26, 2016 8:59 am #44070 input R = 3.3/.000050 = 66000 ohms --- Maybe?????
output R 3.3/.012=275 -- I don't thinkso..........
if assume .1v sink then .1/.012 = 8.33 Maybe ?????? but still a guess
no clue on source drop....1v?
so what happens when only drawing 1ma

- Sat Mar 26, 2016 11:16 am #44085 GPIO Input 66K, yes, that make sense.

GPIO Output 275R, yes, that make sense too : don't forget that Output means driving capabilites, so, anything under 275R will probably damage the GPIO.

If you think about "sinking" instead of "sourcing", the same specs apply, since the load will be across GPIO and VCC instead of GPIO and GND, you still have 3.3V/0.012=275R.

When drawing only 1mA, you simply have higher load resistance, so 3300R.

Don't forget, there is no such "output resistance" terminology, it is the "load" that is the resistance against a driving capability.

- Sat Mar 26, 2016 12:18 pm #44091 The stated maximum output current is really a figure that the manufacturer considers to result in permissible heat dissipation in the chip due to the I2R loss in the output transistor stage. While it's certainly related to the effective output resistance it's not a parameter that would let you calculate the resistance with any accuracy.

However, if you apply a known resistor load and compare the output voltage to the unloaded output voltage you'll measure a volt-drop. Say you measure 3.3v without any load on a pin, then wire a 1K resistor between the output and ground and measure 3.0V at the output you know that the 0.3V "got lost" in the output resistance of the chip. Ohms law says this resistance = 0.3V/(3V/1000R) = 100R

That's when sourcing 3mA and it won't necessarily be linear so try making measurements with a couple of resistors and get a feel for the characteristic. 220R is about as low as you should go. The same can be done for sinking current by setting the output pin low and connecting the load to Vdd.