Electronic – Driving an LED from an 1.8V GPIO pin that defaults to “IN”

driverfangpioled

I have an EspressoBIN board, which has 1.8V GPIO pins with a max current of 8mA.

I'm attempting to drive some status LEDs from them, and a 5V fan, so I'm using 2N2222s with 220 Ohm base resistors connected to the GPIO pins, emitters on ground, and between 0 and a few hundred ohms connecting collectors, load and +5V. (220 Ohm so I'm not losing much voltage to drive the fan)

This works fine, except during boot. Turns out the EspressoBIN, by default, sets the GPIO pins to "IN". In that state they are at 1.8V with about 10kOhm internal impedance. This is enough to turn on the transistors so the LEDs and the fan come on (on half speed).

I can only turn the GPIOs to "OUT" once the board has fully booted (about 60 seconds later).

What's the simplest-possible circuit you can think of that keeps my loads off during boot, and lets me have them switch 5V on/off once the GPIO is set to "OUT"? Note that per vendor, 8mA is the max current I may draw.

Best Answer

Add a resistor to ground on each GPIO, say about 2K. When it is booting you will have about 0.3V and after you set the GPIO to OUT then the pull down will draw 0.9 mA when the GPIO is high.

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Electronic – Driving a 2V LED from a 1V8 FPGA IO pin (MOSFET vs open-drain)

The only reliable and bullet-proof method is to use the MOSFET. I can tell you that without a doubt that the MOSFET will work. But I cannot say the same thing for the open drain pin version.

You may be able to get the LED to light using an open drain pin, but it also might turn on and stay on too. Or the FPGA pin might get destroyed. Or it when you turn it off, it might only get dim and not turn off entirely. A lot of this depends on the FPGA (which you didn't tell us what it is), the exact type of LED, and probably the phase of the moon.

The "signal path" that keeps the LED lit (dimly) when it should be off is: +3.3v -> Resistor -> LED -> FPGA Pin -> ESD Protection Diode -> +1.8v. On the surface this seems unlikely since Vf of both the LED and the protection diode is greater than 3.3v-1.8v. But you have to remember that Vf decreases as the current through the diode decreases. So even in this circuit, there could be a small amount of current flowing. Of course, the question then becomes, "at some small amount of current, will the diode be turned on?" And that question is very dependant on the diode itself.

Using the MOSFET approach has none of these issues. If I were designing a product, I would not take shortcuts and instead use the MOSFET approach.

Electronic – Maintain state of GPIO pin across reboot

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. State holding capacitor.

Wire up a spare GPIO as shown. On reset read the state of the capacitor and set the output appropriately. This will give you a short-term 1-bit memory.

//Pseudo code to go early in boot sequence.
pinPullup(pin) = false;        //Turn off the pull-up.
pinMode(pin) = input;          //Set the pin to input mode, if required.
pdState = pinRead(pin);        //Read the input to get the power-down state.
pinMode(pin) = output;         //Configure as output.
pinWrite(pin) = pdState;       //Restore the power-down state.

You may wish to swap the order of the last two lines (depending on micro) to avoid a momentary blip.

Note that in this configuration the pin can't be used for anything else.

Best Answer

Related Solutions

Electronic – Driving a 2V LED from a 1V8 FPGA IO pin (MOSFET vs open-drain)

Electronic – Maintain state of GPIO pin across reboot

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