Electrical – Microcontrollers (STM32) and GPIO with momentary buttons

Single transistors don't really have any "quiescent current". If you don't feed them some current (or, in case of MOSFETs, voltage), they don't conduct. (A couple of nA leakage current can safely be ignored.)

As @geometrikal stated, these days there are better ICs for your application. Modern port expanders have an interrupt output, so you don't need the array of diodes which your solution would require. This particular IC has a standby current of just one µA, so you don't need to turn it off.

That's perfectly fine...

The short answer is that there is nothing wrong with this approach. It presumes, of course, that you have time to switch and do an ADC conversion (which at 200Hz) you do.

You might want a series current-limiting resistor in line with the gate to protect your MCU driver (if the total gate charge of the N-FET is in the tens of nC, didn't read the datasheet).

If you want a completely "digital" solution:

^{simulate this circuit – Schematic created using CircuitLab}

The component choices are (CircuitLab defaults) approximates, a wide range of parts will work, but it's a balancing act between R3 and R4.

• You need to make the R3/R4 ratio big enough that V(R4) < M1's Vth
• You need to make the R3/R4 ratio small enough that Vsrc-V(R3) > MCU V_IH

...for SW1 "on", MCU Hi-Z

Tuning

Here's a specific configuration that should work (5V source):

Materials:

See "documents" at these links:

• Fairchild FDY300NZ
• Atmel ATTINY24V-10SSU

Targets:

Procedure:

Start with the (GPIO: Hi-Z; SW1: Closed) case:

2. Vsrc -> R3 -> D3 -> R4 -> GND, must yield V(R4) < Vth,M1,min = 2V
3. We need V(gpio) > V_IH = 2.6V
4. This spread determines the minimum Vf,diode (Vfd) we need

Now, look at the (GPIO: Logic-1; SW1: Open) case:

5. We need V(R4) > Vth,M1,min = 1.8V (ideally with some comfortable margin)
6. This determines the upper-bound of the required Vfd

Now, look at the (GPIO: Logic-1; SW1: Closed) case:

1. We need I(R3) < I(OH),max
2. This determines the minimum size of R3 (go bigger for reliability)

Example:

• R3 = 15k
• D3 = 1.6V (forward) = approx 3x 1N4148 in series
• R4 = 10k

Control the FET/LED:

V(gpio) = 5V; V(g) = 3.4V

• PASS: 3.4V > 2V -- FET turns "on"

Read the state of an "on" switch:

V(gpio) = 2.9V; V(g) = 1.4V

• PASS: 1.4V < 1.8V -- FET turns "off"
• PASS: 2.9V > 2.6V -- MCU reads logic 1

Avoid damaging contention:

Switch is "on" AND MCU is driving the GPIO "low"

• PASS: 5V / R3 = 500uA

Power dissipation in the FET

The issue of power dissipation in the FET has been raised by a few commenters. It isn't a problem in this circuit due to the highly non-linear behavior of the LED.

Let's ignore the LED to bound the problem, by considering a worst-case impossible D4 with I(D4) = 20mA but Vled = 0 and R5 = 0 (impossible!). Now all of the power dissipation happens in the FET.

Under these conditions, the power dissipation in the FET can be maximally 100mW or ~1/5 of the maximum tolerable power of the suggested part. So we're safe.

However, you won't see dissipation near that level for any appreciable length of time. The transition time from R4 = 10k is approximately (RQV) = 10k * 1.1n * 3.4 = 37uS overall, but since we only need to move from 3.4V to below 1.8V we can finish in less than half that time.

At 200Hz, that translates into a mere 0.75% to 1.5% duty-cycle or less than 1mW in aggregate.

...and remember we ignored the real power consumers in the path -- the LED and current-limiting resistor (R5). In practice, it is impossible to deliver Vds = 5V to the FET, while Iled = 20mA, and the power dissipation in the FET will be negligible.