Electronic – Choosing between relays and MOSFET driving multiple loads

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.

I really do not think it is possible to regulate the output of the fixed CC driver due to the fact that it will be continually tracking the output current and trying to keep it at 320mA and varying the voltage to do so.

To be able to do it I would suggest investing in new drivers that are constant current but accept pwm input, after a quick search I came up with this driver from Amazon or this one from eBay.

It is very inexpensive and would use a common DC power supply. With the requirement of 5 to 35 V input, with the current set to 350mA x10 would only require a 220AC to 12V@850mA supply, though I would try and find a 1.5A or 2A supply to have some headroom. These drivers appear to directly accept PWM input so no additional circuitry would be necessary for this functionality.