I should have a CTR at around 4000% from the 3.67mA that should be sourced from OUT1, thus resulting in ~147mA

Which would exceed the 150mA dissipation rating, burning your opto and possibly making it short out?

Why not use a MOSFET? I use AO3400 for this sort of thing all the time, works great - just don't forget the flyback diode

The threshold voltage of the BS170 is too high to operate off 3.3V.

if you look at the BS170 datasheet you will see from Fig1 on page 3 that at 3.3v it will only pass about 200 mA which is almost certain not enough to drive the solenoid.

The MOSFET will possibly pass enough current to heat up significantly so be careful operating in this state.

You depict a TO220 package - the only data sheet I could find was for smaller packages - is it really a BS170?

You need to find a device that will turn on adequately with only 3.3v drive or amplify the voltage up to a reasonable drive level.

Older MOSFETs we designed for 10V gate drive, a number of logic-level devices were the designed to be driven by 5V logic. There are even fewer that will work from 3.3v.

You need to know how much current the solenoid requires in order to select the MOSFET.

http://www.mouser.com/ds/2/308/BS170-1118810.pdf

A MOSFET based switching will be ideal solution for you. Basically you need a high switching speed MOSFET with gate control voltage of 3.3 volts and 0v at gate will anyway result in 0v drain voltage.

I had chosen this MOSFET for my project and it worked well for me. The input is coming from a micro-controller GPIO pin having 3.3v voltage level. You can connect your 3v3 control signal directly at this point.

Further, the output is supposed to be connected to inductive loads such as motor or solenoid, in which case freewheeling diode is added on purpose. Since your load is FAN which might be having an inductive motor, adding such diode will be suitable for you.

VCC5 is mentioned for 5v compatible loads. You can directly connect this point to 12v voltage source as the MOSFET is compatible to sustain upto 60v. Read the datasheet for full details.

You sure could. Make sure the 12V supply is able to accept current flowing into it, as well as providing current flowing out.

But it's more efficient to use an "H-bridge" circuit to reverse both ends.
To move in one direction, you put one end at 0V and the other end at 24V.
To move in the other direction, you put one end at 24V and the other end at 0V.

The half-H-bridge is just a pair of switches, one to 24V and one to 0V, so that the output can be either 24V or 0V depending on which one you turn on. Turning on both at the same time creates a short circuit, so don't do that. The H-bridge is just two halves, one for each end of the solenoid.

Of course, the solenoid has to be mechanically designed to work in both directions, too. There's no point moving a solenoid both ways, if one way causes it to bump into an end stop and not move.

The solenoid must use a permanent magnet in the moving piece, because iron that isn't magnetized will be attracted to either a north or south field equally well - reversing the current will reverse the magnetic field but that won't achieve anything.

The Rds(on) is only guaranteed with a minimum drive voltage of 4.0V. 3.3V is less than 4.0V.

Possibly the MOSFET is heating up which typically decreases the Vgs for a given current. If you got the IRLZ44 from some dodgy source it's possible it does not meet specs. Although it's bad design to use an IRLZ44 with 3.3V drive, I would expect it to typically work under nominal conditions. That kind of time constant sounds like thermal or possibly leakage (as in an open or almost open connection to the gate).

There are many, many MOSFETs which are rated with 3V, 2.5V and even 1.8V drive, but few, if any, in a through-hole package. Here is one example, 0.005\$\Omega\$ maximum with 2.5V drive and 0.075\$\Omega\$ maximum with 1.8V drive.

If the MOSFET fails with a drain-gate short it could damage (ie. destroy) the MCU. Electrical noise could reset the MCU. You could use an optocoupler or add some transistors to drive the MOSFET which would provide protection. You might also be worried about the solenoid lock being destroyed if the MOSFET fails on- I think most of them are only rated for very intermittent duty.

I think I have three options:
  (a) 3.7 V li-ion    + boost converter to drive solenoids
  (b) 12  V lead-acid + buck converter to drive MCU
  (c) 3.7 V + 12 V dual-battery system
How to decide between these options?

I've named your options (a)..(c). I've also deduced that these are in order of preference: (a) would be best because the battery's smallest, then (b) then (c) least.

The below explanations don't account for falling battery voltages as they drain down. You'll have to factor that in yourself from the battery manufacturer's data. But the calculation results will each be distinct enough to select your option before that anyway.

For (a), the 3.7 V battery 'batt3V7' has to be large enough for 6 months of the MCU anyway. So you're looking at the additional battery capacity needed for the solenoid.

This extra capacity is: (Isol x 12/3.7 x (100/boosteff) x ton x 3600) Ah

Where: Isol is solenoid average current drawn, across pull-in current to hold-in current (best measured with scope, while loaded with your actual mechanics); 12/3.7 is the step-up voltage ratio; boosteff is the boost converter efficiency as a percentage; ton is the on-time in seconds; 3600 is number of seconds in an hour.

You can then decide if the new capacity of batt3V7 is suitable and practical.

For (b), it's a similar method in reverse, with the MCU loading 'batt12V' and the capacity being the sum of the solenoid and MCU capacity requirements.

The solenoid capacity needed is: (Isol x ton x 3600) Ah

The MCU capacity needed is: (Imcu x 3.3/12 x (100/buckeff) x 3600) Ah

Where: ton is the solenoid on-time in seconds; 3600 is number of seconds in an hour; Imcu is MCU average current drawn including any use sleep modes; 3.3/12 is the step-down voltage ratio; buckeff is the buck converter efficiency as a percentage.

For (c), the size of each battery can be determined by applying the methods shown above.

As has been suggested in the comments: put the solenoids in parallel. In fact your idea where you put then in series with a switch for each won't work. As soon as one switch is open none of them will work.

More important: for outside circuitry you want to stick to a low voltage. It is safe.

I think your concern is that you need ~35Amps to have them all on. You could put two of them in series with one switch per pair. You would then need a 48V transformer of ~17.5 Amps. But you still have the same amount of watts so the transformer will be about the same size (and cost).

What also has been mentioned is that the current will be much higher when you switch them on. If you use a computer, you should program it, so it never switches them all on at the same time. That way you can get away with the 35Amps transformer.

You will end up with a bundle of 87 wires. I don't know how these solenoids are placed. If they are far apart you might put a tiny relay at each so you have one set of power cable and very thin control cables. The cost is cable versus an extra relay.

Ultimately you could place a tiny controller (They come in as small as 6 pins) at each solenoid and use a protocol. (e.g. a UART). You need only three wires then, but that may be beyond your capabilities. (It has the advantage that you can interface the system to your computer using a standard USB-serial adapter. )

There is a common misconception:

The solenoid is physically installed inside the machine, and I have no access to place a flyback diode directly on the solenoid terminals

The source of the flyback, is the element whose effective resistance changes suddenly: the transistor.

The flyback diode must be placed at the transistor.

This is an oft missed point, and I have even seen misinformation claiming the opposite is true. It is not. The fundamental is as above, simple as that, nothing more, nothing less.

There are circumstances where putting the diode at the solenoid works out anyway -- but this is not counterproof, this is simply a tolerable case along a continuum of cases more distant from the fundamental. Namely, the stray wiring inductance between switch and diode, still itself incurs flyback. The amount of flyback might be negligible (say the switch takes some ~ms to transition from closed to open, and the cable is only some meters long -- a few µH, charged to, say, less than an ampere), but this is a mitigative solution to a non-fundamental situation.

When a diode is sufficient (given other operating constraints), and you are free to place the diode anywhere in the system, place it by the switch, and place a bypass capacitor between GND and +V. All three (bypass, switch and diode) are best co-located to minimize stray inductance in the loop between them.

The proof of this fact is underlined by its universality. What if the switch is carrying 100A, not just 1A? What if the cable is 100m long? What if it switches in 10ns? Or 1ns? A switching power supply is not a separate case, based on unique fundamentals; it is a specific case of the same fundamental. Keep the "fast" loop small.

(Mind, the bulk of the above is not so much addressed to the OP; a simple statement of fact suffices for that. It's mainly directed at those who lack understanding on this topic. I'll repeat this as often and as loudly as necessary until the message is received. Sadly this platform doesn't have any sort of "pinned comment" function, so repetition it shall be.)

So, with that covered, it seems the bulk of this question is, where to wire the power supply.

If PCB ground plane cannot be joined with chassis ground, then it would seem the remaining issue is isolation.

Provide a power supply that can be joined to chassis ground, and electronic (logic signal) isolation to couple the signal source to the driver.

A simple phototransistor optoisolator will suffice at this frequency. If a separate power supply is not available, or an external unit cannot be provided (for cost or space reasons, say), a DC-DC isolator module can be placed on the PCB, or the equivalent circuitry built out using a typical flyback/forward controller/regulator and associated components. (Beware EMC issues due to common-mode currents, i.e. switching noise injected into the ground-return path.)

Supply reversal is then an independent issue, easily solved with a polarity protection diode, or if the voltage is low enough that a diode drop would be problematic, a PMOS wired in the usual way. (Pay mind to transient and sudden-reversal protection, if applicable.)

Isolation also solves ground lift, as there simply isn't a closed circuit path so that nothing works, as expected (at least, I would hope no one expects the thing to work when it's wired entirely wrong, whether by accident or otherwise..!).

Illustration of isolated scheme:

^{simulate this circuit – Schematic created using CircuitLab}

Input filtering not shown, REG could be any linear or switching reg, 5V could be 3.3V or whatever else the logic/control supply is, XFMR1 is actually a proper DC-DC converter, SW1 represents the logic source, and maybe M1 has a better gate drive circuit. Maybe current limiting or thermal protection or whatever, too.

Note that this particular arrangement is unfavorable for CM noise purposes (solenoid switching voltage is applied to the isolation barrier), which might not be a problem at the slow rate (particularly with rise time control, i.e. drive the MOSFET slowly, but maybe some other considerations like an R+C snubber between G-D). Mainly, showing it this way to illustrate how, with isolation, you don't much care how it's driven, allowing a cheaper NMOS to be used. Otherwise, making it common-ground with a PMOS in the obvious manner is perfectly fine too.