Electronic – MCU 3.3v to 5v Relay Driver, MOSFET or BJT

The solution you propose is good enough as a basic solution as long as you can guarantee that the FEZ output stays high at all times until you set it low. That's your job to determine.

Add a highish [tm] value resistor from +5V to PNP-base to hold it turned off when "FEZ" is open circuit. Failure to do this will probably lead to "surprises". 10k to 100k will suffice.

Note that as shown FEZ must rise to close to +5V to turn off the transistor. As little as 0.5V below rail ( 4.5V) may lead to turn on. This may or may not be a problem depending on the processor / drive circuit implementation. What works here may fail somewhere esle if this is not taken account of. Changing the 10k to 100k above to a eg 2k2 base to emitter will divide the drive voltage so that you need more negative gping voltage before the transistor turns on. Design to suit.

The solenoid MUST have a reverse diode across it if DC operated to minimise the reactive spike on turn off.

As shown your SSR has a TRIAC output stage. This may imply AC drive or may just be the symbol that you had available. If it is in fact an AC driven solenoid then you may or may not need to "snub" it to prevent reactive energy problems, as for DC.

• If the optocoupler is "zero crossing" switched at turn on and turn off you may not need a "snubber" although, with a highly reactive load current and voltage are nearly 90 degrees out of phase and turning off at a voltage zero crossing will still result in an inductive "kick".

• If the optocoupler is not zero crossing switched then you need a snubber to dissipate the reactive energy. You can't just use a reversed diode as it will conduct on every second half cycle (and there would be "trouble!" :-) ). . One method is to use opposing polarity zener diodes in series, with zener voltages at least slightly above the peak AC half wave voltage. eg for 110 VAC mains coil drive Vzener >= 110 x 1.414 >= 160- V. Say 18- V zeners. ( solenoid terminal 1 - zener 1 cathode, zener1 anode - zener2 anode, zener 2 cathode - solenoid terminal 2.

Powerup handling

While the above solves the problem as presented it MAY not solve the problem as it happens in generic real world cases, and just possibly not in this case either. Here we have allowed ourselves to be "taken captive" by what the processor does at startup, rather than designing what we want to happen. If the processor does not do quite what we want we are then forced to find some other workaround.

An alternative is to design a circuit that is guaranteed to start up properly. This may well not be quite the cheapest and simplest solution and whether it is worthwhile depends on application. I'll not go into any great detail as there are many ways this can be done.

One example only. An NPN driver is used. Transistor base is connected to an electrolytic cap via a diode (possibly a Schottky diode). The capacitor is also charged via a resistor from Vcc. At turn on the capacitor is at 0 V so the transistor base is clamped low by the capacitor and the transistor cannot turn on. The capacitor is charged high by the resistor from Vcc and after a period of probably < 1 time constant Vcapacitor > 0.6V (ignoring the diode effect) and the capacitor has no more effect on operation. What happens at power down needs to be determined.

Various options can be imagined. eg the capacitor could instead be controlled by a "spare" pin and a number of drivers controlled together in this manner. This could be used as a mass "driver disable" at any time at an incremental cost of one Schottky diode per driver.

Whether this or some other scheme is worthwhile is TBD in each case.

An alternative is to switch the supply which is used to control peripherals (such as the SSR in this case). This too could be done with pin control or by an RC delay. Lots of opportunities for thought and playing.

3.3V drive & other real world issues

The MOC3010 optocoupler LED requires 1.5V max forward voltage so 3.3V drive is OK.

Connect PNP emitter to 3V3 (if available) and drive as per above comments.

If there are miscellaneous blips on the input line before things settle down then one of my powerup ideas or something similar is required if false operations are to be avoided.

If response time is not an issue, or if a delay of say 0.5s is acceptable, then adding a "suitably sized" capacitor to the transistor base (whether PNP or NPN) will slow operation. This also slows release time so you need to be aware of overall effects.

Even if operation with no snubber seems to work Ok I would VERY STRONGLY recommend that a snubber be used, as "stuff happens" and sooner or later you are liable to hit peak mains voltage 'just so" and experience an inductive spike of considerable magnitude. In the absence of a snubber the TRIAC itself may decide to step up and dissipate the energy. Whether it does so destructively or effectively just turns on for another half cycle is TBD. This is not the sort of thing that you want your components determining on your behalf. A reliable design is one that is (at least) designed. If things may happen that you have not designed for then you can expect that the unexpected may happen.

I'm not sure why you think BJTs are significantly slower than power MOSFETs; that's certainly not an inherent characteristic. But there's nothing wrong with using FETs if that's what you prefer.

And MOSFET gates do indeed need significant amounts of current, especially if you want to switch them quickly, to charge and discharge the gate capacitance — sometimes up to a few amps! Your 10K gate resistors are going to significantly slow down your transitions. Normally, you'd use resistors of just 100Ω or so in series with the gates, for stability.

If you really want fast switching, you should use special-purpose gate-driver ICs between the PWM output of the MCU and the power MOSFETs. For example, International Rectifier has a wide range of driver chips, and there are versions that handle the details of the high-side drive for the P-channel FETs for you.

Additional:

How fast do you want the FETs to switch? Each time one switches on or off, it's going to dissipate a pulse of energy during the transition, and the shorter you can make this, the better. This pulse, multiplied by the PWM cycle frequency, is one component of the average power the FET needs to dissipate — often the dominant component. Other components include the on-state power (I_D² × R_DS(ON) multiplied by the PWM duty cycle) and any energy dumped into the body diode in the off state.

One simple way to model the switching losses is to assume that the instantaneous power is roughly a triangular waveform whose peak is (V_CC/2)×(I_D/2) and whose base is equal to the transition time T_RISE or T_FALL. The area of these two triangles is the total switching energy dissipated during each full PWM cycle: (T_RISE + T_FALL) × V_CC × I_D / 8. Multiply this by the PWM cycle frequency to get the average switching-loss power.

The main thing that dominates the rise and fall times is how fast you can move the gate charge on and off the gate of the MOSFET. A typical medium-size MOSFET might have a total gate charge on the order of 50-100 nC. If you want to move that charge in, say, 1 µs, you need a gate driver capable of at least 50-100 mA. If you want it to switch twice as fast, you need twice the current.

If we plug in all the numbers for your design, we get: 12V × 3A × 2µs / 8 × 32kHz = 0.288 W (per MOSFET). If we assume R_DS(ON) of 20mΩ and a duty cycle of 50%, then the I²R losses will be 3A² × 0.02Ω × 0.5 = 90 mW (again, per MOSFET). Together, the two active FETs at any given moment are going to be dissipating about 2/3 watt of power because of the switching.

Ultimately, it's a tradeoff between how efficient you want the circuit to be and how much effort you want to put into optimizing it.