1N4733 is a 5.1V zener diode. I had to look that up. There's a proper symbol for a zener diode, different from a Schottky diode. That's a good start, but you still need to limit the current to within the range the 4N35 diode will handle. And you need to watch for wattage ratings as well. I'm no expert on zener's by far. But I think you're reducing the voltage to the 4N35 by 5.1 volts. That is to say that 14V - 5.1V = 8.9V. And at 36V your 4N35 will see 30.9V.I just did more testing - mu Multimeter shows the voltage briefly dropping down to about 9 volts for a brief moment when the motor changes speed. It's hard to know for sure as the multimeter takes a bit to show a voltage stop.
I've drawn my circuit as it currently stands and added in the capacitors as I understand your instructions:
1) is this what you meant by across the resistors ?
2) Given I'm only using .39K resistors - would I still use a 1uf capacitor ?
The resistors at the GPIO end are as Numato instructed.
Thanks
Kent.View attachment 226334
Looking at A data sheet, the 1N4733 is a 5.1V Zener with a max current of 178mA. So your limiting resistor will need to dissipate, as you said, up to 36VDC. So 36 ÷ 0.178 = 202Ω at 6.4W. Your zener won't survive that much power. You can still get 14 to 36 volts dropped down to 5.1V using a much higher value resistor. At 2KΩ, (36 - 5.1) ÷ 2KΩ = 15.5 mA @ 556mW. And that's all just to protect the zener. 556mW is still going to blow the 4N35 diode. So you need a resistor on that line as well. (5V - 1.2Vf{4n35}) ÷ 10mA = 380Ω. However, since this is also subject to a reverse voltage, you still need the 1N4004 diode to protect against reverse voltage because the 4N35 reverse current is only 10µA (0.000,01A).
Now - remember; I said I'm not the expert on zener's so I'll let someone else point out the problems I may be introducing. The numbers I took were from generic data sheets just to get an idea of what you're looking at.
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