I also added a buffer to ease the microcontroller digital output pins. Each pin will control a set of 8 solenoids.
I could see that the out pin drives only one optocoupler; but why do you need the buffer?
You need a latch or a hysteresis circuit because (i) it is imp to prevent the relays from chattering (ii) and a way for the micro to know the current state of the relay.
You don't really need D2 but if the solenoids are widely spaced it would be advisable to connect a diode (like D1) across each individually. If the diode is close to the inductance of the solenoid it will create a short loop for the back-EMF as it switches off. If you use just one diode, there is a risk of spikes radiating from the longer wiring. I think I would add a capacitor across the 12V supply at the solenoid end to help it stay stable as the load changes, maybe 100uF or so. Otherwise it looks OK.
Brian.
1. One micro out pin can happily drive one LED: i guess opto can run even at 5mA. The out pin of the micro does not care what happens after the opto...
2. The opto-coupler provides isolation but not much of a noise immunity. I would have put the opto closer to the micro because the micro side of the opto is the low voltage, noise sensitive, high imped comp. On the right side of the opto the volt and curr are high and imped is low and noise sensitivity is low. But exerts can disagree...
3. Suppose the software on the micro gives some error (hypothetical) and the relay is pulsed 10 times a second; within 10 sec, all the relays can go bad (coils burned) but it will be good to see that the relays are not turned on/off within 1 sec.
4. After a warm boot, the micro should know whether the relays are on or off. As you have too many of them, it is good to know the status of the driver FET. You may need another opto but that may an wise investment. It is good to know whether the FET is on or off.
5. A good driver circuit will act like a toggle; one pulse to turn off and next one to turn on. Two pulses within 1 sec and the sec pulse is ignored. Another pin tells whether the driver FET is on or off.
I smell over-engineering.
You need a single fly-back diode near MOSFET and bypass capacitor. You want to protect the MOSFET, not the solenoid valves. As an alternative to bypass capacitor, a TVS diode across the supply could absorb overvoltage produced by the cable inductance.
Hysteresis avoids switching losses with slow control voltage edges. The MOSFET has however some thermal capacitance so that it's probably not required under usual conditions.
In some cases, shorted solenoid cables or shorted coils may be a possible scenario. If so, you should consider either a fuse or a fully protected smart switch.
Hi,
some calculations:
4.5W at 12V gives 375mA.
375mA x 8 valves gives 3.0A of current.
Let´s accept a 1V drop on the wiring over the 25m.
This means R_wiring = V_drop / current = 1V / 3A = 0.333 Ohms
1m of copper wire with 1mm^2 is about 1/56 Ohms.
A = 2 x 25m / ( 0,33 Ohms x 56) = 2.7 mm^2
--> use a 2 x 4 mm^2 cable.
*****
Power dissipation.
Your chart shows about 40W peak.
But for your swithcing time (about 40us) the average power is about 10W.
Now if you switch ON 100 times per second you may expect 2 x 100Hz x 10W x 40us = 80mW average switching power dissipation. For this you don´t need a heatsink.
From the SOA chart you can see that the MOSFET can withstand 50A x 50V which is 2500W for a 100us signle pulse.
The given 88W maximum power is not the peak power for a short time, but the maximum continous power dissipation.
And with 80mW average power dissipation you can almost see the 100Hz switching pulse like a single pulse. (At least with your peak power dissipation).
*****
Independent if you use the switch close to the valves or close to the microcontroller. (when we consider 100 duty cycle)
Thus - for ease of installation - I recommend to install the switch close to the microcontroller.
use a "logic level MOSFET", a series resistor to the gate, a pulldown resistor to the gate.
No other parts needed to control the MOSFET. No bjt, no optocoupler. (but you need to take care about power path wiring)
You need to protect the MOSFET. As FvM said: One diode close to the MOSFET is sufficient.
I recommend to install an electrolytics bulk capacitor and a fast ceramics capacitor at the 12V node close to the MOSFET.
Klaus
No problem with this.I really wanna isolate tne uC pins and use IRF530. So I kept the opto.
Looks better now.I took all the PSUs and the circuits to the left side.
At the far end there is only solenoid-valves now.
I added TVS at vthe far end.
I didnt get where I should place the diode.
Hi,
No problem with this.
Looks better now.
* Put D1 where now TVS1 is as freewheeling diode. At switch OFF it prevents the energy stored in the solenoids to be pushed back to the driver circuit.
Most of the stored energy will be dissipated in the solenoids.
The main remaining problem now is the stray inductance of the 25m cable. It will cause a fast high voltage peak at the MOSFET. This may kill the MOSFET.
Thus you need a fast high voltage protection. It needs to be installed close to the MOSFET. Whether you use a fast zener or a TVS is not that critical.
Only one is needed. In schematic the position of D3 is correct.. across drian and source. TVS1 is at the wrong place.
The energy stored in the stray inductance of the cable is low. Even at ideal fast turn OFF it will see max. 3A for some 100ns. If this is within the SOA of the protection device, then you are safe.
With real switching times the current is less.
No need for high current rating, no need to take care about heating.
Klaus
Can you really turn on/off these relays 100 times /sec? These are mechanical solenoid valves that control fluid under pressure. The relays may not respond (and the flow will not be controlled) at all at 100 Hz frequency...
With 100Hz I don't expect the valve to mechanically move, but one can activate the relay with 100% duty cycle, then -depending on valve type one may reduce duty cycle to 30% to reduce heating and safe power...
Good point.But should not C2 be in parallel to the zener?
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