Hello all,
So to give a brief summary, I'm developing a control board for some of my milling machine's peripherals. The bit that has been giving me trouble is a solenoid valve for the misting system. I've gone through a few tests a few times now that seemed to work, and then ordered a PCB with the changes, but problems seem to keep coming up (hence I'm asking here now as I'm a bit tired of it).
The hardware
The output voltage of the GPIO on my mill's controller (MicroMill 2000HD/LE)seems to be 5mA @ 3.3V (I was told 5mA @ 5V by the manufacturer, but this turned out to be wrong. Waiting on further responses from them...). I found out that the parallel port pins of my mill PC was being passed directly through to this GPIO port, with a 1K pull-up to 5V (diagram). It seemed to happily source ~53mA when quickly shorted, but I presume somewhere in the range oif 10mA is safer.
The solenoid I am using operates at 12V DC (7W), consuming ~600mA.
The schematic
Here's the schematic. The OUT1 pin is 3.3V. On the right side, the optocoupler should control the solenoid connected to J9.
I decided to use an optocoupler (CPC1301GR) in order to control the solenoid with the GPIO, as this one has a low input control current. My choice of resistor (900ohm) should draw ~3.67mA. I am also aiming to isolate the peripherals (e.g. solenoid) from the rest of controller, hence another reason why I chose an optocoupler.
The results
Unfortunately, it's misbehaving (as you may have guessed). If I briefly toggle OUT1 (thus not fully saturating the solenoid coil), the optocoupler is able to 'shut off'. However, leaving OUT1 high for a few moments (thus saturating and actuating the solenoid valve), then setting OUT1 low, causes the optocoupler to remain conducting (across pins 3/4) until I remove power.
I am noticing now in the optocoupler datasheet in the "CTR vs. LED Current (IF)" graph, I should have a CTR at around 4000% from the 3.67mA that should be sourced from OUT1, thus resulting in ~147mA that can be drawn across the optocoupler pins 3/4 (if I have understood the datasheet correctly). However, I don't understand how the solenoid could be actuating with this little current (unless the optocoupler has become damaged at this point).
So obviously my design is bad, and I should redo this. So I am wondering, how should this circuit actually be properly designed (i.e. did I miss anything important?), or should I consider an alternative approach not using an optocoupler? If any more info is needed, I would be happy to provide.
Thanks for reading all this, and any help/advice is appreciated!
Some updates: I have updated the design to this using some advice on designing around parallel port data pins. The Microcap sim of the design with OUT1 high can be seen here as well.
From playing around with it so far, this looks like it will be a more reliable design.
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
Current limiting in the FAQ and Wiki:
https://www.reddit.com/r/AskElectronics/wiki/faq#wiki_power
https://www.reddit.com/r/AskElectronics/wiki/faq#wiki_current_limiting_resistors
https://www.reddit.com/r/AskElectronics/wiki/design/leds
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Can I power the solenoid from a separate 6–9 V supply and still use 3.3 V control?
What MOSFET works best to drive a 12 V solenoid from 3.3 V logic?
How do I orient the flyback diode across the solenoid?
I'm new to this electronics thing and am a bit stuck with trying to find the parts for a simple circuit.
I have a 3.3V DC line from my micro-controller that I'd like to use to switch a 12V DC 250mA solenoid valve. My understanding is that my micro-controller is not going to supply the current required for operating the valve, so I wanted to supply the power from a 12V Li-ion battery. But because these circuits don't share a common ground, I need to separate them? I was looking at a relay or opto-isolator for this, but I can't seem to find anything that looks appropriate for these currents. Speed is not important, but I'd like this to be pretty low-cost.
Am I on the right track here? Is there a more effective way for looking for components than asking google for a 3.3V to 12V opto-isolator? Is there a better way to do this voltage conversion?
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\
maximum with 2.5V drive and 0.075\
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.
Although it is stated that it is a "logic-level gate drive" where "Gate Threshold Voltage: 1<VGS<2" in the mosfet datasheet, mosfet is acting weirdly with a gate voltage of 3.3V.
It is specified for Vgs of 4V, but not 3V3.
Why does it take 10 seconds for the current to rise to 1.8 amps (Gate: 3.3v)? While working perfectly with a gate voltage of 5.0v.
Most MOSFETs have negative tempco on their threshold voltage, and positive tempco on RdsON.
My guess is that there isn't enough voltage on the gate to turn it on fully. You can confirm this by measuring Vds and Id, which will give you RdsON. If it is much higher than the specified datasheet RdsON, then it's not turned on enough. So the MOSFET works in linear mode, and as it heats up, its threshold voltage decreases, which means it turns on more, and current increases. If current increases over 10 seconds, you should feel the MOSFET get hotter.
You either need a MOSFET that will turn on with a Vgs below 3V3 (so, another MOSFET). If you have a 5V power supply on your board, you can use a 3V3 to 5V level converter/driver, for example a 74HCT logic gate, to drive this MOSFET with 5V.
I'm looking for a solenoid that I can control from an Arduino pro mini 3.3V. The solenoid would be used as a means to open a springloaded hinged box. I'd want that solenoid to pull in for a short time freeing the lid to open. I'm thinking the 'movable pin' part should be about an inch or two long and have at least 0.5 inch travel. Can this be powered by one of the io pins or will I need to use a relay to provide power? I know nothing about solenoids and their power consumption so perhaps what I'm planning is a bad idea.
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
Do you have an actual schematic of your circuit? The graphic you showed has the gate on the middle leg of the MOSFET's package, that isn't the case for the MOSFET whose datasheet you linked. Values of your pull-down resistors would be helpful too.
Without further information, I would say the threshold voltage is too large to be turned on by the 3.3V logic. The max threshold voltage of your transistor under tested conditions is 2V, this isn't a huge amount of leeway.
Measuring 12V between drain and source doesn't mean that the solenoid should be activated. If you measure across a 12V battery whose terminals are unconnected you will measure 12V. The transistor is not turned on so there is effectively an open circuit between the drain and source, you will measure 12V but not current will flow.
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.
Could you use an opamp to raise and lower S2 to control the flow of current through the solenoid?
Thats fine, there are a few problems. One is overvoltage from the solenoid actuating. If it simply to turn the solenoid on and off, then an opamp would be an expensive way to do this.
You can use hit and hold if circuits if you need different current/voltage levels: How do I use a hit and hold circuit with a solenoid?
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.
Definitely the second option: cheaper, smaller, requires a single battery charger. A DC-DC converter to power the MCU is smaller because the MCU takes less power than the solenoids. Also, the DC-DC converter that powers the MCU isolates from the MCU the voltage dips as a solenoid is turned on and the high voltage spikes as a solenoid is turned off. (In the first option, when you turn on and off a solenoid the MCU power supply is affected, which may reset it.)