Timeline for Driving a solenoid via a digital output
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| May 4, 2023 at 20:56 | review | Close votes | |||
| May 19, 2023 at 3:04 | |||||
| May 4, 2023 at 20:53 | comment | added | Troutdog | For a low-power circuit like this, there would be no difference, assuming the appropriate engineering was done in both solutions. High side and low side are both valid approaches. | |
| May 4, 2023 at 20:47 | comment | added | user2063586 | @Troutdog So I should expect no noticeable difference between a high-side driven solenoid-valve and a low-side driven solenoid-valve? The differences seem to be more in the implementation of the digital output MCU (sinking vs sourcing implementations). | |
| May 4, 2023 at 20:43 | comment | added | Troutdog | Any high side circuit has a low-side equivalent. Low side is often preferred due to the wide range of N-channel devices available. High side requires the use of a P-channel device or a charge pump for the gate driver. The flyback diode does the same thing in either case by dissipating the energy stored in the inductor when the switch (FET) is turned off (this also limits the voltage spike). | |
| May 4, 2023 at 20:39 | comment | added | user2063586 | It's hand drawn because I just combined the drawings from two different manuals and I figured it'd be easier to hand draw it, that's why there's dotted lines in the drawing. | |
| May 4, 2023 at 20:37 | comment | added | Andy aka | The first diagram is hand-drawn and unlikely to be from a manual for something. So, if you "have no idea what the actual MOSFET configuration is" then who made that drawing? So, given it can't work being driven from an MCU, the rest of your question about MOSFET protection is null and void. | |
| May 4, 2023 at 20:27 | comment | added | user2063586 | @Troutdog I have never done high-side switching on an inductive load, I'm use to low-side switching. I just realized that the flyback diode is the same thing as the clamping diode. I had though you would want different protections if you switch high-side vs low-side, but it looks like the flyback diode would have the same impact (e.g. clamping the output to GND). Is there any performance draw backs for switching between high-side and low-side? | |
| May 4, 2023 at 20:22 | comment | added | Troutdog | OK. I think I understand. The usual method to protect the FET is to place a flyback or catch diode across the inductive load (cathode towards the positive side). The diode will clamp the negative spike to one diode drop below ground in this case. It may be the case that there is protection inherent in the circuit already. A 33K resistor is too large to act as protection here. It may be there simply to drain all charge out of the circuit when off. | |
| May 4, 2023 at 20:15 | comment | added | user2063586 | @Troutdog Two separate pieces of hardware. | |
| May 4, 2023 at 20:14 | comment | added | Troutdog | @user2063586 What do the dotted lines signify in your drawing? | |
| May 4, 2023 at 20:10 | comment | added | user2063586 | @Andy aka I have no idea what the actual MOSFET configuration is -- that drawing is referencing a figure from the manual for the digital output, which I've attached in the second figure. The actual circuit does indeed supply 12V at above 25 mA, it just might be drawn wrong here and I have no way of knowing what the actual implementation is because I didn't design the digital output device. I'm just trying to demonstrate a switching MOSFET. | |
| May 4, 2023 at 20:04 | comment | added | Andy aka | Your driver circuit is fundamentally wrong in the way you have used your MOSFET. It's a source follower and you'll be lucky to get 2 volts out of the source when the MCU drives 3.3 volts. You might get 3.5 volts out if the MCU drive level is 5 volts so, there's no need for a 12 volt supply because, and trust me on this, you'll get nowhere near 12 volts at the source, not in a hundred years. There is absolutely no-chance that you can drive 25 mA into your solenoid with that circuit. | |
| May 4, 2023 at 19:42 | comment | added | Criticize SE actions means ban | I suppose the 20ms includes the reaction time of the mechanical parts of the solenoid, not just the electrical inductance, so it's probably lower than your estimate. (No problem, though) | |
| May 4, 2023 at 19:38 | answer | added | Criticize SE actions means ban | timeline score: 1 | |
| May 4, 2023 at 19:36 | comment | added | user2063586 | @Troutdog see above edits. I've attached the MCU diagram. There's really no electrical datasheet for the solenoid. I've just modeled it after the fact that it has a 20ms response time and requires 0.35W to operate. | |
| May 4, 2023 at 19:35 | history | edited | user2063586 | CC BY-SA 4.0 |
The schematic I've drawn up is just derived from the driver ciruict schematic. There's no datasheet on the solenoid -- only a brief section that describes the response time and power requirement.
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| May 4, 2023 at 19:24 | answer | added | rdtsc | timeline score: 1 | |
| May 4, 2023 at 19:08 | comment | added | Troutdog | Agree with previous comment. Do you have a datasheet for the solenoid? The problem with this arrangement is that the FET may not stay ON as the voltage across the load increases. | |
| May 4, 2023 at 19:02 | comment | added | periblepsis | Why are you using a source follower NFET here? Do you have a datasheet for the relay? (Just curious about it.) | |
| S May 4, 2023 at 18:57 | review | First questions | |||
| May 4, 2023 at 19:45 | |||||
| S May 4, 2023 at 18:57 | history | asked | user2063586 | CC BY-SA 4.0 |