I need to control 3 separate gas solenoid valves in my lab. A few relevant points for this project are:
Only 1 valve will ever be open at a time (controlled by the code).
All three solenoid valves are the same model.
All three will be powered by a 24 V, 3 A DC power supply.
I am attaching a drawing of the circuit I planned to make, a long with the two relevant data sheets. I believe the 1n4001 can only handle 1 amp of current, but I'm actually not too sure what current I should be targeting for this diode (I realize the solenoid will use 2 A, but I'm not sure what current I need to rate the 'flyback' diode at). The IRLZ34N can handle 68 W, and I'll be at ~ 48 W, so I think this should be OK.
Anyway, I just wanted to check if anyone on this forum thinks there is a major issue with this circuit before I go building it.
TwoChain:
The IRLZ34N can handle 68 W, and I'll be at ~ 48 W, so I think this should be OK.
Here you're mixing up things.
The IRLZ34N is rated to dissipate up to 68W (that's going to require some interesting heat sinking to deal with), which is totally unrelated to how much power it can switch: it can handle up to 30A continuous, and a maximum voltage of 55V (that'd mean it could switch 1,650W).
Connect Arduino GND directly to the power supply GND (that's probably done through your buck converter already), and separate wires to your solenoids. That helps keeping things stable.
For diodes, you indeed may need to look for a higher rated one.
Grumpy_Mike:
I would put the FET gate pulldown on the Arduino pin. As you have it, you get a potential divider action reducing the drive voltage.
Reducing 5V drive to 4.93V is completely not worth worrying about. Logic level FETs are rated for 4.5V of drive,
you can make the pull-down 100k if you want.
Grumpy_Mike:
I would put the FET gate pulldown on the Arduino pin. As you have it, you get a potential divider action reducing the drive voltage.
As a rule of thumb the fly back diode needs to be rated at the same current that the solenoid is.
Thanks for pointing this out. I had completely overlooked it. While I understand what Mark is saying about this having minimal impact on the functionality, I will revise it as you suggested .
The IRLZ34N is rated to dissipate up to 68W (that's going to require some interesting heat sinking to deal with), which is totally unrelated to how much power it can switch: it can handle up to 30A continuous, and a maximum voltage of 55V (that'd mean it could switch 1,650W).
Connect Arduino GND directly to the power supply GND (that's probably done through your buck converter already), and separate wires to your solenoids. That helps keeping things stable.
For diodes, you indeed may need to look for a higher rated one.
Thanks for your message. I am a bit confused about two points (well... probably many more...!).
You say that I am mixing up the rating of the MOSFET, and I'm quite sure you're right (I seem to have a lot of confusion regarding these datasheets). However, I don't see how I could switch 30A at 55 V. Based on a previous discussion I had on this board (Power dissipation on MOSFET datasheet - General Electronics - Arduino Forum) , it was my understanding that 30A OR 55V are the max. You cannot have both together. If I were to switch a circuit operating at 30A and 55V, how could the MOSFET handle 1650 W? It's max rating is only 68 W. In my circuit, I'm switching 24 V and 2 A, so 48 W, which I would think is OK.
What do you mean by "separate wires to my solenoid"? In the circuit, each solenoid does have it's own wire, right?
For my diode, I will use 1N5400 to handle the 2 A.
Your circuit looks good as is but following are a few observations and suggestions.
Since the avalanche rating of your FET is 16A, 8 times the solenoid current rating you do not need the diodes. If you do the calculation you will have about a 0.07 Volt drop across the Source Drain of the Fet with a 2A load. Multiply that by 2 (current of the solenoide) by the voltage drop (0.07) and you get about 0.14 watts of heat to dissipate. Since the junction to case is 2.2C/W it will barely get warm.
I would recommend changing R4 R5 and R6 to the 50 Ohm range or less. This will help the enhancement of the FET and keep it cooler. With your existing circuit assuming 5V at the output of the micro you will get about 4.926v Gate to Source minus the ground lift voltage. Unless you are in a very warm area no heat sink is needed.
Expect a turn on glitch when the micro turns on as the default is to tristate the outputs which will let them drift high. The pull down resistors (R1 R2 R3) need to compensate for the leakage. I would use 4.7K but what you have is adequate. The first thing I do in my code is set the outputs and never assume what they will be.
However, I don't see how I could switch 30A at 55 V.
Well you are right in practice you can’t.
You see FET data sheets are a bit like tabloid headlines, they seem to promise more than they deliver. Each one vying for a bigger number than the competition.
By ignoring thermal conditions you can get a good power dissipation figure. If you look closely a lot of figures are for very narrow pulses.
Wvmarle knows this but is illustrating your mixups with an extremely example, to emphasis the point.
Of course running a part at it's limits is always a bad idea. I wouldn't go near 55V or 30A, especially the current as that's what gives you the power dissipation.
The on resistance of this FET is about 35 mOhm. At 30A the power dissipation would be 31W. That's a lot, hard to deal with for long.
Now let's be a bit more realistic, say 15A, 24V. That's well within spec and within reason for this part. Power dissipation would be some 8W, perfectly manageable with a proper heat sink. That's the power dissipation. You see this is far below the 68W rating.
Don't mix that up with the power switched, which is a lot higher: 360W.
gilshultz:
Your circuit looks good as is but following are a few observations and suggestions.
Since the avalanche rating of your FET is 16A, 8 times the solenoid current rating you do not need the diodes. If you do the calculation you will have about a 0.07 Volt drop across the Source Drain of the Fet with a 2A load. Multiply that by 2 (current of the solenoide) by the voltage drop (0.07) and you get about 0.14 watts of heat to dissipate. Since the junction to case is 2.2C/W it will barely get warm.
I would recommend changing R4 R5 and R6 to the 50 Ohm range or less. This will help the enhancement of the FET and keep it cooler. With your existing circuit assuming 5V at the output of the micro you will get about 4.926v Gate to Source minus the ground lift voltage. Unless you are in a very warm area no heat sink is needed.
Expect a turn on glitch when the micro turns on as the default is to tristate the outputs which will let them drift high. The pull down resistors (R1 R2 R3) need to compensate for the leakage. I would use 4.7K but what you have is adequate. The first thing I do in my code is set the outputs and never assume what they will be.
Good Luck and have fun!
Gil
Hi Gil,
Thanks for your very helpful comments. I will consider these points in my final design.
