I'm building a musical instrument using thirty JF-0530B solenoids, rated 6VDC and 300mA. But I've discovered when I power them directly with 6VDC they draw ~1.5A, because they only have a resistance of ~4 Ω (they all do, except for three that appear to be shorted/bad).
(The values above are as measured with my multimeter, like so – just to make sure I didn't mess something up here.)
My concern is that I'm going to damage the MIDI decoder board, which switches the solenoids via ULN2803 chips (Darlington arrays with embedded flyback diodes) that are only rated up to 500mA per output.
The odd thing is, before discovering all this I tested the solenoids with the decoder – and not only does it work at 6VDC, it works at 12VDC when the solenoid pulls nearly 3A! (I just used the wrong supply.) Maybe I just got lucky here because the solenoids are only powered for a fraction of a second.
I'm not familiar with solenoids, but isn't 4 Ω a low resistance for a 300mA-rated solenoid? Did I get a bad batch or do I misunderstand how to use them? I'm trying to ask vendors for the resistance of the 12VDC version of this solenoid (which I'm thinking of using instead) but no success yet. I've seen this thread (among others) but that seems to have boiled down to misuse of equipment. Another forum suggested I was seeing inrush current, but this seems to apply only to AC solenoids.
Thanks in advance for any thoughts!
Best Answer
I can only find data for the 12 V version of the JF-0530B solenoids. These appear to be the type used in vending machines, etc., where they typically have short "on" times. I suspect that the voltage and current ratings are not simultaneous: you can run at 6 V (or 12 V in the case of the datasheet version) for a while or at 300 mA continuously. For the latter condition the voltage will be reduced.
Assuming the frame size and power rating is the same then we can use the temperature rise versus time curves from that data sheet. If we use the rated voltage and your measured resistance then we can calculate the power dissipated in the coil: \$ P = \frac {v^2}{R} = \frac {6^2}{4} = 9 \; W \$. That would give you a temperature rise as shown by the curve 1. After 30 s continuous on you would have a coil temperature rise of 50°C.
Figure 1. Graphs from solenoid datasheet.
Some of these applications require a good pulse of current to overcome inertia, etc., but not so much to hold the solenoid in position for the time required. It is possible to charge a capacitor to 6 V through a resistor to give a good initial "kick" to the solenoid but limit the continuous current to a safe value.
simulate this circuit – Schematic created using CircuitLab
Figure 2. C1 charges up to 6 V via R1 and provides a short pulse current when Q1 turns on. The current will rapidly fall off to the value limited by R1 and the solenoid's internal resistance.
You need to read and understand the stroke details on the graph on the right too.
ULN2803
Yes, they're not adequate for this task. Even if you parallel the outputs you'll likely exceed the chip total current limit. (Remember that all the current has to pass though the internal gold wire to the common pin.) Look for a suitable Darlington transistor.
Ideally you would be able to share the one RC pair between all the solenoids but this depends on the RC time values selected as it would have to recharge adequately between firings.
You might also need to prevent the actuator damping the ringing. The piano mechanism does this by propelling the hammer to the string but it's allowed to fall back to a rest position if the note is held. This allows the string to vibrate undamped. You may succeed by just getting the pulse length right or you may need a separate hammer mechanism.