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I have a 3/4" tubular pull-type solenoid from Magnetic Sensor Systems, model number S-15-75-24. The datasheet can be found here: http://www.magneticsensorsystems.com/solenoid/tubular/s-15-75.asp

I need help understanding what I am looking at so I can either use a different solenoid I have from this company, or use the knowledge gained when acquiring another. I have no background in electronics, and at this point just trying to understand how to properly actuate the solenoid.

From what I can tell, the solenoid can take 9.2 volts maximum based on the model number and the "24" being representative of 24 AWG wire. When applying a Duracell 9 V battery (this exact model https://www.duracell.com/wp-content/uploads/2020/02/9V-Duracell-Plus.pdf) the solenoid will not actuate like I would expect a solenoid to do; however, pulling forces are present when the battery is present and the plunger will not fall out until the battery is removed.

From what I understand by reading other posts on this forum about solenoids is that the amperage of a 9 V battery is not enough to drive this solenoid to its fullest potential. The problem is that I do not understand how to read the datasheet in order to purchase a battery capable of actuating the solenoid, if multiple 9 V batteries in parallel is needed, or if batteries will even work.

If someone can explain what I'm looking at on the datasheet from MSS (provided above), it will allow me to have a better understanding when dealing with solenoids, whether it be from the same brand which I have multiple different models, or from different brands entirely; and how to properly go about using solenoids in safe and efficient manner when applying different power sources.

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The datasheet shows a resistance of 1.05Ω, which is indeed too low for a standard PP3 (9V) battery. It would have to deliver I = V/R = 9V/1.05Ω = 8.57A.

A series 4-cell lead-acid or 3-cell lithium-ion battery would seem to be more suitable, or a 10A+ bench supply. (Use lithum-ion batteries with care; I do not suggest using one without a battery management system to control charging and min/max voltage.)

Note that the solenoid dissipates 9V * 8.57A = 77W while energized at this level, which will heat it up quite quickly. Keep on-times short and infrequent.

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  • \$\begingroup\$ Looking at Energizer lithium battery data.energizer.com/pdfs/l522-1119.pdf the battery has 1000 mA. My understanding is that 1000 mA is 1 Amp. I assume 8.57A is needed to power the solenoid, but how does it draw 8.57 Amps from the battery when the battery can only provide 1 Amp. Or is it that the lithium battery is the best of bad options if using 9v batteries, while only using roughly 11.6% (1 / 8.57) of the solenoids potential? \$\endgroup\$ Commented Aug 1, 2023 at 3:01
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    \$\begingroup\$ That's the point: the battery voltage won't be nominal, and current will be well beyond the rating. The actual value will be somewhere inbetween (perhaps 4V and 4A), but that datasheet doesn't provide enough information to determine what exactly it will deliver. Anyway, it won't do it for very long, and may catch fire in the process. You need much bigger batteries. \$\endgroup\$ Commented Aug 1, 2023 at 3:05
  • \$\begingroup\$ Thank you. One last question, You mentioned using lithium batteries with a battery management system. Is that only when charging the batteries, or in the theoretical circuit while the battery is in use to prevent too much power being drawn and causing a fire? \$\endgroup\$ Commented Aug 1, 2023 at 3:17
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    \$\begingroup\$ Both, yes. Lithium ion is dangerous if over-discharged. \$\endgroup\$ Commented Aug 1, 2023 at 3:56

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