Hi guys, i want to ask how to reduce the heatsink produced by the push-pull solenoid ( let's just say i'm gonna use this kind of product : https://www.adafruit.com/products/413 )
Or you can see below :


And it says that it will work with 24V supply and will have current about 250 mA. And if i'm not wrong with my calculation, the power dissipation would be around let's just say 24V*250mA = around 6 watt. And i think it would be hot if i'm gonna control it with my microcontroller.

What should i do to decrease the heat produced by solenoid without using a heatsink ?
Any idea ?

Oh yeah, and i want to ask, how to maintain the current if the solenoid it's in normally open (NO), and i want to close the door by giving it logic high, it will pull the current (250 mA or even more) all the time when i closed my door. How to make it more efficient without pulling the current all the time ?

Thanks

 

You can't reduce the heat caused by 6 watts but you can reduce the temperature.
Blow a fan on it. Is that better than using a heatsink?
Redesign the system so you can use a solenoid that is normally off.
Use a solenoid that doesn't need a heatsink to survive.
Power it and find out how hot it gets.

• 24 DC operation (you can use 9-24 DC volts, but lower voltage results in weaker/slower operation)

Read the instructions and see that you can use 9V.

 

Use a bi-level drive.

One pulse of 24 volts to pull in the armature, then switch to a lower 'holding' voltage after the solenoid pulls in.
This can also be done with PWM, 100% duty cycle to pull in, then drop down to lower duty to hold.
Some pinball flipper solenoids actually have two windings, pull and hold, the hold winding having much higher resistance.

This actually works well because the magnetic circuit is closed when the plunger is inside, this makes a high holding force possible with lower current.

Solenoids are terrible power hogs anyway, consider a motorized actuator if power is a concern.

 

A cheap car alarm with a magnetic lock had a solenoid with a PTC in series. Ok I guess if it was only operated infrequently to allow the ptc to cool.

 

A cheap car alarm with a magnetic lock had a solenoid with a PTC in series. Ok I guess if it was only operated infrequently to allow the ptc to cool.

I worked on a car with the lock switch stuck, "on". The PTC saved the car from burning. It cut the continuous current by enough for the battery to last 3 days. This seems like a good solution compared to trying to fit a fuse that will allow the locks to work and still protect the wiring from a stuck or shorted lock circuit.

 

You can't reduce the heat caused by 6 watts but you can reduce the temperature.
Blow a fan on it. Is that better than using a heatsink?
Redesign the system so you can use a solenoid that is normally off.
Use a solenoid that doesn't need a heatsink to survive.
Power it and find out how hot it gets.

• 24 DC operation (you can use 9-24 DC volts, but lower voltage results in weaker/slower operation)

Read the instructions and see that you can use 9V.

Yeah, i agree with you sir. I think it's better to use heatsink and combine it with sir Sensacell idea.
And about the redesign the system that worked in normally off, i'm afraid, that in my town here, we can't predict when will the main electrical shut-off. So if from the power plant got an electrical failure, my town would be totally dark (just home without a genset generator). And if it happened, because of the normally off, the solenoid would be in closed state and we'll be locked inside. But that's a a good idea to make it more efficient.
Thank you sir for your information.

Use a bi-level drive.

One pulse of 24 volts to pull in the armature, then switch to a lower 'holding' voltage after the solenoid pulls in.
This can also be done with PWM, 100% duty cycle to pull in, then drop down to lower duty to hold.
Some pinball flipper solenoids actually have two windings, pull and hold, the hold winding having much higher resistance.

This actually works well because the magnetic circuit is closed when the plunger is inside, this makes a high holding force possible with lower current.

Solenoids are terrible power hogs anyway, consider a motorized actuator if power is a concern.

Thank you sir for your information. It really helps a lot !

 

One way to get two level drive is to put a resistor in series with the solenoid and then an appropriately sized capacitor across the resistor. Did this once on some stepper motors and cut total power in half without loosing responsiveness.

 

Find a solenoid rated for continuous duty, they are made to take the heat. Add a bell crank to the lock mechanism, so the lock is open when the solenoid is "off" instead of on.

 

Expanding on Dick's post, use a capacitor and resistor in series with the coil, choose a resistor of half the coil resistance or slightly more ....

 

One way to get two level drive is to put a resistor in series with the solenoid and then an appropriately sized capacitor across the resistor. Did this once on some stepper motors and cut total power in half without loosing responsiveness.

Expanding on Dick's post, use a capacitor and resistor in series with the coil, choose a resistor of half the coil resistance or slightly more ....

View attachment 102909

Thanks sir for both of your idea. But i'm wondering, how does the capacitor in parallel with resistor and both of them are in series with the solenoid, cut the total power ? Let's just say when the switch turned on, the current will go from supply through the solenoid --> switch (let's just say mosfet logic level or BJT) --> charge the capacitor. And then, if the switch is turned off, the capacitor will discharge and the current will go through the resistor that parallel with the capacitor. How can this cut the total power ?

Find a solenoid rated for continuous duty, they are made to take the heat. Add a bell crank to the lock mechanism, so the lock is open when the solenoid is "off" instead of on.

Wogh, that was a nice idea sir. Thank you for your information !

 

When the solenoid is powered up , the capacitor will be discharged and will put the maximum current into the coil, then the capacitor charges up and the resistor takes the current down to its minimum holding level, this will reduce the heat/power in the coil.

 

Hi guys, i want to ask how to reduce the heatsink produced by the push-pull solenoid ( let's just say i'm gonna use this kind of product : https://www.adafruit.com/products/413 )
Or you can see below :
View attachment 102869

And it says that it will work with 24V supply and will have current about 250 mA. And if i'm not wrong with my calculation, the power dissipation would be around let's just say 24V*250mA = around 6 watt. And i think it would be hot if i'm gonna control it with my microcontroller.

What should i do to decrease the heat produced by solenoid without using a heatsink ?
Any idea ?

Oh yeah, and i want to ask, how to maintain the current if the solenoid it's in normally open (NO), and i want to close the door by giving it logic high, it will pull the current (250 mA or even more) all the time when i closed my door. How to make it more efficient without pulling the current all the time ?

Thanks

There are various circuits floating about the web, the solenoid needs most of its rated voltage to pull in, but will hold on a lower voltage once in.

The circuits are usually a one shot voltage doubler type thing. A capacitor is charged to Vcc in the resting state, various isolating diodes allow a transistor to move that capacitor temporarily from parallel to Vcc, to in series to feed the solenoid.

 

When the solenoid is powered up , the capacitor will be discharged and will put the maximum current into the coil, then the capacitor charges up and the resistor takes the current down to its minimum holding level, this will reduce the heat/power in the coil.

How do we know sir if the solenoid is powered up and the capacitor will be discharged 'at the same time' ? Because solenoid works like a capacitor isn't it ? The differences between capacitor and inductor is that, capacitor keeps the voltage while the inductor maintain the current. Both of them are exponentially being charged ( capacitor by the voltage, and inductor by the current, as we can see from the graph ( i got it from the internet http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html ):

And my thought would be like this sir :
This is when the switch turned ON :

And this one when the switch turned off :


Am i correct sir ?
Thank you by the way for replying my question. It really helps me a lot to understand these kind of things

There are various circuits floating about the web, the solenoid needs most of its rated voltage to pull in, but will hold on a lower voltage once in.

The circuits are usually a one shot voltage doubler type thing. A capacitor is charged to Vcc in the resting state, various isolating diodes allow a transistor to move that capacitor temporarily from parallel to Vcc, to in series to feed the solenoid.

Is this the circuit i found it from internet that you mean sir ?


Thank you sir for your information

 

Your pictures are correct but there another function. After the capacitor is charged the solenoid continues to get current through the resistor. That is how the solenoid stays on at low power.

 

Your pictures are correct but there another function. After the capacitor is charged the solenoid continues to get current through the resistor. That is how the solenoid stays on at low power.

You mean state like this sir :



When it has the same polarity ( solenoid and capacitor ) ?
Thank you sir for your help. I will learn a lot from it

 

Yes.
Yes.....

 

Yes.
Yes.....

Thank you sir for the knowledge ^^

 

Hi guys, i want to ask how to reduce the heatsink produced by the push-pull solenoid ( let's just say i'm gonna use this kind of product : https://www.adafruit.com/products/413 )
Or you can see below :
View attachment 102869

And it says that it will work with 24V supply and will have current about 250 mA. And if i'm not wrong with my calculation, the power dissipation would be around let's just say 24V*250mA = around 6 watt. And i think it would be hot if i'm gonna control it with my microcontroller.

What should i do to decrease the heat produced by solenoid without using a heatsink ?
Any idea ?

Oh yeah, and i want to ask, how to maintain the current if the solenoid it's in normally open (NO), and i want to close the door by giving it logic high, it will pull the current (250 mA or even more) all the time when i closed my door. How to make it more efficient without pulling the current all the time ?

Thanks

Just a quick note - that solenoid you linked to is now being sold in a 12V not 24V version. So, new ones will be 12v.

 

Hi guys, i want to ask how to reduce the heatsink produced by the push-pull solenoid ( let's just say i'm gonna use this kind of product : https://www.adafruit.com/products/413 )
Or you can see below :
View attachment 102869

And it says that it will work with 24V supply and will have current about 250 mA. And if i'm not wrong with my calculation, the power dissipation would be around let's just say 24V*250mA = around 6 watt. And i think it would be hot if i'm gonna control it with my microcontroller.

What should i do to decrease the heat produced by solenoid without using a heatsink ?
Any idea ?

Oh yeah, and i want to ask, how to maintain the current if the solenoid it's in normally open (NO), and i want to close the door by giving it logic high, it will pull the current (250 mA or even more) all the time when i closed my door. How to make it more efficient without pulling the current all the time ?

Thanks

I would try it in your application to see how much it will heat. Design your linkage for full stroke of the solenoid. Your power rating is the work available. It won’t all go to heat.

 

Would not the Texas Instrument DRV120 be an easy way to provide for high pull in and then low hold in for a solenoid?

It looks to me that it would be a simply way to reduce the heat by reducing the voltage being put through the solenoid.

The datasheet is here