The voltage rating shouldn't be a concern to you because you are using only 9V. Look that it can conduct 4A safely and what is its gain value (hfe) at this current. You need to know that because the PIC will supply a base current of Ic/hfe, so it has to be able to supply this base current. It would be better to find a darlington one which will have a much greater hfe than the others. Have a look in ebay, there are many. Also, you can consider a MOSFET like this one here:

http://www.nteinc.com/specs/2900to2999/pdf/nte2985.pdf

 

Thanks to all of you, I managed to get my solenoid working.

However I have noticed one funny phenomenon.

My 9V battery pack which consist of 6 AA batteries sometimes can get very hot suddenly. If I have been using it for a long time, maybe it is normal.

But I can feel the heat just 3 minutes after I have inserted the 6 batteries, why is it so??

Is the battery pack spoilt? ( This is what I suspect, but I only have one, so can't confirm yet)

FYI, it is working normally now. The "heating phenomenon" occurred 2 times already)

Pls advise.

 

Dear all,

I am still confused. How do we calculate the value of the resistor that limits the base current?
As Audioguru mentioned,
"The BD679 darlington transistor has a 2.5V max voltage loss when its collector current is 1.5A and its base current is 30mA.The 1k resistor in the circuit limits the base current to only about 2.9mA. Reduce the value of the 1k resistor to 120 ohms for a base current of 20mA then any BD679 will work".
How do we calculate the base current when the value of the resistor is 1k? why is the base current 2.9mA?
Why is that if we change the value of the resistor to 120 Ohms, the base current is 20mA??
Why is it that any BD679 will work? Are there many kinds of BD679?
Kindly advise

Thanks and Regards,

 

Dear all,

I have measured the current for actuating the solenoid actuator, and the measured current is only 550mA. This is very different from what I have calculated in the previous post, where the voltage 9V is divided by the resistance of the solenoid actuator which is 8 Ohms, and the calculated current is 1.125A.

May I know which one is correct?

Kindly advise

Thanks and Regards,

 

Dear all,

I am now working on my final year project. Now I have a problem, I need to activate a 9V solenoid from my microcontroller's (Basic Stamp 2) pin which supplies only 5V. How can I do it?

One idea is to connect the pin to a transistor, so the transistor can turn on the solenoid for me. Will it work? If it works, what kind of transistor can I buy?

hi ,
my advise is you use "IRF640" insted of trasistor.because the current carrying is very high in IRF.use 100E resistor in gate to drive direct from microcontroller

Thanks .

 

hi
My idea is to use IRF640.It is a mosFET ,it is current carrying is high.Use one 100ohm resistor in gate of MOSFET to drive direct from microcontroller

 

the easiest way is to use a solid state relay like R1-1A0550.. then connect the relay to the microcontroller's pin and the to ends of the 9V will be controlled by the relay...

I've used this in one of my projects..

 

Dear all,

Actually I have succeeded in using the BD679 to control my solenoid actuator. Thanks for the alternate suggestions as well.

However I am quite confused as to how to calculate the value of the resistor to limit the base current, and why is that resistor needed?

Also, when I measure the current needed to activate the solenoid actuator with a 9V battery cell, my reading is 550 mA. But when I use calculation, that is, (voltage = 9V)/ (resistance= 8 Ohms), it is 1.125A. So actually what is the current needed to activate my solenoid actuator?

Kindly advise

Thanks and Regards,

 

Hello,

You will need a higher current when powering the solenoid.
This current will fall very soon to the hold current.
This higher current is called "inrush current".
See this page for more info on solenoids and actuators.
http://homepages.which.net/~paul.hills/Solenoids/SolenoidsBody.html

Your calculation of the current will be work as you do not take the Vce in account.
The current will be (Vcc-Vce)/Rcoil => (9-1)/8 =1 A
(Vce is taken from the datasheet)

Greetings,
Bertus

 

Hi, thanks alot for your reply,
I have some doubts which I hope you can help me.
you said
"You will need a higher current when powering the solenoid."
So you mean the current that I measured to be 550 mA is actually not the current to power up the solenoid? Then what current is this suppose to be?
I have used a transistor BD678 which is capable of supplying collector current of 4A, so I think it is ok.

"This current will fall very soon to the hold current."
May I know what is the hold current?

Kindly advise.

Thanks and Regards

 

Hello,

Here is a quote from the page I gave you:

With the plunger fully out, the inductance is low, and so the current will be quite high. As the plunger moves in, the inductance increases, and the current falls. This high initial current is called the "inrush current" since it only last for a short time until the plunger is fully in. The inrush current is useful, because it allows a large current to start with which generates a large force to get the plunger going. Once the plunger has pulled in, less force is required to just hold it there, and conveniently, there is less force because the current is now lower.

Greetings,
Bertus

 

Hi Bertus,

Thanks alot!