How to Use Transistors to Switch Electromagnet Voltage from 12V to 9V?

I have a 12V battery connected to an electromagnet, and I need to be able to change this input voltage from 12V to 9V using a transistor. I was thinking of something that changes the voltage path to a more resistive one? I am just starting out with electrical engineering, so if someone could help clarify this problem it would be great!

What current does the solenoid draw?
If the battery is lead acid it is useable over a 10-14V range typically, so if you definitely need 12 to 9V you would have to use a switch mode power supply. If the solenoid will work satisfactorily at lower voltages, then none of the above will be an issue, and the simplest circuit is variable resistor, but since you have not specified a solenoid current, I have no way of knowing what power rating would be needed, it may well be impractical. Next option is an emitter follower. The output voltage can be adjusted with a variable resistor. You have to be aware of power dissipation in the pass transistor (again, you need to specify the solenoid current), and put a diode across the solenoid to clip the high reverse voltage flyback if the power is turned off, damaging your control circuit.
Now there are a number of problems with this simple circuit, which mean that additional parts are needed to overcome them. With a fixed load resistance, and no need for high precision (temperature stability etc) you would probably get away with it if the current is low enough.
Article here https://learnabout-electronics.org/PSU/psu22.php covers some of the basics.
Quite a bit to consider for such a simple request!
There are any number of integrated circuit solutions available as well, including monolithic if the current/power is low enough.
Cheers,
Richard

Hello
This is a logic solution, but you can project many similar.

If you want 12 volt:
- gate A: HIGH
- gate B: LOW
- gate C: LOW

If you want 9 volt:
- gate A: LOW
- gate B: HIGH
- gate C: HIGH

Cheers
Giovanni

Thanks for your detailed response Richard, but I've found an alternative way to approach this after consulting a few others. Thanks for bringing up the idea with diodes for backcurrent, very useful!

By the way, if the solenoid turn off time is critical, you need to allow the back emf to swing to a much higher voltage to reduce it, so an RC snubber or zener diode rather than a typically used silicon diode is necessary,
cheers,
Richard

i need some diagram. can anyone help?

HI, big_pablo

Here is a simple and efficent circuit i have designed for you.
try this and let me know the feed back good luck .

V.Sambath kumar

Sambath Kumar - Can you upload your schematic? Thank you

Hi Mourizio di Paolo Emilio,
First of all thank you very much for your feed back ,In my circuit d1 acts as a polarity protection mainly for the ic and the in put current is also limited to 6amps by d1. d3 protects Q1 and d4 protects the relay driver transistor from back emf.the max rly can go ony a max 120ma to 150ma not more than that.
so you can use the same 9v for some other ckt also. To avoid chattering of the relay a 100mfd 40v cap should be used parallel to the rly coil since high magnetic field if produced across the rly coil that is why i mentioned 40v rating for the cap min value of the cap should be 47 mfd 40v.if you are using a rly driver transistor use a 100mfd 25v cap from the base of the transistor to ground thisalso helps to avoid chattering of the relay. i have allready uploaded theschematic earlier any way i will do it again.
Regards V.Sambath kumar.

Hi
let me remind you one more thing for d1 it is better to use a schotty diode because the forward drop will not be more than 0.4 volts .you can use MBR1545 .It is a 15a 45v silicon rectifier.in the schematic the voltage rating of 47 mfd cap is showen as 25v it should be 40v.