# 555 to Power a Solenoid

Discussion in 'General Electronics Discussion' started by Angus, Jul 5, 2014.

1. ### Angus

29
0
Jul 4, 2014
Hello,

I am using a 555 timer to open and close a 2W solenoid valve.

To give enough power to the small solenoid - I want to use a transistor as a switch (completely ON/ completely OFF) - so it takes its power of the main 9V (battery) supply.

Could somebody suggest a transistor and resistor I should use - coming out of PIN# 3 on the 555.

Many thanks - Angus

File size:
10.4 KB
Views:
115
2. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Hi Angus and welcome to Electronics Point

A 2W solenoid at 9V will draw 222 mA. So you should use a transistor rated for at least 300~400 mA. Depending on where you live, one of the following will be readily available:
Those are listed in order of decreasing current gain.

These all have a current gain of 100 or more, which corresponds to a base current of 2.22 mA, but to saturate the transistor you should feed several times this amount of current into the base; I suggest around 10 mA.

With a 9V supply, about 2V dropped in the 555's output (they don't pull high very well) and 0.7V base-emitter voltage, you'll have about 6.3V across the resistor. From Ohm's Law:
R = V / I
= 6.3 / 0.01
= 630 ohms

Closest lower preferred values would be 620 ohms or 560 ohms.

3. ### Harald KappModeratorModerator

11,813
2,752
Nov 17, 2011
Welcome to our forum, Angus.

2W at 9V = 220mA.
Any medium power transistor is suitable. You can use for example this selection guide.
A popular transistor is the 2N2222. Its DC gain at 20mA collector current is 40ish which translates to a base current of 5.6mA. Make this 8mA to ensure the transistor is well saturated.
The 555's output voltageg when powered from 9V and sourcing 8mA is worst case 9V-1.7V=7.3V (datasheet).
Base-Emiter voltage of the 2N2222 is ~0.9V at 200mA collector current.
This leaves a voltage drop across the base resistor of Vr=7.3V-0.9V=6.4V. 64V/8mA=800Ohm. 820 Ohm is a good choice from the E-series. Up to 1kOhm wil work, too (a bit less of base current is no problem since the 8mA are over-specified anyway).

A note on the side: The 555 is specified for 200mA output current. If you lessen the powr of the solenoid to 1.8W or less, you can drive it directly from the 555 without the need for the additional transistor.

4. ### Harald KappModeratorModerator

11,813
2,752
Nov 17, 2011
Kris, youbeat me by seconds.
To the op: Kris has assumes a DC current gain of 10 whereas I have used a very conservative alue of 40. This does not mean either of us is wrong. It is a matter of interprettaion of the datasheet since the gain for 200mA is not explicitly stated. And Ive been very conservative.

5. ### Angus

29
0
Jul 4, 2014
Hello - thankyou very much for your quick replies - very impressive - I spent hours last night trying to make sense of transistor data sheets - and now I've cut to the chase very quickly!

So thanks again for all your expertise.

Angus

6. ### Angus

29
0
Jul 4, 2014
Also - would it be advisable to put a diode on the outlet of the 555 to protect it, when the solenoid de-energises. Thanks.

7. ### kpatz

329
88
Feb 24, 2014
Put the diode across the solenoid coil, reverse biased (cathode on + side of coil and anode on - side).

8. ### Angus

29
0
Jul 4, 2014
Yes - will do thanks

9. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
You already have the diode across the solenoid in your diagram in post #1.

10. ### Angus

29
0
Jul 4, 2014
True - was just wondering if it should be directly of PIN#3? But all good now - I'll keep the diode around the solenoid. Thanks.