I'm in need of some help with a circuit I've put together. I am brand new to electronics, just recently took it up as a hobby and am an extreme novice.
I have assembled a timer circuit using 555 timer chips. I have two monostabe circuits in series. The first is set for a one second pulse. When it finishes it triggers the second and that is set for a six second pulse. I have them both feeding the same output with diodes on the outputs to prevent backfeeding the opposite circuit. The whole thing is triggered by an infrared sensor. The trigger is through an npn transistor. The object is to have the pulse stay on as long as an object is detected and then six seconds after object is gone. The circuit works perfectly when I use a voltmeter or light bulb to test it. My problem is the the final output is to a solenoid. When I hook it up to the solenoid it does not work correctly. The solenoid keeps retriggering or fails to stop triggering when the object is gone. When I tested it with a voltmeter I get a residual voltage across the solenoid. The final output of the circuit is 12 volts and the residual volatage is approx. 9 volts. PS; I do have a protection diode installed across the solenoid. Do I somehow have to do something to drain the voltage left over in the coil. Anyone have any suggestions for a beginner.

 

Can you provide a schematic of the circuit you are using to facilitate this discussion?

hgmjr

 

What is the current rating of the solenoid, or the resistance of the coil?

A bipolar 555 can source/sink up to about 200mA of current with a 15v supply. CMOS versions are much less; an LMC555 can source 10mA, or sink 50mA.

Inductive loads such as solenoids and motors need reverse-EMF protection diodes, otherwise the output of the timer circuit will be subjected to voltages far in excess of what it was designed for. Which direction the diodes go depends upon whether you are supplying the solenoid with current, or ground.

If you are supplying the solenoid with current (sourcing), then connect the anode of a diode to pin 3 of the 555, and the cathode to the ungrounded terminal of the solenoid's coil. Additionally, place a diode across the terminals of the solenoid's coil, cathode to the ungrounded side.

If you are supplying the solenoid with a ground (sinking current) then the diodes are connected exactly the opposite of above. See the attached partial schematics (no timing diagram given)

Note also that if pin 5 is unused, it should have a small capacitor on it.

Note also that inductive loads place a heavy load on the supply, which may cause the Vcc of the timer to drop very low, re-setting the timer circuit. To prevent this, use a capacitor across the 555's supply pins (Vcc and ground). A minimum value to use is 0.1uF, but you may requre much higher, depending upon how long your wires are to your power supply and how heavy the load is.

 

Thanks for the response. In answer to Your question about the solenoid. It draws 130 milliamps at 12 volts.
I'm sourcing the solenoid, I checked and the diodes are installed correctly. I'm going to try Your suggestion and put a capacitor across the supply pins. How do I calculate the size of that? As I mentioned the solenoid draws 130 milliamps at 12 volts and the cord on my power supply is 6' long. Also in case it matters the supply is 12 volts, maximum 1000 milliamps for this project. Any further suggestions? Thanks again for the help.

 

I put a 10uF capacitor across the supply pins on the timer chip and it solved the problem perfectly. I took a gamble on the size, but figured that You saying that I may need more than the .1uF capacitor meant that a larger value wouldn't do any harm. Can't thank You enough for the advice.

 

Glad you got it fixed

10uF was a good guess for that length of wire, with just that load.

I'd recommend connecting a 100uF cap where the power connects to the board, and 0.1uF caps across the Vcc/Vdd and ground/Vss pins of each IC. This will help a great deal to keep "noise"/transients under control.

The shorter the leads are to your power supply, the smaller cap you can use for the "main" one. It's the combination of inductance and resistance of the wires between your power supply and the load, and the fact that your load was a heavy transient that caused the timer to reset. The 10uF cap absorbed much of the transient until the current flow in the wire could catch up, and re-charge the cap.