Solenoid valve (version 2)

[Roff]

I tried again and didn´t work.
There are two thinks i'm not doing:
1st: i'm not putting diodo zener
2nd: i don't know

There are no zeners. The diodes are Schottkys, but could be replaced by 1N4148s, now that I know how much current your solenoid requires. If you omit the diodes, you risk destroying your transistors. The diodes absorb the flyback voltage that will otherwise occur each time the solenoid current is interrupted.

 

[Roff]

I ran a simulation on the all-NPN version, and added some parts as a result. See notes with schematic (above) for details. The PNP/NPN version is probably preferable, unless there are no PNPs on your continent. :lol:

 

[kybert]

Hi

Please dont see this as hijacking the post....

Im doing the exact same thing, except my solonoid is 5V, 2.2W (0.45A). Im going to use the NPN/PNP method unless mosfet would be cheeper?

What transistors and diodes would i need?

 

[Roff]

Hi

Please dont see this as hijacking the post....

Im doing the exact same thing, except my solonoid is 5V, 2.2W (0.45A). Im going to use the NPN/PNP method unless mosfet would be cheeper?

What transistors and diodes would i need?

You are driving a latching solenoid also?
I would use the MOSFETs. The bipolars would require about 40ma of base current each, which you can't get from 4000 series logic. For MOSFETs, just choose some with Ron < 0.2 ohms. You should use Schottky snubber diodes.

 

[leonel]

??

At a few weeks ago i did your circuit (see attachement) and everything works ok till i notice one thing...
Every time i changed my voltage input for example 12V to 11V (simulating my battery discharge) an impulse comes associated in my "picture input", making my solenoid change state.
My "picture input" comes from a hef4538 output... (I already tried to isolate my 4538 output with transistors, RC circuits to ignore small signals, ...) and nothing...
I don't know what to do??!!

 

[Roff]

At a few weeks ago i did your circuit (see attachement) and everything works ok till i notice one thing...
Every time i changed my voltage input for example 12V to 11V (simulating my battery discharge) an impulse comes associated in my "picture input", making my solenoid change state.
My "picture input" comes from a hef4538 output... (I already tried to isolate my 4538 output with transistors, RC circuits to ignore small signals, ...) and nothing...
I don't know what to do??!!

How fast are you changing the voltage input? You should not change it more rapidly than you would expect your battery to discharge.

 

[leonel]

At the beginning i was turning off my Digital Lab (my voltage went from 12V to 0V very rapidly), and i thought as you... but know i'm changing with Digital Lab voltage regulator (more or less 0.3V in one second).

 

[Roff]

At the beginning i was turning off my Digital Lab (my voltage went from 12V to 0V very rapidly), and i thought as you... but know i'm changing with Digital Lab voltage regulator (more or less 0.3V in one second).

It will not be a problem at 0.3V/sec, and your battery should not discharge that rapidly anyway. Does your simulation show a problem?

 

[leonel]

Even if i drop my voltage rapidly i cannot have this problem...
Imagine if anybody remove one battery!? there are a voltage drop at least of 1.5V
Do you have any suggestion?

 

[G8RPI]

Mosfet alternative

Hi,
To save parts in Ron H's circuit and use bipolar transistors, you can use a dummy full H bridge (Ron's is a full H). Replace the top two transistors (M1 & M2) with resistors (D1 & D3 are not needed either). Ideally you should use a coil rated for half the supply and a resistor equal to the coil resistance, but as most coils will pull in at lower voltage and you are just pulseing the output, you can use a resistor of about 1/5th to 1/10th the coil resistance. With both lower transistors off, both ends of the coil are at 12V so no current flows. If one transistor is on current flows through the coil and resistor and through the other resistor.The circuit uses MUCH more current when it's on but it's on for such a short time it's average current is very low.
You can use Mosfet's or bipolar transitors with this design but the bipolars would need to be high gain (darlington) devices, don't forget to add a reisistor in series with the base. A little thought about the logic may allow the use of just the CD4070 as you have only two transistors to drive.

Robert G8RPI.

 

[Roff]

Even if i drop my voltage rapidly i cannot have this problem...
Imagine if anybody remove one battery!? there are a voltage drop at least of 1.5V
Do you have any suggestion?

If you have eight 1.5V cells in series to make 12V, and someone removes one of them, the voltage will not drop by 1.5V. It will go to zero. You should probably design it so this can't happen.
The CD4070 logic thresholds are more than 3 volts from the rails with a 12V supply, so a 1.5V change should not cause a problem. If you are REALLY worried about this, add two Schmitt triggers (CD40106) in series between C1 and the CD4070. Remember to connect all unused inputs to one of the power supply rails.

 

[leonel]

You're right, my voltage goes to zero (i don't know what i was thinking )
I have to be a "little" worried... Imagine this case:
in a toilet, my valve is closed (no water flows) and somebody remember to remove one battery! What happens - my water starts flowing (because there was a impulse to open my valve). This situation only stops when somebody put again my batteries...
I already tried to put CD40106, but nothing... Still appears one impulse when i drop my voltage rapidly to zero (simulating removing the batteries).
I already removed 4538 to put a 555 and the problem is the same.

 

[Roff]

You're right, my voltage goes to zero (i don't know what i was thinking )
I have to be a "little" worried... Imagine this case:
in a toilet, my valve is closed (no water flows) and somebody remember to remove one battery! What happens - my water starts flowing (because there was a impulse to open my valve). This situation only stops when somebody put again my batteries...
I already tried to put CD40106, but nothing... Still appears one impulse when i drop my voltage rapidly to zero (simulating removing the batteries).
I already removed 4538 to put a 555 and the problem is the same.

What is the purpose of the monostable (4538 or 555) in your circuit? I don't recall any mention of it previously.

 

[leonel]

Sorry to be out...
The ideia of 555 is when my PIR sensor detects my hands make a temporization of 2 seconds and make a pulse. This pulse is my circuit input.

 

[JimB]

If the water is turned on when the supply is disturbed, this does not seem to be a very "fail safe" idea.
I think you need to have a plan to ensure that the water is off if the circuit or its supply fails.

JimB

 

[leonel]

I agree with you JimB that's way i'm a litte worried...
Now it's happening a very strange thing:
When i turn off my power supply everything works ok, but when i change a little my power suplly my output change state. More, with batteries my output it's always changing state.
The circuit i have is the one in the figure without relay. The 4538 output is my input in the circuit discussed previously.