Hi everyone,

I'd like to bring this up because I have the same task, but the implementation must be a little different. I am an amateur and not at all a professional in electronics, so there must be something that I am unaware of.

I am making a circuit that is based on the one two posts earlier which works well. The difference is that I want the 555 IC to be disconnected from power when the output impulse finishes. My application is the start of a servo-motor seated on a water valve. The motor needs an initial impulse of about 1 sec to start, then it keeps itself powered on by a switch until a certain position. I don’t want to keep the 555 IC powered on because the motor must never ever false start due to some surge. The circuit I’ve built is shown in the picture.
When one presses the button, the timer is powered on and produces a positive impulse on its output that opens the transistor which locks the timer powered on. After a second, when the timer runs out and impulse finishes the transistor closes and disconnects the IC from power. The circuit then waits for the next button press. This is how it is supposed to work.

For the purpose of troubleshooting, I’ve replaced the motor with a lamp keeping the same current. In fact, when I connect the circuit to power, the timer starts immediately and the lamp lights full. After a second, the timer runs out and the lamp continues lighting in half. Consecutive button presses do not restart the timer. There must be a current leakage through the transistor, but I’ve no idea why and how to prevent it. I’ve tried to connect a diode in different positions in the output circuit and the only position shown in the picture gives some effect. The lamp dims more during the off-cycle, but it lights anyway.

I am asking dear members to point out my mistake. I know how to use a 2-pair relay for this and intentionally not doing so because I need the schematics to be as compact as possible. I’ve researched a lot of resources on 555 and haven’t found any example where the IC is kept self-powered by a transistor only until the timer runs out, so please help me.

Mod: Link to old thread.E
https://forum.allaboutcircuits.com/threads/555-monostable-self-triggering.64555/post-442710

 

I suspect that you need to add a resistor between the base and emitter of the transistor, perhaps 200Ω to 1k.

 

I think you will always get leakage through the 555 then the transistor B-E junction.
The 555 is not disconnected when not in use. The output is still connected to the transistor.
Maybe use an opto isolator, with the LED driven (via a resistor) from the 555 output, and the opto transistor in the +ve supply to the 555.
The start button will be across the opto transistor. Just use the power transistor to switch the motor, and the opto transistor the 555.
Add a reverse diode across the motor too to help limit transients.

 

This schematic will not work. This is because the 555 unit acts as a "resistor" feeding the base of the transistor from Vcc irrespective of whether the 555 output is Hi or Lo; except that in the Lo state, it is a higher value "resistor".


Use another transistor to cut off the 555.

 

I would change two things and try again:

1. Change to a CMOS 555, such as the LMC555.

2. Change the transistor to a FET, such as the 2N7000.

As above, there needs to be a resistor between the 555 output and GND. It could be as low as 1 K, but with a CMOS part I suspect that 10 K would be enough.

ak

 

Hi everyone,

I'd like to bring this up because I have the same task, but the implementation must be a little different. I am an amateur and not at all a professional in electronics, so there must be something that I am unaware of.

I am making a circuit that is based on the one two posts earlier which works well. The difference is that I want the 555 IC to be disconnected from power when the output impulse finishes. My application is the start of a servo-motor seated on a water valve. The motor needs an initial impulse of about 1 sec to start, then it keeps itself powered on by a switch until a certain position. I don’t want to keep the 555 IC powered on because the motor must never ever false start due to some surge. The circuit I’ve built is shown in the picture.

What happens if there is a false start?

How is the motor powered after initial start?

 

Here's the LTspice simulation of a 555 one-shot that uses a high-side transistor switch to turn off the power at the end of the period.
That avoids any problems with trying to remove the power on the ground side.

Note the 3MegΩ is a little large to use as the timing resistor with a standard 555 timer.

 

Thanks a lot to everyone who responded so quickly. I will try the suggestions and give comments on every post once I have some free time.
A little more details to make it clearer. I added another switch (S2) to the schematics that is mechanically associated with the motor gear. S2 switch is a "normal-on" that closes the motor circuit after some angle of rotation once its button is released by a cam. Then the motor keeps going until the next cam that breaks the circuit.
Please note that this is a hobby project and I use pretty obsolete electronic components that I have on hand. Also, I'd like the schematics to be as simple as possible because it will be placed in a very confined space around the motor (no PCB.)

 

What happens if there is a false start?

The mechanism will operate 24/7 and it would be disadvantageous if the valve got opened or shut unexpectedly.

 

Note the 3MegΩ is a little large to use as the timing resistor with a standard 555 timer.

According to my datasheet, a resistance of up to 10M is OK. The IC gives a stable enough impulse duration with my RC values.

 

S2 switch is a "normal-on" that closes the motor circuit after some angle of rotation once its button is released by a cam. Then the motor keeps going until the next cam that breaks the circuit.

So If my understanding is correct, you don't need a one-shot, just a latch.
The push-button sets the latch to start the motor and the cam resets the latch to stop the motor.
The 555 can be used as a latch, if that's all you have.

 

Thanks a lot to ................................... be as simple as possible because it will be placed in a very confined space around the motor (no PCB.)

The BD135 will never turn OFF completely. There will always be a current feed to the motor and the BD135 will neat up in the OFF condition.

 

I suspect that you need to add a resistor between the base and emitter of the transistor, perhaps 200Ω to 1k.

Tried this one. A value of as low as 150 Ohms turns off the transistor and lamp. But anyway, there is a current of 4mA flowing through the output of the IC and this resistor. The IC feeds itself through the output pin. 4mA is what it eats normally.

 

This schematic will not work. This is because the 555 unit acts as a "resistor" feeding the base of the transistor from Vcc irrespective of whether the 555 output is Hi or Lo; except that in the Lo state, it is a higher value "resistor".

I am afraid I don't fully understand your picture. Is that how it is? What is "I1 I"?

 

The BD135 will never turn OFF completely. There will always be a current feed to the motor and the BD135 will neat up in the OFF condition.

Well, it does. When the base is on the GND, the collector-emitter current is about 1uA (according to my multimeter).

 

Here's the LTspice simulation of a 555 one-shot that uses a high-side transistor switch to turn off the power at the end of the period.
That avoids any problems with trying to remove the power on the ground side.

Thanks. As I said earlier, the IC appears to be indeed feeding itself through the ground side. I will try to lock it from the Vcc side. I will try to reduce your schematics, though.

 

I will try to reduce your schematics, though.

What does that mean?

 

What does that mean?

I mean use fewer elements. Otherwise, it will be a spider web without a PCB.

 

I mean use fewer elements. Otherwise, it will be a spider web without a PCB.

Well you are welcome to do that, but my circuits generally have a minimum of parts for proper operation.

It's not that hard to build on a vector type board.

 

The difference is that I want the 555 IC to be disconnected from power when the output impulse finishes.

I hope you realize that none of the solutions in this thread actually disconnect the timer from power.
There is always a conductive path thru the timer. So...a glitch can always false trigger the timer and cause the output to bounce. Maybe change your target solution to address the surge problem instead.