555 Monostable

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zcacogp

Joined Mar 10, 2015
8
Hello,

I'm new here, and have signed up as I am planning to build a timer circuit. It's a monostable, based on a 555 chip.

I want to produce an output that is high (+5v, with +5v being the circuit voltage) for around 4 seconds when an input voltage is applied. The input voltage will be +5v as well, but the input voltage will remain high for a long period of time (minutes or hours), and I need the output voltage to only go high for 4 seconds before going low again.

With apologies for the hand-drawn diagram, I think this is the circuit I need to build. It seems that the time of the circuit is 1.1*C1*R1, so 4MOhms and 1MicroFarad will give me a little over four seconds, which is what I need.



However I have two questions;

1. Will this circuit work given that the input pulse is much, much longer than the output pulse?
2. How do I make this triggered by a rising input pulse? It seems that 555's work on falling-edge pulses, which I can't give it. This would be terminal 2 on the timer (with a ? on the diagram).

(Background, should it be relevant; I have bought a bluetooth audio receiver to use in the car. It works well, but needs to be turned on by holding a momentary switch in for about four seconds whenever it is plugged in - just plugging it in won't turn it on. The momentary switch connects a terminal in the receiver to live.

I could mount the thing in the car such that I press the button whenever I want to use it, but that's inelegant and I'd like to make it happen automatically.

I'll run the thing off an ignition live via a USB supply (+5v), so I want the 'turn on' button to be pressed for four seconds whenever the ignition live (i.e. supply) goes live. The supply will clearly stay live for some period of time, but I don't want the output to stay live for more than 4 seconds as that will make the device reset.)

Thanks,


Oli.
 

Dodgydave

Joined Jun 22, 2012
11,277
To make the 555 trigger on power up, you need to put a resistor/ capacitor to pin 2 for power up so it will

trigger automatically, like this 555.png
 
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Thread Starter

zcacogp

Joined Mar 10, 2015
8
Ron, Dave,

Thanks for your answers.

Ron, that's a helpful overview of the 555 but I'm not sure I followed all the detail, but thanks for the link anyway.

Dave, Thanks for the explanation of the power-up trigger. I'm not sure I understand why that would work; when there is a +ve signal on pin 2, current will cease to flow through the 10k resistor and the 100nF cap will start to charge very quickly. How does that provide a falling-edge to trigger the 555 timer? Surely this will provide a falling edge at the point that the input to pin 2 drops back to 0v?


Oli.
 

Dodgydave

Joined Jun 22, 2012
11,277
When powered up ,pin 2 is held low long enough to trigger the timer, then pin 2 is high as the cap charges up, its a classic power up trigger circuit tried and tested .
 

Reloadron

Joined Jan 15, 2015
7,480
Ron, Dave,

Thanks for your answers.

Ron, that's a helpful overview of the 555 but I'm not sure I followed all the detail, but thanks for the link anyway.

Dave, Thanks for the explanation of the power-up trigger. I'm not sure I understand why that would work; when there is a +ve signal on pin 2, current will cease to flow through the 10k resistor and the 100nF cap will start to charge very quickly. How does that provide a falling-edge to trigger the 555 timer? Surely this will provide a falling edge at the point that the input to pin 2 drops back to 0v?


Oli.
Actually it doesn't provide a falling edge or negative trigger. Since the resistor and capacitor form an R*C network on pin 2, the trigger pin, when power is applied pin 2 sees a low. Pin 2 will remain low as the capacitor charges. This initial low is long enough for the 555 to trigger in the mono stable mode. So what we get at power up is a single mono stable pulse out of the 555 the duration of which is set by the R*C timing network connected to pins 6&7. That's it. Following that initial pulse the circuit is done until power is removed and reapplied. That is why it works. When using such circuits I generally use a 10K Resistor and 1.0 uF capacitor but the 0.1 uF as shown in the link should also work.

Ron
 

Thread Starter

zcacogp

Joined Mar 10, 2015
8
Ah, thanks Ron and Dave - that makes perfect sense when you put it like that. I guess it relies upon having the +5v rail starting at 0v as well, which I hadn't thought of. Thank you.

I have a fairly limited collection of capacitors, so I guess using a bigger one in that position will simply mean it stays low for longer.

However I have just discovered a problem which may make the whole thing impossible. The aim of the circuit is to turn on a bluetooth receiver when power is applied by mimicking the holding down of a power button. However the receiver has a battery in it, which means it stays on for a period of time after the power has been disconnected. (I don't know how long this is - probably several hours). If, while it is on and running on the battery, you hold down the power button a second time then it turns the device off. This means that if the car ignition is turned on while the device is running from the battery and the 555 circuit does it's stuff, it will actually turn the device off - which isn't good.

I guess the answer is to introduce a test such that the 555 circuit won't act as a one-shot if the receiver is already on. I presume the 'reset' pin could be used for that, although I can't find a point on the receiver that goes high when it is on, so I don't know how to get the data for the test. If anyone has any suggestions I'be interested to hear them - thanks!

Thanks again for your help.


Oli.
 

Wendy

Joined Mar 24, 2008
23,408
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