timer 555

[audioguru]

If you don't install the diode across the relay's coil then the 555 will be destroyed by the inductive very high negative voltage spike that occurs when the relay is switched off.

 

[Hero999]

Also, make sure that the diode is connected in reverse or the 555 is be destroyed by the large current drawn.

 

[Fahime]

Clear. Thanks alot.

 

[Fahime]

I see that the IC 555 gets hot when it's running. I guess it's because of the half wave rectification. So I replaced it with a full-wave rectifier. But it's getting warm yet. Actually not as hot as before. Is it a clue of a problem in the circuit or not? Should I use a voltage regulator or it's not the matter with getting warm?

 

[Nigel Goodwin]

I see that the IC 555 gets hot when it's running. I guess it's because of the half wave rectification. So I changed it to a full-wave rectifier. But it's getting warm yet. Actually not as hot as before. Is it a clue of a problem in the circuit or not? Should I use a voltage regulator or it's not the matter to get warm?

More likely it's getting warm because you're feeding the realy directly from it!.

 

[Fahime]

More likely it's getting warm because you're feeding the realy directly from it!.

Dear Nigel,
What do you mean of "feeding directly"?

 

[Nigel Goodwin]

Dear Nigel,
What do you mean of "feeding directly"?

Not using a driver transistor.

 

[stevez]

Fahime - is the current required by the relay coil more or less than the amount that pin of the 555 can supply? If it the coil requirements do not exceed the limits then it's not a problem but if it's close or exceeds the ability of the 555 then the driver that Nigel suggests is an appropriate next step.

 

[em2006]

Check and post the resistance of relay's coil (use am ohmeter).
Check and post the supply voltage of 555 (pin 8).

 

[Fahime]

Check and post the resistance of relay's coil (use am ohmeter).
Check and post the supply voltage of 555 (pin 8).

The resistance of relay's coil= 400 ohms
the supply voltage is alternating between 13V and 15V.(It's not too smooth)
I also measured the current that relay draws. I=27 mA
I looked for the allowable current that can be drawn from 555 in the datasheet but I couldn't find any. Is it at most the mentioned 15mA in the 3th page of the datasheet that I attach it?

 

[em2006]

The resistance of relay's coil= 400 ohms
the supply voltage is alternating between 13V and 15V.(It's not too smooth)
I also measured the current that relay draws. I=27 mA
I looked for the allowable current that can be drawn from 555 in the datasheet but I couldn't find any. Is it at most the mentioned 15mA in the 3th page of the datasheet that I attach it?

In the 3th page of the datasheet you saw that the 15mA is the maximum supply current, when the supply voltage is 15V and RL=infinite (with no load)
In the 1st page of the datasheet you can see that the output can source or sink up to 200mA.
If the resistance of relay's coil= 400 ohms and supply voltage is 15V, the output current is no more than 35mA (you have diode series coil relay), you measured the current = 30mA =OK. This current is not dangerous for output of the 555 (pin 3).
Check also, if the timing resistor (pin 7) is not too low (must be over 1KOhm)

 

[Fahime]

Check also, if the timing resistor (pin 7) is not too low (must be over 1KOhm)

Dear em2006,
The timing resistor is a 1Mohm potentiometer in series with a 100Kohm resistor. And the timing capacitor is 220uF.
The idling current is 12mA(I measured with a amperemeter) and the 555 gets warm even with no feeding load. Is it natural?

 

[em2006]

I think not.
Did you tried to replace the IC with another one?

 

[Fahime]

I think not.
Did you tried to replace the IC with another one?

Yes, I tried it with 2 more new ICs. They get less warm but they do so.

 

[audioguru]

The max allowed output current from an LM555 is 200mA. Its operating current when it has a 15V supply, no load and its output is low is a max of 15mA.

The max power dissipation without a load is 15V x 15mA= 225mW which will make it warm. Your load current of only 27mA is so low that it adds only 38mW to the heat in the LM555.

 

[em2006]

...The max power dissipation without a load is 15V x 15mA= 225mW which will make it warm...

That's it!
So, a solution is to decrease the supply voltage, to decrease the power dissipation when the output is low.
But, in the our last schematic, is not voltage on the circuit in stand-by. In the normal working, the output is high, and the load current is
27mA. Supply current when output high, is small, and it adds a small heat.

Fahime, is this the final schematic wich you tried (attach) ?
If not, try it, and decrease the supply voltage to 12V.

 

[audioguru]

The output of a 555 has a 1.4V loss when it is high and the load current is low. The diode in series with the output adds another 0.7V loss so the 12V relay gets only 9.9V and it might not work.

The output low loss from a 555 with such a low current load is only about 0.2V which would be much better.

 

[em2006]

The output of a 555 has a 1.4V loss when it is high and the load current is low. The diode in series with the output adds another 0.7V loss so the 12V relay gets only 9.9V and it might not work...

I agree.
Actually, is not dificult to select a relay type according disponible voltage, because ussually, Operating Range of relays is (0.7...1.5)*Nominal voltage, or, better, to use a 9V relay with a series resistor.
Another posible way to use a 12V nominal voltage relay, is to reduce voltage drop eliminating (shorting ) the series diode, and keeping parallel diode only.
The series diode is necessary for inductive load having a very high Q. Of, course, is preferable to keep both diodes to fully isolate output terminal of 555 from the negative voltage wich inductive load generates, to avoid aplying about (-0.6V) accros the output of 555.

 

[Fahime]

Fahime, is this the final schematic wich you tried (attach) ?
If not, try it, and decrease the supply voltage to 12V.

Thanks for your shematic, Em2006. My last circuit is the same as yours. But I can't figure out the operation of C4 that you inserted between V+ and ground!? I don't have it!

 

[Fahime]

2 more questions:
1) Can I use a 555 for timing 20 hours?
2) Please turn to page 3 of the datasheet for IC555 that I have already attached. What is the meaning of (trigger voltage =5V if Vcc=15V) that is mentioned in the electrical charactristics table? Does it mean that in my circuit the voltage of pin 2 is too high when Capacitor charges ? should I use a voltage divider before the capacitor? (Because when the capacitor charges it's voltage reaches 15V and is much more higher than 5V!!??)