Relay Turn OFF Delay

[Suraj143]

Guys for a door EML I need to switch on a relay for 5 seconds (no need accuracy) when the door open button pressed.

Whats the easiest way of doing without using NE555 timer?

 

[ljcox]

Here is a 60 second timer that will do what you want.

You can reduce the time to about 5 sec by reducing C1.

0.56 uF should give you about 5 sec.

Note that CMOS can be damaged by static charges. So you need to take electrostatic precautions when handling CMOS ICs.

Also note that CMOS has a very low quescent current, so it can be supplied from a battery if necessary.

The relay will consume energy from the battery while it is operated, but the current will be virtually 0 while it is released.

 

[davenn]

or something a bit easier without worrying about static sensitive chips

cheers
Dave

 

[ljcox]

Static sensitive ICs are not a big problem.

You just touch something that is earthed before starting work.

Your circuit does not have any hystersis & will be subject to variations due to temperature & transistor characteristics.

 

[colin55]

What is "hystersis" and what does it matter in the circuit presented???????

 

[ljcox]

The circuit proposed does not have properly defined threshold levels.
The transistor starts to turn on when the Base - Emitter voltage reaches about 0.5 Volt & is fully on when it reaches about 0.7 Volt.

So the delay time will depend upon the transistor's current gain (β or hFE) & Base - Emitter characteristic - both of which are temperature dependent. (Hence the delay is temperature dependent.)

Also, since the Collector current rises slowly & the Collector voltage falls slowly in proportion (according to Ohm's Law), the power dissipated in the transistor rises from zero to a maximum when the Collector voltage is at about Vcc/2 (6 Volt in the case of the circuit shown above) & then falls amost to zero when the transistor is saturated.

You therefore need to be careful that the transistor dissipation does not exceed its limit when the collector current is in the active region (ie. not in cut off or saturation). The collector current will pass through the active region when it is rising & later when it is falling (ie. when the relay is to be released).

The relay will operate when the collector current reaches the relay's operate current level & later release when the current falls to the relay's release current level.

Hystersis is provided by a positive feedback mechanism that clearly defines the threshold levels.

When the capacitor voltage rises to the upper threshold, the device output switches rapidly.

The same occurs when the capacitor is discharging but at the lower threshold voltage.

If you look at page 3 of the data sheet of the 40106 attached, you will see that, for a supply voltage of 10 Volt, the positive threshold is typically 5.9 Volt & the negative threshold is 3.9 Volt.

Therefore, the output will change rapidly when the capacitor voltage rises to 5.9 Volt and will change rapidly later when the capacitor voltage falls to 3.9 Volt.

Thus the transistor (not shown in my diagram) that switches the relay is rapidly switched on & later off so there is no danger of it being damaged by excessive power dissipation while in the active region.

 

[colin55]

You are talking about minute things, that don't matter.

 

[ljcox]

We may have to agree to disagree on that.

 

[Suraj143]

Hi guys thanks for the input.

ijcox thanks for your input.But it uses another chip.Anyway I'll work on that circuit in my another application.

@ Dave
That's the one I was looking.Note that the user won't press & hold the switch.He/she just press the switch quickly & release it.At that time the relay must be ON for at least 4-5 seconds.During this time power to the EML will be released. So the door will be released.

I was thinking whats the purpose of that 0.1uF capacitor transistors base to GND?

 

[davenn]

Hi guys thanks for the input.

ijcox thanks for your input.But it uses another chip.Anyway I'll work on that circuit in my another application.

@ Dave
That's the one I was looking.Note that the user won't press & hold the switch.He/she just press the switch quickly & release it.At that time the relay must be ON for at least 4-5 seconds. During this time power to the EML will be released. So the door will be released.

yup exactly !! just a quick press of the pushbutton switch will charge up the C1 capacitor and start the timed power on cycle. I had originally done this cct for a guy's car, who wanted a light on delay before it swtiched off. The pushbutton switch in the above cct, was in his cct, another relay which would supply a pulse to C1 when the ignition key was turned.

he, like you, wasnt worried about exact timing just wanted X seconds give or take a second or 2. I vaguely remember him going for ~ 2000uF which gave ~ 12 sec

cheers
Dave