[LectronFanModerator]

Hi everyone !

Today, we will have a look at the Lectron relay its behaviour and properties.

The relay itself has a coil resistance of 309Ω.
Thus, if we would connect it to a 9V battery, the current flowing yhrough it would be 9V / 309Ω = 29mA.

A relay has a turn on current and a turn off current.
This means that the relay will turn on at a certain current and that it will keep on untill this current is too low to keep the relay energised.

We will do now an experiment that demonstrates how these properties can be studied.

Be sure to have a fresh battery of 9V to conduct the experiment !

After assembling the circuit, turn P1 completely CCW (counter clockwise).

Then, when keeping S1 pressed, turn P1 CW (clockwise) untill the relay energises and the lamp turns on.
Read now the voltage, indicated on the meter. This will be about 6,5V.
The turn on current is thus 6,5V / 309Ω = 21mA.

Now, turn P1 CCW untill the relay de-energises and the lamp goes off.
Read now the voltage on the meter, it should be about 2,5V.
The turn off current is thus 2,5V / 309Ω = 8,1mA.

Try this experiment over again to fully understand the principle.

We’ll continue now how to make use of this relay property.
Set P1 till the lamp just turns on (don’t forget to press S1 while doing so !).
Leave P1 at  this position and press and release S1 and you’ll notice that the relay goes on and off according to S1.

Press now S2 (and keep pressing it) and then, press S1.
You’ll see that the relay keeps off.
When you look at the diagram, you see that S2 shunts R2. So, when you press S2, the relay is in series with R2 (220Ω).
The relay current is now too low to energise the relay. (calculate it once !).

Now, see what happens when you release S2 (S1 is still pressed !), the lamp goes on !
And it keeps on, even when you press or release S2 !
What happens now, is that once the relay is energised, the current through R2 and the relay is high enough to keep the relay on. (Look once above for the turn off current).

Release all switches now.
Turn P1 fully CW.
Press S2 (and keep it pressed down) and press S1 : the relay keeps off. Release the switches.
Press now S2 & S3 (and keep these pressed down) and then S1.
The relay energises and the lamp turns on !
How is this possible ?

Can you figure out what happens ?

This phenomenon is widely used in circuits to keep the relay currents low.

Post your comments below !

Many greetings

[MichaelKeymaster]

Hi LectronFan,

What a great in depth study demonstrated in such a practical manner!  Thank you!!

With S3 closed, doesn’t that allow what ever charge was on the capacitor to flow and energize the relay?

[LectronFanModerator]

<p style=”text-align: left;”>Hi all !</p>
The solution of the capacitor is as follows :

At first, C1 is completely discharged.

When pressing S1 and S3, C1 starts to charge very quickly through the relay coil and T1.
This means that a brief high current will flow the relay coil (capacitor property).
This high current causes the relay to be activated and the lamp turns on.

As soon C1 is charged, this high current stops but the relay will still be energised through the lower current flowing through R2 (the relay needs less current to be in the hold position).

Noticed how we used T1 as an emitter follower to adjust the current through the relay circuit ?

Many greetings

[Author]

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