[LectronFanModerator]

Hi all,

In previous experiment, we saw how the Schmitt trigger reacts when a certain charging level of the capacitor is reached.

Now, we will test how much current is needed to turn the Schmitt trigger permanently in the ON state.
After building the circuit, set the potentiometer so that the Schmitt trigger keeps on.

Some explanation :
The 1st transistor in the Schmitt trigger needs a certain current to turn on.
When the voltage across the 100µF capacitor is large enough, current will flow through the B-E junction of this transistor, turning it on.
Now, a current will also flow through the 47Ω resistor inside the Schmitt trigger module. When this current is larger than the charging current through the 100µF capacitor, it will start to discharge the 100µF capacitor.

This causes a loop where the Schmitt trigger will be switching from ON to OFF, making the lamp blink.

We will make use of this phenomenon to create a pulse generator in the next experiment.

Many greetings

[MichaelKeymaster]

Always the leader in Lectron System circuits!  Thank you Frank!

[Author]

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