Electronics > Beginners
Transistor activation / deactivation voltage in
(1/1)
renzoms:




The capacitor was allowed to discharge completely when the transistor was activated with the 3.4 V.
The capacitor is allowed to charge again when the conduction from the 5 V line through the resistor to the ground of the IC is open circuited because the transistor deactivated.
The transistor must deactivate at 0V to have the 5 second cycle. In the case that the transistor deactivated at 1.4 V across the capacitor, the 5 s cycle would be reduced to the time it takes to charge the capacitor from 1.4 V to 3.4 V.
So my question!:
If the transistor was activated at 3.4 Vin (3.4 V across the capacitor) and allowed the capacitor to discharge through it, then the voltage at the capacitor would be decreasing. V in of the transistor would not be 3.4 V anymore..
There is the mystery.
If The transistor activates at 3.4 V, then when does it close as it has current running through it from the capacitor to ground. The voltage across the capacitor would be 1 V at some time 't' , therefore the V in of the transistor would be 1V, which is less than 3.4 V, which is when it activates.
Does it deactivate at 0 V in - ~0 V in AFTER it has activated (opened/activated), and only activate at 3.4 V in?
A second question: If I replaced the cap with a resistor the IC would switch rapidly correct?
The schematic on the top left is the one I redrew on the wide ruled paper.

Super thanks.
atmfjstc:
The problem description doesn't show the circuit between Vin and the transistor's base, so you can't make any assumptions as to what the logic for turning it OFF is.

The only thing that is clearly mentioned is that the transistor will be turned ON when Vin reaches 3.4V. We can safely assume that the circuit will not turn the transistor back off as soon as the voltage goes a bit under 3.4V, since then the circuit would just oscillate around that point and never discharge the capacitor. We can also conclude, given the wording used, that the transistor will be turned on fully (i.e. into saturation).

There are many possible circuits that would achieve such an effect. Depending on which is used, you can make the circuit turn off the transistor at any point you wish... 1V, 2V, etc., or never turn it off at all.

If you replace the cap with a resistor.. initially Vin will be the result of the voltage divider. If it's <<3.4V obviously nothing will happen and the circuit will stay that way forever. If it's more than that, the transistor will be turned ON and basically pull Vin close to 0 almost instantly. What happens then is again unspecified. As before, depending on the circuit, it might be turned off and restart the cycle at high speed (thus the whole thing becomes an oscillator), or it might just stay open and keep drawing current through the top resistor.
renzoms:
Thanks I think that is the best reply to a post I have made.
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