Hi fellas
look at these circuit:

It's supposed to make a switch to cut off the current after a small delay. the resistor is for limiting the current and capacitor is for making delay. like the two above circuit. but the question is:
Does the both circuit work like each other? I'm thinking about circuit 1. Does it work properly like circuit 2?

 

Although neither circuit has a load in the collector to prevent the transistor from burning up ...

Both circuits do the same thing, but #2 does it better. In circuit #1, the end of the turn-off transition will last much longer because there is nothing to assure that the base current will go to zero, and as the current gets lower and lower it changes more and more slowly. theoretically, the transistor never turns off because an exponential curve is asymptotic - it never reaches its final value.

In #2, the resistor pulls the base voltage up above (Vcc-Vbe), forcing the transistor into a true off state. The final base voltage still never reaches Vcc in theory, but it is way past what it takes for cutoff.

Note that for both circuits the collector current turn-off can be pretty slow, depending on the R-C time constant and the size of the capacitor.

ak

 

Additionally:
In circuit #2 the capacitor gets charged at turn-on through the base, and you should have a resistor in series with the base to limit the current so you don't damage the transistor.

 

with the circuit 1 you may have the issues in the noisy HF waving on supply rails (even if r1 is the highest possible)
... same with the circuit 2 -- there was an override to that however . . .

 

Assuming the same value components (R1 & C1) in both circuits, circuit #2 will charge the capacitor rather quickly. It's charged through the EB (emitter / base) junction. R1 does almost nothing for the circuit. I said "ALMOST nothing". A whole lot depends on the component values; and while most of the current to C1 will be through the EB junction, some current DOES pass through R1. So it DOES have some effect.

In circuit 1 C1 is charged through the EB junction then through R1. The delay will be much longer (assuming high values for R1 & C1). Just how long it will take depends on their values. AND as AK said, the transistor will never reach full cut-off.

And to reiterate, it would appear you forgot to add a load on the collector.



Current will travel along the minor path until C1 charges sufficiently to reduce the current pull on the base of Q1 (2N3906) transistor. When the base current drops sufficiently the major current will be cut off. However, this won't act like a switch it will act like a dimmer. As the base current drops the EC (emitter / collector) current will drop too. If R-load is a light bulb (low power) then it will dim till it extinguishes. Whereas a switch will be full on until it is full off. No in-between's.