What will happen when i would turn this circuit on?

 

Whatever would happen, you won't know it until you first try to guess it yourself. After you have made an effort on solving your problem, you will receive help from the AAC community.

 

Whatever would happen, you won't know it until you first try to guess it yourself. After you have made an effort on solving your problem, you will receive help from the AAC community.

Ok ill try. Im not so good with electronics so im gonna make a guess. So if you press the switch the the led lights up because you allow current through the transistor which lights the led. But only for a few moments because there is a capacitor so the led would slowly dim and after that the led wont light anymore because the capacitor is charged.

 

With the capacitor fully discharged you'll get an initial surge of current in the transistor and diode. Then there would be an extended interval over which the capacitor charges with ever diminishing current flow in the transistor & diode. Presumably the transistor eventually turns off as the capacitor charges towards ~6V. What happens in the end would probably depend upon component leakage currents and the diode dynamic resistance at decreasing forward current.

 

O_O umm i didn't get that

 

That's what you said in essence. So you got the idea.

 

Hi aartt,

Can you see why the circuit as drawn might lead to either a dead transistor a dead diode or both - once you close the switch?

 

nope i don't i really don't know anything

 

nope i don't i really don't know anything

Well you must know something. Are you doing a course in electronics? If not, why have you posted your question? Interested in electronics as a hobby?

 

I know really nothing about electronics exept for the extreme basics like what is a resistor but i dont get a transistor. I am doing this for a hobby and i probably can't take a course because im too young

 

Maybe you should do some reading first. Electronics needs a basic theoretical background before you can start building circuits.

Try the e-book of this site. You can find links in the top of the page.
Is english your native language?

 

Hi aartt,

Can you see why the circuit as drawn might lead to either a dead transistor a dead diode or both - once you close the switch?

Well you must know something. Are you doing a course in electronics? If not, why have you posted your question? Interested in electronics as a hobby?

I've taken a few basic courses in electronics and I'm curious to know the answer

 

I'm not seeing how the damage is done to either the diode or the transistor.

Can someone explain?

When capacitor is empty, the capacitor act juts like a short circuit.
So when we close the switch the current is start to flow.
But there is no device in the circuit that can limit the amount of initial current that start to flow, when the switch is close.

 

Yes - it's a debatable point as to whether there is sufficient energy let through during the current transient to cause any damage. With no series limiting resistance the initial current peak will be quite high [perhaps a couple of amps] but of quite short duration. With only a single shot repeated at long time intervals, the risk of damage would probably be low. If the switching was done at a high repetition rate then the risk of damage might be higher.

I was trying to add some 'spice' to the problem.

 

Spice might be helpful here, preferably a version supporting a "smoke alarm". (Sorry folks, but I just couldn't resist that one.)

Seriously though, something is likely to go pop unless that capacitor is pretty small, or the battery has lots of internal resistance.

A few hundred ohms resistance in series with the LED would make this the basis for a reasonable experiment, rather than an accident waiting to happen.

 

I tried your suggestion of a simulation.

With a 2N2904 a 1N4148 and 1uF cap with 6V DC I get a current greater than 1A for approx 15usec. IFSM for the 1N4148 is quoted in one data sheet as 2A for 1usec. Using the (I^2)*t concept this would mean the diode is possibly at risk even with a single shot.

It's hard to judge with respect to the BJT. Again one data sheet quotes Ic(max) of 600mA. But for what time that can be safely applied is not clear.