EveryCircuit - Why Cant It Find A Solution

10 years ago

I realy dont know just get rid of pover source on led and anode of led place on output and catode to ground pover source on 2 v

ihughes24

10 years ago

Get rid of 9v battery, turn led round and connect to ground. If you don't want to blow up led add a resistor in series with led. Hope that helps.

wyoelk

10 years ago

Ihughes hit it on the head!

sygic

10 years ago

Just change the polarity of the 9v voltage source.*reverse it.

FarhanRahaman

10 years ago

>>ihughes -it works but if you try to adjust the voltage a bit more it cant gind a solution

hurz

10 years ago

Interesting, its one more of 555 inaccurate output model. You made everything correct! I m not asking what this circuit is made for and why this way. Its just correct. However, 555 seems to be again a bit nasty. Cuz the low level out is a few hundred of millivolt below of zero volt, which cuz an overload and Cant find solution. This is again a combi problem of 555 and LED convergence. Use a simple 9V bulb and its working fine.

hurz

10 years ago

What also helps is increase the LED voltage by 0.3V to 9.3V

thebugger

10 years ago

Just remove the 9V supply, delete the LED and get a new one with standard parameters, add a series current limiting resistor and connect it to ground... Ah also reverse the polarity. LED anode to the output, through a resistor and cathode to ground

hurz

10 years ago

Buggzy, this tip is to boring don't you see the real problem of this actual configuration? Keep the 9V and light the LED up if 555 goes to low!

thebugger

10 years ago

The 9V source and the LED is the problem. Remove the 9V supply replace the LED with one with normal parameters (2V/20mA) and add a current limiting resistor in series.

thebugger

10 years ago

By the way i see what you mean with the LED but it's irrelevant, because in reality it would behave normally.

FarhanRahaman

10 years ago

>>thebugger&hurz-Thanks a lot!

JAYlab

10 years ago

Be more specific guys cause u don't seem to solve the problem

JAYlab

10 years ago

Umh remove the 9v source n turn your diode to face the other side then set its voltage to 8.something then run it

JAYlab

10 years ago

Or put a resistor in front of a diode but don't forget to turn it

hurz

10 years ago

The initial circuit has two different voltage source in use. 10V and 9V ! Whatever we can simplify here, the original request would like to drive the LED against a 9V source. Suppose the 10V is a weak source and only the 9V can handel the 20mA or more LED! And from logic it should lightup while 555 output is low! Dont do a complete redesign if you dont know what the requirements are. Or at least see the circuit as is and the request!

FarhanRahaman

10 years ago

Who are you saying this to,hurz?

hurz

10 years ago

To all who are going to change your circuit

FarhanRahaman

10 years ago

So this is a software bug and in reality, I think this circuit is going to work normally.

hurz

10 years ago

I would be careful by calling anything a software bug before fully understanding

hurz

10 years ago

Anyway, what is this for a LED in reality?

thebugger

10 years ago

Ahh now i get it, duh. The 555 timer will just provide the voltage drop from for the 9V supply to ignite the LED. In reality it should work (though I'd add a current boosting stage to do it) while here in EC it doesn't work. Anyway i don't think 555 are supposed to work like that, dropping current through them from the output. I'm positive the problem will be fixed if he adds an external transistor to handle the drop and not 555 itself.

hurz

10 years ago

555 is made to handle source and sink of current. It can even sink more than source.

CrazyBound

10 years ago

Remove the 9v source voltage then use a 50 ohm resistor and change the LED to 8v instead of 9v. Place the 50 ohm resistor and the LED (anode side) on the output and connect the cathode side to ground and boom works great.