THEONE Posted December 8, 2012 Author Posted December 8, 2012 (edited) One more thing, is it possible to calculate the time delay by a given capasitor - resistance ?? It looks like the god don't want to do any timer circuit.... I bought a new 9v battery and i test it some times, but after that it doesn't looks to work.... it always the LED lights when i just connect the battery.... Edited December 8, 2012 by THEONE
BengalFlair Posted December 9, 2012 Posted December 9, 2012 You can get an approximate idea about delay with following equation -Delay =1.2* R*C sec. (assuming that the circuit activates at average 70% charging of capacitor).(Where R= resistance in ohm, C = capacitance in farad)For this circuit, total R =1k+100k =101k= 101000 ohms and C =220uF = .00022FSo max delay =1.2*101000*.00022 sec= 26.6sec. (approx) First close or short S1 (switch across capacitor) then put on the supply switch S2 and now to start the timer open or remove the short of S1.Check for damaged transistors or mosfet and replace with a new one if it still would not work.
THEONE Posted December 18, 2012 Author Posted December 18, 2012 (edited) I have checked everything and they works well except the PNPs cause i haven't got any spare of them. So i guess the PNPs be damaged... With and old one battery the circuit works well but when i add a new 9V battery it looks something be damaged and i guess PNPs. Also that i want to tell is that the original circuit has been made for a 3v supply not for 9v, so i guess maybe we must change the resistors.... I am not an expert but i guess... BengalFlair surely will know... Also the mosfet becomes hot with a new 9V battery, is that a problem ?? Edited December 18, 2012 by THEONE
BengalFlair Posted December 18, 2012 Posted December 18, 2012 I have tested the modified circuit up to 12v with no problem so there might be a wrong in your assembling.I have just noticed from your posted picture (27 July 2012) that you have connected the LED (load) directly to the circuit but it is not desirable as an LED must have a resistor connected in series to limit the current through the LED, otherwise it will draw huge current and will damage or burn out almost instantly. The mosfet also may be affected due to this current and excessive current must heat up the mosfet. (New battery = more power = more heat)Typical current through a red LED is about 20mA.So first check or replace it with a new one and add a 390 or 470 ohms resistor in series with it for a 9v supply. I think it will solve your problem.
THEONE Posted December 18, 2012 Author Posted December 18, 2012 (edited) Maybe that is why also the 555 circuit doest work for me a lot... a also then add a LED as a load... I will try to add sometime else... Like a small electric motor...I will try again... but i am afraid that the PNPs be damaged... Edited December 18, 2012 by THEONE
THEONE Posted December 18, 2012 Author Posted December 18, 2012 BengalFlair you are just AWESOME !!!!!!! :D It works again!!! thanks a lot !!! you are just PERFECT !! :)
THEONE Posted December 18, 2012 Author Posted December 18, 2012 BengalFlair i have one more idea...I checked the Amp of the new 9v battery and it was round 1.55A and after i checked the current to the load of the circuit and it was 0.82 so i was wondering if it would be better to use a 6v relay in case of the mosfet
BengalFlair Posted December 20, 2012 Posted December 20, 2012 It would be possible with 6-9v relays having coil resistance around 200 ohms or higher but for lower than that, Q2 (bc558b) should be replaced with a higher current pnp transistor. A diode should be used parallel to the coil in reverse direction for protection of Q2.
THEONE Posted December 20, 2012 Author Posted December 20, 2012 (edited) mmm... i have 2 6v relays... can you please remake the circuit ?? If you have time... The relay where i have, their coils are under 99 ohms... Edited December 21, 2012 by THEONE
BengalFlair Posted December 22, 2012 Posted December 22, 2012 Here is the circuit diagram of the timer with relay. Maximum collector current of bc558b is 100 mA but the required current to activate a 6v relay having coil resistance 100 ohms or lower is about 60 mA or higher. So it would be better to replace Q2 with a higher current pnp transistor for better stability of the circuit. I have tested the circuit with a c100 pnp transistor but I think other equivalents would do well.
THEONE Posted December 22, 2012 Author Posted December 22, 2012 Thanks a lot again bengalflair.... i wish i have your knowledge about this subject...
beny Posted September 23, 2013 Posted September 23, 2013 @theone, I have read this thread and bengalflair has actually helped with complete sincerity and awesomeness. I have to applaud the effort of bengalflair who has replied to all the questions without saying a word. It's people like him that make this world a better place. Secondly, on the issue of using 6V relays with a 9V source, I'd say it's a little unadvisable as the current that will flow through the relay will be more than it can handle and if you have to keep the relay turned on for larger periods of time, it means that it's definitely going to create an issue sooner or later. Best go with a higher resistance relay of a 9V relay which obviously will be the best option for you. Cheers.
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