Usually the voltage at the base of transistors will not go lower than -6 to -8 V. Lower voltages will exceed the typical avalanche breakdown voltage of the base to emitter diode. This is also the reason why the time periods of such astable multivibrators tend to get shorter or the frequencies tend to get higher, the higher the operating voltage is. It is therefore not recommented to exceed 5 V operating voltage, especially in the case of larger capacitors (muliple 100 µF). The energy of the charge can evtl damage the base emitter diode.
How is this possible without damaging the capacitors? As I understand it C1 and C2 are switching polarity and the capacitors are polarized (often electrolytic capacitors are utilized in Astable multivibrators).
Ordinary aluminium capacitors can withstand a reverse voltage of about 1V to 1.5V without taking damage. Since the maximum reverse voltage is well below 1V everything is fine.
@@MirceaPricop Thanks. This is the answer that i was looking for. btw how did you get to know that? just by figuring out or someone told you? either way thanks.
@@kurtti1043 Haha glad it helped after such a long time. I remember being confused by the same problem and researching it online, it's stated on Wikipedia: en.m.wikipedia.org/wiki/Electrolytic_capacitor#Reverse_voltage
@@MirceaPricop If i change the values on capacitors C1 and C2 does it make any difference how long the leds are going to stay on? Can the capasitors have charge over 0.7V since the transistor Q1 starts to conduct 0.7V and C2 starts to discharge and C1 start charging. Can i make the discharging slower if i put resistors between emitter and ground? also if i want to change the chargin time i can change the 2 resistor in the middle, right?
@@kurtti1043 Yes, the duration each LED stays on is directly proportional to R3*C1 and R4*C2, so changing either capacitors or resistors in the middle will influence timing. I think adding a resistor on the discharge path will cause the output to fade out over time instead of instantly swap, but I never tried that out in practice
Thanks, this helped me take a nap!
Usually the voltage at the base of transistors will not go lower than -6 to -8 V. Lower voltages will exceed the typical avalanche breakdown voltage of the base to emitter diode. This is also the reason why the time periods of such astable multivibrators tend to get shorter or the frequencies tend to get higher, the higher the operating voltage is. It is therefore not recommented to exceed 5 V operating voltage, especially in the case of larger capacitors (muliple 100 µF). The energy of the charge can evtl damage the base emitter diode.
I finally got it
How come all the videos on youtube using this circuit use only bipolar transistors? Why not use mosfets or igbts instead?
Because the whole circuit works on the principle of the base to emitter forward drop of 0.7V/ using RC timing..
Mosfet is different property
@Colin Mitchell Have you tried to build this circuit with a mosfet? I have with a IGBT and it works great.
How is this possible without damaging the capacitors? As I understand it C1 and C2 are switching polarity and the capacitors are polarized (often electrolytic capacitors are utilized in Astable multivibrators).
Ordinary aluminium capacitors can withstand a reverse voltage of about 1V to 1.5V without taking damage. Since the maximum reverse voltage is well below 1V everything is fine.
@@MirceaPricop Thanks. This is the answer that i was looking for. btw how did you get to know that? just by figuring out or someone told you? either way thanks.
@@kurtti1043 Haha glad it helped after such a long time. I remember being confused by the same problem and researching it online, it's stated on Wikipedia: en.m.wikipedia.org/wiki/Electrolytic_capacitor#Reverse_voltage
@@MirceaPricop If i change the values on capacitors C1 and C2 does it make any difference how long the leds are going to stay on? Can the capasitors have charge over 0.7V since the transistor Q1 starts to conduct 0.7V and C2 starts to discharge and C1 start charging. Can i make the discharging slower if i put resistors between emitter and ground? also if i want to change the chargin time i can change the 2 resistor in the middle, right?
@@kurtti1043 Yes, the duration each LED stays on is directly proportional to R3*C1 and R4*C2, so changing either capacitors or resistors in the middle will influence timing. I think adding a resistor on the discharge path will cause the output to fade out over time instead of instantly swap, but I never tried that out in practice
Cool but annoying circuit lol