Serious error in that diagram! - you should have protection diodes on the transistor bases, with a 9V supply it's likely to cause reverse Vbe breakdown of the transistors.
No, it's the charged capacitors being switched across the 'be' junction that causes the problem - with a 9V battery you might just be lucky and they survive, but long term reliability is suspect. Certainly the LED's will help to some extent - but for two extra diodes it's a poor risk.
Thanks again. what about changing the supply voltage. Let it be just 3 volts.
is the problem still exist? (considering "Cry Wolf" asked for the simplest circuit)
Thanks again. what about changing the supply voltage. Let it be just 3 volts.
is the problem still exist? (considering "Cry Wolf" asked for the simplest circuit)
6V or lower would be fine - it's around 7V that transistors start to break down. For higher voltages just add 'catching diodes' across the base and emitter of the transistors, cathode to the base (for NPN transistors). These limit the reverse Vbe to 0.7V.
6V or lower would be fine - it's around 7V that transistors start to break down. For higher voltages just add 'catching diodes' across the base and emitter of the transistors, cathode to the base (for NPN transistors). These limit the reverse Vbe to 0.7V.
This will seriously screw up the circuit. The capacitors need to discharge through the base resistors. With shunt diodes added, the frequency will be much higher. Collector risetimes will also suffer.
I sometimes add one diode in series between the junction of the two emitters and GND. This works if you don't need good logic "0" levels. Alternately, you can put diodes in series with the bases.
Serious error in that diagram! - you should have protection diodes on the transistor bases, with a 9V supply it's likely to cause reverse Vbe breakdown of the transistors.
Well, OK, but a novice might want to be aware that the circuit posted above oscillates at about 8Hz with the diodes in series with the bases, and about 39Hz with them shunting the bases. Also, in the latter case, peak repetitive collector and diode currents can exceed one amp, depending on transistor beta - speaking of dead BC108s.
Transistor manufacturers publish datasheets that list the absolute maximum voltages and currents allowed for each transistor. The reverse-biased emitter-base junction of a BC108 transistor has a max voltage rating of 5.0V.
In the multivibrator circuit, with a 9.0V battery and 1.8V LEDs, the cathodes of the LEDs are at +7.2V. Each capacitor charges at a separate time with its positive end at +7.2V and the negative end at +0.6V, so the capacitor's charge is 6.6V. When a transistor conducts, its collector goes to about +0.2V which causes the capacitor connected to it to try to instantly have its positive wire at +0.2V and its negative wire at -6.4V, causing the emitter-base junction of the turned off transistor to have avalanche breakdown (like a zener diode) because the voltage exceeds the max allowed of 5.0V. The avalanche breakdown is bad for transistors and ruins them after some time.
We have been talking about adding diodes to the transistors to block the excessive negative voltage.
Thank you for explanation...
I think I should make some changes to the circuit:
-The power supply: 6V or less (I suggest 5 volts)
-Maybe we should change the old BC108 transistor (what about 2N2222) - or a newer equivalent..
Any advice to improve the circuit ?? (with explanation, please) Thanks..