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I set up this astable multivibrator on a breadboard. Sometimes it works and flashes on and off it, and sometimes it does not flash on and off - the LED stays on.

Is there any reason or specific condition to trigger this on/off?

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Power is from three AA batteries (nearly 4.5 V.)

Updated on Nov 9, for difference in flash frequency in Multisim and breadboard (multisim) enter image description here

Updated on Nov 9, for difference in flash frequency in Multisim and breadboard (breadboard) enter image description here

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3 Answers 3

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Is there any reasons or specific condition to trigger this on/off?

The theory of operation is,

  1. At \$t=0\$ all transistors are off and capacitor is discharged.
  2. The 4.5 V battery and 470k resistor slowly applies a tiny current (\$\dfrac{4.5 V}{470k\Omega}=9.57\mu A\$) to the 9013 NPN transistor base by charging the capacitor. When the 9013 base reaches about 0.6 V, it has just enough current to "activate" and short its E-C pins. The capacitor has about 0.6V (in reverse) across it.
  3. The 9013 collector is now at battery negative potential, so pulls a large current* through the E-B junction of the 9012 PNP transistor. This also causes the 9012 to short its E-C pins, so full battery voltage is supplied to the LED and it illuminates. This current will be about \$\dfrac{4.5V-2V}{100Ω} = 25mA\$, so the LED will be very bright.
  4. The capacitor had (negative) 0.6 V across it, but now 4.5V is supplied from the right side 9012 - so the 9013 base is supplied 4.5-0.6 = 3.9V for an instant. This quickly depletes the capacitor, and the 9013 base is once again near 0.6 V. But this time, the LED is lit and *both transistors are dissipating current, so the battery voltage is lower than 4.5 V, so there isn't enough current supplied by the 470kΩ resistor to keep the 9013 conducting and it switches off. The cycle then repeats.

This circuit is not ideal in many aspects. First, it depends on the battery having a high-enough internal resistance that it's voltage will drop enough when "on" to allow the 9013 to "open" and the cycle to reset. Use a different type of battery (or a "fuller" or "emptier" battery) and it won't blink at all.

Second, when the 9013 turns "on", this allows a large current to flow through 9012 E to B, then 9013 C to E. This is essentially a diode junction-drop (in the 9012 E to B) so 0.8 V or so. While on, this "diode" is directly across the battery - so Amperes could flow given big batteries. Such current needs limiting, likely by adding a resistor. But adding said resistor will influence circuit operation...

So to answer your question, the design of the circuit is dependent on the exact transistors used, exact voltage and internal resistance of the batteries, temperature, sunlight, amount of wind, phase of the moon, etc. There is no single "fix" that will make this design work correctly 100% of the time for 100% of conditions. There is a closely related circuit though, the astable multivibrator which is much more robust. Such a multivibrator, combined with a lamp and two big 6V batteries, powered the blinking amber road barrels in the US for decades. Their operation is largely unaffected by environmental conditions.

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  • \$\begingroup\$ 1，I also made the closely related astable multivibrator as you advised, it works fine each time. However many circuits to be done has this complementary astable multivibrator, so I have to make this work if not 100%, but with higher success rate. So any advice to improve this success rate by replacing any component? \$\endgroup\$
    – ankyxia06
    Sep 28, 2022 at 5:08
  • \$\begingroup\$ In my experiment book. It includes this complementary circuit for a lot of experiments. So it is unavoidable. any way to achieve more successfully. Shall I replace the battery with constant current/voltage power source or add some decoupling capacitor or etc to realize it. looking forward to any good advice. \$\endgroup\$
    – ankyxia06
    Sep 29, 2022 at 9:51
  • \$\begingroup\$ thanks rdtsc, I worked this out with constant DC output source equipment, which is connected to household AC power) instead of batteries, and it worked everytime. and it also works fine for lithium batteries instead of these in the picture. both start to oscillate @2.8V. So the batteries selection is a crucial factor. \$\endgroup\$
    – ankyxia06
    Oct 8, 2022 at 9:38
  • \$\begingroup\$ hi rdtsc, regarding "The capacitor had (negative) 0.6 V across it, but now 4.5V is supplied from the right side 9012 - so the 9013 base is supplied 4.5-0.6 = 3.9V for an instant. This quickly depletes the capacitor, " I did simulation in Multisim and found 9013 never has 3.9V and capacitor never depletes capacitor during review video frame by frame. is it too fast to be video taped? can you give more details on this paragraph? \$\endgroup\$
    – ankyxia06
    Nov 7, 2022 at 10:21
  • \$\begingroup\$ Even simulations of this circuit will be tricky, as most simulators simplify some aspects for easier (and faster) numerical convergence. This circuit is dependent on things which are often omitted, such as battery internal resistance, leakage resistance, etc. But you can see in the answer of @Antonio51 the voltage across the capacitor as the red trace. This should be close to how it would look in reality. The only way to truly "see" this in operation is to use an oscilloscope. Such a tool would greatly assist development. \$\endgroup\$
    – rdtsc
    Nov 7, 2022 at 13:32
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I made some simulations to see all parameters of this "oscillator".
As pointed out by @rdtsc ... It shows all the phases of oscillation (including start phase).
Used a supply with a "ramp" rise for the starting phase.
Hope this help.

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  • \$\begingroup\$ hi Antonio, I have did connect to oscilloscope but could not see ramp rise at the beginning, any idea why, or any parameter setting could affect this, can you post your Oscilloscope setting? \$\endgroup\$
    – ankyxia06
    Nov 9, 2022 at 10:42
  • \$\begingroup\$ Ramp rise at starting is made by simulator. If you want exactly reproduced it in the real world, you must use a power supply with a 'programmable' rise time. As power supply current is 'low', an opamp ( something as an integrator ...) could be used. \$\endgroup\$
    – Antonio51
    Nov 9, 2022 at 10:55
  • \$\begingroup\$ For sure I did this in simulation, I can only see periodic abrupt rising but not the ramp rise at the beginning. \$\endgroup\$
    – ankyxia06
    Nov 10, 2022 at 10:20
  • \$\begingroup\$ Have you used a "step" generator as a power supply? Rise time (and some other things) can be defined in a step generator. \$\endgroup\$
    – Antonio51
    Nov 10, 2022 at 10:26
  • \$\begingroup\$ no, I used a DC power in simulation. \$\endgroup\$
    – ankyxia06
    Nov 14, 2022 at 11:02
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The problem looks familiar. The solution will be the same as in the previous case. 470 kΩ is at the lower end of the workability. First, we replace the 470 kΩ resistor with a larger one. The usable range will be 600 kΩ to 3.3 MΩ. Transient current limiting resistors are added to the base circuit of T1 and in series with the 47 uF capacitor. After changing the 470 kΩ, the frequency can be adjusted with the capacitor. With the modifications, the astable will be able to operate in a fairly large voltage range (from 3 V to 12 V).

At a higher voltage, the current of the LED will be proportionally higher, this must be taken into account. At 4.5 V, the LED current is 26 mA.

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  • \$\begingroup\$ please tell what Vc connects? and what connects to out? \$\endgroup\$
    – ankyxia06
    Nov 7, 2022 at 9:40
  • \$\begingroup\$ hi csabahu, I did on breadboard and virtually in Multisim (R1=1000k ohm), they both oscillate, but virtual circuit oscillates slower than quickly flashing(once less than 1 second) circuit in breadboard. what causes this difference in oscillation frequency? others are the same as your schematics. \$\endgroup\$
    – ankyxia06
    Nov 7, 2022 at 10:24
  • \$\begingroup\$ The voltages Vc and OUT are relative to ground. It would be nice to see your drawing exactly, because there couldn't be a big difference between the simulated value and the real one. I hope you changed the 680 kΩ resistor to 1 MΩ and 1kΩ remained unchanged. The parameters of the first (NPN) transistor may be important for the simulation. The color (forward voltage) of the LED also affects the frequency. \$\endgroup\$
    – csabahu
    Nov 7, 2022 at 13:59
  • \$\begingroup\$ updated my post with dwg and multisim snapshot. yes changed to 1M ohm. no 1k ohm after capacitor. sorry NPN and PNP brand is different since no 9012 or 9013 in database. \$\endgroup\$
    – ankyxia06
    Nov 9, 2022 at 10:37
  • \$\begingroup\$ The result of the simulation and the measurement will never be the same, if the transistors are different, the value of the capacitor is not accurately measured. The forward voltage of the used LED is not known. Since the frequency here depends on all parameters, the simulation can basically only be enough to determine the operability. In the case of precisely known components, the obtained result will be closer to reality. \$\endgroup\$
    – csabahu
    Nov 10, 2022 at 8:11

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