# Is this an astable multivibrator?

In this circuit the LED flashes on and off, however could it be classified as an astable multivibrator?

• Do you know what an astable multivibrator is? – dirac16 Nov 5 '17 at 5:35
• yes but a lot of them seem to all mirror one particular set up, so I wasn't sure whether you could have something else and still be one. – Lisa Ban Nov 5 '17 at 5:47
• Lookip def’n: astable – Tony Stewart Sunnyskyguy EE75 Nov 5 '17 at 5:52
• If the LED is typical 20-mA one, then it will flash. Just once, because it will pop. – next-hack Nov 5 '17 at 8:04
• Yes, it oscillates so YES it is an astable. || That cct is usually intended to operate on Vbat in 1.5 - 3v range. As next-hack implies, the LED on-current will be high. It may still operate but will be "suboptimal". – Russell McMahon Nov 5 '17 at 10:49

This circuit is a complementary astable multivibrator. In fact, the circuit is just a two-stage command emitter amplifier with positive feedback add by the capacitor.

At the beginning of a cycle, $Q_1$ opens via $R_1, R_2$ base current. $Q_1$ collector pull to low the $R_3$ resuistor and $Q_2$ base. So, $Q_2$ base current starts to flow. And this opens $Q_2$ transistor and $Q_2$ will try to pull his collector high (to Vsupply). At the same time $C_1$ starts to charge (very quickly, because of a very large charging current, almost a short circuit).

And the capacitor will be charged until his voltage reaches around $V_{LED} - V_{BE1} \approx 3V - 0.7V \approx2.2V$

At the end of a charging phase$C_1$ capacitor (Charge=0A) no longer provide any current to the $Q_1$ base. The $Q_1$base current is now only coming from $R_1$ and $R_2$ resistors. And this current is too low to maintain saturation in $Q_2$. And this is why $Q_2$ immediately after $C_1$was fully charged comes out from the saturation region into the linear region. And voltage at $Q_3$ collector starts to drop. And this drop in $Q_2$ collector voltage is feed back via $C_1$ onto the $Q_1$ base. And the voltage at the $Q_1$ base drops from $0.7V$ to $-2.2V$. And this negative at the $Q_1$ base immediately cut-off $Q_1$ and $Q_2$.

$C_1$ capacitor starts the discharge phase via $R_1 , R_2 , D1$

And when the capacitor voltage reach $0V$ ($0V$ at $Q_1$ base) the capacitor will start the charging phase but this time in opposite direction (in reverse direction, but the charging current will still flow in the same direction as before). The voltage at $Q_1$ base starts to increases from $0V$ towards $Vsupply$ due to the $C_1$ capacitor charging. And as this voltage reaches $Q1$ transistor forward voltage ($0.6V$) $Q_1$ starts to conduct the base current. So $Q1$ and $Q2$ opens again. And the cycle repeats itself.

• Now compute how likely it is that this circuit will work with randomly selected junkbox TO92 NPN BJTs, if you have to replace the pot with a fixed resistor. ;) – jonk Nov 5 '17 at 17:27
• @jonk Good question. But in real life, it will be better to use your version of the circuit. – G36 Nov 5 '17 at 19:02