Simple question: how to find the frequency of oscillation of this circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

My attempt at answering the question is the following:

Suppose we are observing the circuit at when the voltage on the capacitor is zero. Q1 is ON and Veb is equal to 0.8 roughly. Applying KVL and neglecting base current through R2 (assume voltage at R2 equal to 10/11*9=8.18V) we get that voltage drop on R5 should be 0.02 V which implies an emitter current of Q1 of about 1.81 micro ampere. Assuming the collector current is roughly equal to the emitter current we can see that the capacitor needs 0.295 seconds to get to a voltage 4.5+0.8V which is going to turn on Q2 and Q3 for discharge. But that time gives me a frequency of about 3.43 Hz which is 1 order of magnitude off of what I read on the oscilloscope (45 Hz). Why is this so?

EDIT: Vcc is 9V

  • Well. Could you double-check your calculation for the \$Q_1\$ emitter current? I think you might be off by around an order of magnitude. I find closer to \$25\:\mu\text{A}\$ (assuming less than \$600\:\text{mV}\$ of \$V_\text{BE}\$ drop.) – jonk Aug 18 at 20:31
  • Yes, assuming less than 600mV things check out, I get a frequency which is reasonably close to the one measured. With 600mV I get about 41 Hz, if it is less then the frequency goes up. I think I should measure the base emitter voltage drop. – mickkk Aug 18 at 20:41
  • Well, there is no possible way a BJT will present \$700\:\text{mV}\$ when the emitter current is so very, very low. Even if it were \$700\:\text{mV}\$ (which it will not be), I'd still get \$12\:\mu\text{A}\$ and not that tiny number you posted. I'm pretty certain you were an order of magnitude off in your estimate of the voltage drop across \$R_5\$. – jonk Aug 18 at 20:44
  • I measured the voltage drop on the base emitter and indeed it is 513 mV which, in combination with your help, explains everything. Thanks! – mickkk Aug 18 at 21:02
  • I guess Vcc=12 I get Ic= 56uA and 51Hz on Falstad but very sensitive to Vcc – Tony EE rocketscientist Aug 18 at 21:06

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