In fact, current Ib(Q2) is sum of currents of minority charge carriers Ie(Q2) and Ic(Q2),
earlier stored in base and flowing from base region under impact of reverse
base bias. At 8.5ns base region is devoid of stored charge and Q2 is OFF.

View attachment 251243

when vb goes down to -5v, why Vc not goes up to 6v?

 

when vb goes down to -5v, why Vc not goes up to 6v?

Capacitor C2 is charging through R4.
At 8.5ns, when Q2 is OFF, I(C2) = 27.1mA.
Then Vc(Q2) = V(V1) - I(C2) * R4 = 6000 - 27.1 * 200 = 580mV

ADDED:

I think Vc may go up as charging goes on.

Play million times slower and you will see it (right picture above).

 

Capacitor C2 is charging through R4.
At 8.5ns, when Q2 is OFF, I(C2) = 27.1mA.
Then Vc(Q2) = V(V1) - I(C2) * R4 = 6000 - 27.1 * 200 = 580mV
View attachment 251307

I think Vc may go up as charging goes on.

 

friends, it is said the solid line of circuit is discharging line, is it right? I think it is encharging of C1.

 

friends, it is said the solid line of circuit is discharging line, is it right? I think it is encharging of C1.

See slightly before 1.2s.
Q1 starts ON, C1 starts discharging, C2 starts charging.

 

See slightly before 1.2s.
Q1 starts ON, C1 starts discharging, C2 starts charging.
View attachment 251469

I see, thx very much

 

See slightly before 1.2s.
Q1 starts ON, C1 starts discharging, C2 starts charging.
View attachment 251469

when Vab goes down zero, does discharging turn to encharging?

 

when Vab goes down zero, does discharging turn to encharging?

that voltage across the capacitor briefly passes through zero, and then the cap does start charging in the opposite polarity very briefly. Of course, an actual capacitor may perform differently than a perfect simulation capacitor.

 

that voltage across the capacitor briefly passes through zero, and then the cap does start charging in the opposite polarity very briefly. Of course, an actual capacitor may perform differently than a perfect simulation capacitor.

thx. I think capacitor C1 for real ,it's right side need to be 0.7V to on Q2, as how the simulation do. so the difference is not much,am I right?

 

when Vab goes down zero, does discharging turn to encharging?

When absolute value of V(a,b) is decreasing, then C1 is discharging.
When absolute value of V(a,b) is increasing, then C1 is charging.

 

When absolute value of V(a,b) is decreasing, then C1 is discharging.
When absolute value of V(a,b) is increasing, then C1 is charging.
View attachment 251519

that's great. thx.

 

[QUOTE
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it is said the two transistor in the below circuit may be both in saturation and the circuit doesn't osilate, though of very low chance.


I wonder how the stableness comes into being, and how it holds. I mean if noise with positive feedback can break it.