transistor Oscillator

[muhammad59]

how C2 chargie through R3.while R3 connect positive (+v) ....
i mean R4 can charge capacitor(C2) because positive plate on R4 side not on R3 side

https://www.dummies.com/how-to/content/electronics-components-oscillator-circuits.html

 

[MikeMl]

Note I use different numbering for the resistors and capacitors.

I purposely make C2=2*C1, so you can see the different time constants of (R2+R3)*C2 vs (R1+R4)*C1.

Look at V(b1). When the collector V(c2) switches high-to-low, that causes V(b1) to be forced to ~-5V, so Q1 is turned off, meaning that it's collector V(c1) is pulled high. C2 is discharging through R2 in-series with R3, so the discharge time-constant is (R2+R3)*C2. V(b1) continues to discharge (and then begins to charge) towards V(b1)=0.65V, at which point Q1 begins turning-on, causing V(c1) to rapidly decrease, which is coupled to the base of Q2 V(b2) through C1, causing Q2 to begin turning-off, and the multivibrator flips to the other state.

Since C1 is smaller than C2, it takes about half the time for it to flip again back to the first state.

Since R3<<R2, and R4<<R1, you can just about ignore R3 and R4 when computing the two-time constants. Normally, R1*C1=R2*C2, but I made those asymmetrical to show which time constant is which.

Follow the voltages at V(b1), V(b2) and V(c2). Can you predict what V(c1) would look like?

 

[AnalogKid]

how C2 chargie through R3.while R3 connect positive (+v) ....
i mean R4 can charge capacitor(C2) because positive plate on R4 side not on R3 side

The charge in a capacitor changes (increases or decreases) when the voltage across its two terminals changes. There is nothing special about either end. Just because a capacitor has a preferred polarity to the charge has nothing to do with how the charge gets in there.

ak

 

[spec]

Not fundamental to the question, but this circuit (differentiating multivibrator) suffers from a number of problems. The two main ones are:

(1) Tendency to exceed the VEBO of the transistors. In this case the maximum allowable VEBO for the 3N3704 is -6V which equals the magnitude of the supply line.

(2) If the supply line rise slowly at turn-on, both transistors will turn on and saturate and the circuit will not work.

spec

 

[MikeMl]

...
(2) If the supply line rise slowly at turn-on, both transistors will turn on and saturate and the circuit will not work.

Ah, the "metastability" argument. Because of the thermal noise and gain in the circuit, it is extremely unlikely that it will happen...

 

[spec]

Ah, the "metastability" argument. Because of the thermal noise and gain in the circuit, it is extremely unlikely that it will happen...

For a slow supply line increase it is bound to happen. Once the transistors saturate there is no gain and that is that- the third state.

The integrating multivibrator is much better behaved and does not have a third stable state.

spec

 

[AnalogKid]

The integrating multivibrator

Integrator and comparator in a loop?

ak

 

[spec]

Integrator and comparator in a loop?

ak

Yes, that is what I meant. In the differentiating multivibrator there are effectively two comparators, the VBEs of the two transistors. You can make an integrating multivibrator with transistors as well as comparators, opamps, and DIACs etc.

spec

 

[MikeMl]

For a slow supply line increase it is bound to happen. Once the transistors saturate there is no gain and that is that- the third state....

How slow do you want the power supply ramp up? This is the absolute worst case: R1=R2+-0, R3=R3+-0, C1=C2+-o, Q1=Q2 perfectly, no noise except a bit of math noise due to floating point math, and I am ramping up V(cc) very slowly. Anything that unbalances a "real" circuit makes it even more likely that it will start oscillating. Note it starts right at ~2.8s (when V(cc)=~600mV, regardless what I do to the circuit (within reason).



If you can build a multi that doesn't start when power is applied, I want to see it...

 

[spec]

How slow do you want the power supply ramp up? This is the absolute worst case: R1=R2+-0, R3=R3+-0, C1=C2+-o, Q1=Q2 perfectly, no noise except a bit of math noise due to floating point math, and I am ramping up V(cc) very slowly. Anything that unbalances a "real" circuit makes it even more likely that it will start oscillating. Note it starts right at ~2.8s (when V(cc)=~600mV, regardless what I do to the circuit (within reason).

View attachment 99234

If you can build a multi that doesn't start when power is applied, I want to see it...

Hmm. Try with base resistors of 10K rather than 100K. 100K does not really give enough base drive to saturate the 2N3704 according to the data sheet.

spec

 

[MikeMl]

Hmm. Try with base resistors of 10K rather than 100K. 100K does not really give enough base drive to saturate the 2N3704 according to the data sheet....

When Rb = 10*Rc, I consider the transistor saturated...



ps, I just noticed that C2=2*C1. I made them equal, and that has only the slightest effect on the start-up behavior. It still has started every time....

 

[spec]

My error Mike. I missed the circuit in your sim (10K base resistors). Instead I was going by the circuit of your #2 post (100K base resistors). It would be interesting to try the circuit in real life. It is my contention that with slowly rising power lines that both transistors will saturate, and thus have no voltage gain. And due to the slow rise of the supply line, no significant voltage will be fed to the bases by the capacitors.

As a separate point, do you agree that there is a stable state where both transistors are fully on and saturated?

spec