Help Figure Out This Circuit?, series multivibrator

[Roff]

How it works

Assume that initially, both capacitors are discharged (zero volts across them). Both transistors will be off. As the capacitors charge through their respective series resistors, the base-emitter junctions and the diode will eventually become forward-biased. Collector currents will cause the collector voltage of Q1 to drop, and the collector voltage of Q2 to rise. Each of these couples to the base of the other transistor through a capacitor, causing the transistors to turn on even harder (positive feedback). Both transistors saturate, discharging the capacitors rapidly. Transistor current gain is too low in saturation for this state to be sustained, so the transistors begin to turn off. Again, positive feedback causes them to snap off rapidly, and the cycle starts over.

 

[ericgibbs]

I have to practice my English!

I tried what Ron said (tampered the capacitors a bit to make the sawtooth last longer) and it works well. I might add it as an "Appendix". I have attached the output from the PNP emitter.

Could you explain to me how this circuit actually works? (capacitors charging, then transistor goes on, etc)? Thanks

Do you think i suggest taking the output between the NPN collector and the PNP base (it was the best graph) or would it be 'theoretically' wrong?

hi,
What are you trying to do, generate a square wave of equal mark/space or a sawtooth waveform of equal ramp up/down.?

What voltage level do you require.?

 

[Roff]

hi Ron,
I expect he will have to buffer any 'off circuit' load, so that he dosn't load the timing.

But what I meant was, you won't even see sawtooths at the emitters unless you add resistors as I described.

 

[ACHIEVEIT]

hi,
What are you trying to do, generate a square wave of equal mark/space or a sawtooth waveform of equal ramp up/down.?

What voltage level do you require.?

It is an exercise i have to do for a university course. That's why i asked about the step-by-step procedure.

 

[Roff]

It is an exercise i have to do for a university course. That's why i asked about the step-by-step procedure.

Who chose the circuit you are using? Was it you, or the professor?

 

[ACHIEVEIT]

But what I meant was, you won't even see sawtooths at the emitters unless you add resistors as I described.

The only problem i noticed when adding the resistors was that the pulse from the collector almost turned into a sawtooth as well. All in all, what would you say this circuit does? It produces pulses (more like "impact" pulse - is this how it's called?) but with a slight modification it can produce sawtooth? What about the sawtooth between NPN collector - pnp base? Is it any good?

 

[ACHIEVEIT]

Who chose the circuit you are using? Was it you, or the professor?

The professor and i guess that he just took some circuits from that web site because each student gets a different exercise and we are more than 50.

 

[ericgibbs]

The only problem i noticed when adding the resistors was that the pulse from the collector almost turned into a sawtooth as well. All in all, what would you say this circuit does? It produces pulses (more like "impact" pulse - is this how it's called?) but with a slight modification it can produce sawtooth? What about the sawtooth between NPN collector - pnp base? Is it any good?

Adding a resistor in the emitter circuit will slow the capacitor discharge period, so the square will become a 'sawtooth' shape.

 

[ericgibbs]

hi,

As a simple exercise, remove the diode connecting the emitters and connect a 2K7 resistor in place of the resistor, this should give an approx equal mark/space of higher frequency.
You can increase the value of the two capacitors to reduce the freq.

 

[ACHIEVEIT]

Thanks for the explanation.

Something in general, do you think analog circuit design has good job opportunities nowadays? I am into tube amplifiers, i read books whenever time allows but i know that they are obsolete with only a few uses (guitar amps for example). Is analog circuit design any good these days or should i go with something like vlsi?

 

[ACHIEVEIT]

hi,

As a simple exercise, remove the diode connecting the emitters and connect a 2K7 resistor in place of the resistor, this should give an approx equal mark/space of higher frequency.
You can increase the value of the two capacitors to reduce the freq.

You mean in place of the diode?

 

[ericgibbs]

You mean in place of the diode?

Yes for a test,

I do not use a circuit simulator.

 

[ACHIEVEIT]

I tried it but i don't know if that's correct. I have attached the result for you to see. Is this what you expected? The blue line is the emitter of the npn and the green the emitter of the pnp

 

[ericgibbs]

I tried it but i don't know if that's correct. I have attached the result for you to see. Is this what you expected? The blue line is the emitter of the npn and the green the emitter of the pnp

It looks as though its so high a freq that the simulator cannot resolve it

Increase the value of both caps by a factor of 10

Measure the collector not the emitters

 

[Roff]

The only problem i noticed when adding the resistors was that the pulse from the collector almost turned into a sawtooth as well. All in all, what would you say this circuit does? It produces pulses (more like "impact" pulse - is this how it's called?) but with a slight modification it can produce sawtooth? What about the sawtooth between NPN collector - pnp base? Is it any good?

I suspect he wants the sawtooth with respect to ground, and not as a differential signal as you are proposing (NPN collector - pnp base). The bases do have sawtooth waveforms. They just have pulses added to them.

 

[ACHIEVEIT]

It looks as though its so high a freq that the simulator cannot resolve it

Increase the value of both caps by a factor of 10

Measure the collector not the emitters

I changed the caps to 1u and even 10u but the result was exactly the same. Am i doing something wrong?

 

[Roff]

hi,

As a simple exercise, remove the diode connecting the emitters and connect a 2K7 resistor in place of the resistor, this should give an approx equal mark/space of higher frequency.
You can increase the value of the two capacitors to reduce the freq.

The transistors still saturate, so the resistor does not effectively get multiplied by beta, as I surmise you were thinking. In fact, in my simulation, the circuit won't oscillate if that resistor is above ≈2.2k.

 

[ACHIEVEIT]

I suspect he wants the sawtooth with respect to ground, and not as a differential signal as you are proposing (NPN collector - pnp base). The bases do have sawtooth waveforms. They just have pulses added to them.

Yes you are right. He told me though that i should expect sawtooth from the collectors. That was misleading as well but i guess he must have taken it for another circuit (he selected so many after all) + he didn't really have time when we spoke

 

[ACHIEVEIT]

Guys, thank you so much for your help.

 

[ericgibbs]

The transistors still saturate, so the resistor does not effectively get multiplied by beta, as I surmise you were thinking. In fact, in my simulation, the circuit won't oscillate if that resistor is above ≈2.2k.

I am aware what the effect of saturation of transistors will have on β, the suggestion of a 2K7 resistor test, was to make the discharge paths the same, in order to produce an approx equal mark space ratio so that the OP could observe the effect of the diode in/out.

It would be then an easy task to increase the values of the two timing caps to give a freq required by the OP.

Can you say what the mark/space result was with < 2K2 and the *10 caps, I dont use a circuit simulator.