Astable Multivibrator Frequency Calculation

[foobag]

http://img840.imageshack.us/img840/3351/capturepak.jpg

is it me or is C, capacitance value supposed to be given for me to find out the frequency of oscillation?

 

[berkeman]

http://img840.imageshack.us/img840/3351/capturepak.jpg

is it me or is C, capacitance value supposed to be given for me to find out the frequency of oscillation?

Maybe they just want you to get an answer in terms of C? The value of C does affect the frequency.

 

[foobag]

ok in that case from my textbook

Fo (frequency of oscillation) = 1/(2R1C1*ln(1+(2R4/R3)) based on the circuit alone without the R2 and zener diode emphasis. My book says that the frequency depends only on the external components and unaffected by Vsaturation.

So my question now is the zener diode voltage that is given is essential in calculating the 2nd part of the question asking for Vamplitude and Vrms for the square wave, am I correct? Or does this affect the frequency of oscillation?

 

[berkeman]

ok in that case from my textbook

Fo (frequency of oscillation) = 1/(2R1C1*ln(1+(2R4/R3)) based on the circuit alone without the R2 and zener diode emphasis. My book says that the frequency depends only on the external components and unaffected by Vsaturation.

So my question now is the zener diode voltage that is given is essential in calculating the 2nd part of the question asking for Vamplitude and Vrms for the square wave, am I correct? Or does this affect the frequency of oscillation?

I would think that would depend on the value of Vzeners versus Vhysteresis. Can you just write the KCL equations for the circuit and show us the answers that you calculate?

 

[foobag]

Considering that the op-amp comparator is basically an inverting schimtt trigger

i found that Vhysteresis is = +/- (R4/(R4+R3))*Vsaturation

so Vhysteresis = +/- (10/18.7)*10 = +/- 5.35 Volts

Accordingly, Vzeners = +/- 7.5 limiting for the output of the circuit.

Does this mean, there will be no clipping, and that the output will saturate only to +/- 5.35 volts considering the inverting type schmitt trigger present?*EDIT, the Capacitance value was given by the prof as C = 47 nF so I have everything to calculate the frequency now I believe

 

[foobag]

any feedback?

 

[foobag]

could someone assist me with part 2 please? thanks

 

[berkeman]

i found that Vhysteresis is = +/- (R4/(R4+R3))*Vsaturation

That's not quite right. The Zener clamp comes between the saturated opamp output and the feedback for hysteresis...

 

[foobag]

That's not quite right. The Zener clamp comes between the saturated opamp output and the feedback for hysteresis...

does that imply I use the zener diode voltages as Vsaturation?

 

[berkeman]

does that imply I use the zener diode voltages as Vsaturation?

What does the combination of R2 and the ZD do to the output voltage?

 

[foobag]

well R2 is there to stabilize the zener diode output voltage with its value of 1k, which is considerably less than all of the other resistors

 

[berkeman]

well R2 is there to stabilize the zener diode output voltage with its value of 1k, which is considerably less than all of the other resistors

I would have used different words, but yeah, the series resistor drops the voltage difference between the opamp output and the ZD clamp voltage.

 

[foobag]

so does this answer the part about calculating the hysteresis, and 2nd part of the problem solving for Vamplitude?

 

[berkeman]

so does this answer the part about calculating the hysteresis, and 2nd part of the problem solving for Vamplitude?

Sorry, can you please refresh my memory on what questions you still have left? You know what the output voltage is going to be, and that is the input to the hysteresis voltage divider, right?

 

[foobag]

Sorry, can you please refresh my memory on what questions you still have left? You know what the output voltage is going to be, and that is the input to the hysteresis voltage divider, right?

right, but this time ZD voltage plays a role in calculating the input to the hysteresis voltage divider so Vout = (10/18.7)*7.5 = 4.01 volts amplitude

previously I used the op-amp saturation of 10 volts for the hysteresis. This implies there is no clipping of the square waves generated then, am I correct?

or RMS = 2.836 Volts?

 

[berkeman]

right, but this time ZD voltage plays a role in calculating the input to the hysteresis voltage divider so Vout = (10/18.7)*7.5 = 4.01 volts amplitude

previously I used the op-amp saturation of 10 volts for the hysteresis. This implies there is no clipping of the square waves generated then, am I correct?

or RMS = 2.836 Volts?

I'm not sure what you mean by clipping -- the output of the opamp/comparator is a square wave, as it the ZD clamped output waveform. You are correct that the output is a square wave of +/-4V (so Amplitude = 4V). How do you calculate the RMS value of a square wave?

And in finding the frequency of oscillation, you will need to factor in the hysteresis voltage and the output voltage, to find out how long it takes each half-cycle to recover from hitting the far hysteresis point until the comparison reaches the near hysteresis point. Sketching out the waveforms would probably help to be sure that you get it right.

 

[foobag]

I calculated the wrong RMS for a sine wave, for a square wave it is just the amplitude, am I correct, so both RMS and Amplitude are the same for a square wave.

as for the frequency oscillation, my textbook mentions that it depends only on external components, not the saturation voltage/hysteresis. is this wrong?

 

[berkeman]

I calculated the wrong RMS for a sine wave, for a square wave it is just the amplitude, am I correct, so both RMS and Amplitude are the same for a square wave.

as for the frequency oscillation, my textbook mentions that it depends only on external components, not the saturation voltage/hysteresis. is this wrong?

Correct about the RMS. For DC or for a square wave, the RMS value equals the amplitude (just the amplitude, not the +/- peak-to-peak double amplitude...).

When the hysteresis voltage is a small percentage of the overall oscillation, then it can usually be neglected. But when it is large, like in this question, I believe it has to lower the operating frequency some, because it takes longer for the voltage swing to reach the opposite trip point.

I could be wrong though, so it would be good if you sketched out the waveforms, with Vh labelled, and see if the wider Vh takes longer for the RC to slew between...

 

[foobag]

thanks for your feedback, i really appreciate it!