Note that the first circuit involving nearly ideal opamp based integrator will source out sawtooth waveform of high quality on the output of the opamp.

The second circuitry will produce sawtooth-like waveform on the capacitor but distorted by usually quite apparent 1-exp(-t/RC) factor, however it does not voltage hysteresis circuit, so it is simpler.

thank you very much dear Nikolay Pavlov

Dear Tariq Kanaan sir,

Nikolay Pavlov sir gave a good explanation for the question. I will add one more point to it.

The noise immunity of first circuit will be higher than that of the second circuit due to the usage of bistable multivibrator.

Dear Tariq Kanaan ,

welcome!

The circuit in (a) needs two op amps and more circuit elements but its operation is more superior than the circuit in (b). The waveform control ability of Figure (a) is much better than in circuit (b) because of the linear integrator.

Best wishes

Thank you very much Dear prof. Abdelhalim abdelnaby Zekry

thank you Dear dr. Amit Das

Hello dear Tariq Kanaan ,

I can see that the respectable researchers have already given quite satisfying and informative answers to your question.

Nevertheless, I want to make a different contribution.

The second (bottom) circuit can still be used to obtain a triangular wave signal if we replace the resistor R with a current source controlled (switched) by the square wave output (i.e. the current source will also be square-wave; switched between a negative and a positive value). Being charged/discharged by a fixed current, the capacitor C will have a voltage signal increasing/decreasing linearly with time (i.e. triangular).

Note: Current mirroring may be needed to maintain a sourcing current for a positive vO and sinking current for a negative vO.

Best regards...

Both circuits use a bistable module. The difference is that in the second one this block is the op-amp itself.

The choice will depend on what you need from your circuit. If I need only a square wave, I would choose the second one because it is simpler. If I need both square and triangle wave, I have to choose the first one. If noise immunity is important, I should consider using the first one, as observed by Amit Das .

A lot of thanks and gratitude for your smart contribution all.

Antonio Carlos da Costa Telles Ali Zeki Abdelhalim abdelnaby Zekry Amit Das Nikolay Pavlov

thank you sir Ali Zeki

could you please add circuit diagram that used current source on it and mention any sources to read?

one more question please, how can we determine the maximum frequency of the out put signal for circuit in figure (a)?

best regard

thank you sir Amit Das

both circuits generate squares wave. so, could you please explain me why the circuit with bi-stable is more noise immunity than the circuit (b)?

and one more question please, how can we determine the maximum frequency of the out put signal for circuit in figure (a)?

Best Regards dear.

From an application perspective: the first circuit generates a square wave and a triangular wave and both of them are available at low impedance nodes (=almost ideal voltage sources). The second generates a square wave and a piecewise exponential waveform, and just the square wave is available at low impedance.

thank you dear sir Paolo Stefano Crovetti

could you please add circuit diagram that used current source on it and mention any sources to read?

one more question please, how can we determine the maximum frequency of the out put signal for circuit in figure (a)?

best regard

Dear Tariq Kanaan sir,

1) Yes both the circuits produces the square wave but the noise immunity in first circuit is more due to the usage of bistable multivibrator (Bistable multivibrator in general are more noise immune due to two threshold voltages: larger hysteresis voltage indicates more noise immunity).

2) The frequency of the generated square wave depends upon the circuit elements (R and C) but the maximum frequency of the opamp will be limited by the slew rate of the opamp.

Dear Tariq Kanaan ,

I am sorry for responding late. I was busy with some issues.

You asked me 2 questions. I will answer them below:

1) See the attached circuit schematics (I have mirrored the opamp vertically, to ease my drawings). Fig.1 and Fig.2 shows two conceptual methods to realize switched currents. I am sorry for the non-standard representations, but by using the "control" arrow from the output, I wanted to represent that the current source will be switched ON and OFF for HIGH and LOW vO levels, respectively.

(I used a circle attached to the arrow for the lower current source in Fig. 1, which aims to represent that the current source will be switched ON for LOW vO level, and vice versa)

In both Fig.1 and Fig.2, C is charged with a fixed current I and discharged with a fixed current I, as vO is switched between its HIGH and LOW levels.

I am suggesting a simple realization of Fig.1 in Fig.3. Assuming HIGH and LOW output levels close to VCC and -VEE, respectively, this practical realization can avail current source behavior of the pnp and npn transistors even when the capacitor voltage gets close to VCC and -VEE, respectively.

Besides this advantage, this implementation has at least 2 problems: i) the output resistances of the transistors may not be sufficiently large, so the currents may not remain fixed as C is charged/discharged, ii) the current values are beta-dependent. For an improvement in both, we can insert resistors in series with the emitter terminals. However, this will limit the max./min. voltages that the transistors can operate as current sources (due to entering saturation earlier). Nevertheless, a satisfactory trade-off point can still exist, especially if we don't aim a very large amplitude for the triangular wave.

Unfortunately, I cannot point out to a resource for these realizations. I remember I was solving similar circuits for the students when I was an assistant but I could not find a book where these realizations are implemented as I did here. As I mentioned, one can also use current mirrors (sometimes with an inverter). Actually there are many circuits which make use of extra digital circuits as well for this kind of implementations, which can be found in many books, but I aimed to keep the conceptual circuit simple (1 opamp + 2 transistors).

2) Regarding the max. output signal frequency of Fig.(a) of your schematics, I can say the rise/fall time of the hysteretical/bistable circuit will have one of the dominant effecta. On the other hand, the maximum output current available from the opamp may also be important, because the charging/discharging of C is performed by the opamp (Note: The same current is also drawn from the bistable circuit; so it can also be a bottleneck).

There are other issues like the bandwidth of the opamp, and the fact that opamp can be regarded as an integrator (because of dominant pole creation for stability), whose details are difficult to mention here.

As you see, these issues are usually difficult to answer very neatly with a couple of short answers only. It's easier when you come up with a specific circuit and a specific problem, along with some sufficient background about the topics. I am sorry if I made it more complicated but this is how it usually is. :)

Best regards...

Dear Tariq Kanaan

the differences between the two circuits were all mentioned in the previous answers.

Please find attached the solution of the circuit in (b) as an example of current controlled nonlinear bipole (with negative resistance) loaded with a capacitor.

Dual treatment can be carried out for voltage controlled nonlinear bipole (with negative resistance) with an inductor as a load.

The calculation are useful to find the period of the oscillation of the multivibrator (available also in https://en.wikipedia.org/wiki/Multivibrator), but also to question the alleged superior noise immunity of circuit (a).

As you can see from enclosed Figures the operation is completely equivalent to that of circuit in (a) except for the exponential voltage wave instead of the triangular wave.

Dear prof. Simone Orcioni , Thank you very much and I am very grateful for your great answer, but could you clarify some points for me please:

sincerely thankfully

Tariq Kanaan

> In page1, in your attachment, how do you solve the the differential equation?

> and how do you determine the shape of current?

The attached image image shows the part of document where I solved it.

It is a differential equation with costant coefficients and separable variables. It can be solved by direct integration.

Tariq Kanaan

> In page2, in your attachment, why do the v(t) shape is sawtooth and is different

> from current shape?

During the limit cycle, the current goes between i11 to i12 when the bipole is in the (1) region and between i22 to i21 when the bipole is in the region (2). The voltage is proportional to the current (the characteristic is picewise linear) and in the same regions it goes between the values v1 to v2 and v2 to v1, in the same regions of operation, as highlight in the attached figure.

Tariq Kanaan

> 3) according to Wikipedia link that you mentioned, we can implemented

> multivibrator circuits either by using transistor or by using op amp. in which

> application we use transistor to implement multivibrators and why?

The are many reasons, the same for which opamp have replaced discrete transistors in many others circuits, like amplifiers, filters, etc. ...

thank you dear Simone Orcioni

what really confused me is the plot of current vs time and how it instant changed for I12 to I22 and form I22 to I21 .but, after i reanalysis the circuit and equations i figure out that were boundary conditions.

thank you again sir Simone Orcioni

what i want to know by the third question is: does the multivibrator with transistor circuit faster than the circuits with op amp so that we can use in application with high speed like Radio Frequency circuits ?

best regards sir

Dear Tariq Kanaan

The most important difference is that in figure (a) you will have access to both triangle and square circuits simultaneously. It is important when you want to create a function generator. In the circuit of figure (b) you will have only square wave at the out put with the advantage of power reduction since only one operational amplifier is needed.

Best wishes

Nima

Read below.

Thank you sir Dr.Josef your contribution was really useful.

Best wishes

Dear Tariq Kanaan,

The primary difference between both circuits is how the capacitor charges and discharges.

In the 1st circuit, the capacitor is charging and discharging with a linear current, while in the 2nd circuit, the current is non-linear.

Best wishes.

thank you for all answers sirs.

dear Apurbaranjan Panda could you please give me the reason why n the 1st circuit, the capacitor is charging and discharging with a linear current, while in the 2nd circuit, the current is non-linear?

thank you very much