Hello people. I am trying to understand circuit before I embark on constructing them. Please the explanation in the circuit in the attached is that the input of the triangle-sine wave convert must be \[ \frac{\pi*5U_d}{2} \]. Who can explain the reason for such assertion? Thanks in advance

 

There are 5 pairs of diodes that limit the triangle peak voltages to resemble a sinewave.

 

There are 10 diodes in the sine shaping circuit, five for each half-cycle (polarity) of the waveform. For each polarity, the five diodes interact with the three resistors such that as the output voltage gets higher and higher, the diodes conduct one at a time, effectively adding the resistors in parallel to reduce the gain of the circuit.

The image is so small that the resistor values are difficult to read. At low voltages, R14 and R10 set the gain at 1. When the output voltage is above three diode-drops (approx. 1.8 V), the left three diodes conduct and pull R13 into being in parallel with R14. This reduces the gain of the circuit for this portion of the triangle wave, decreasing the slope seen at the output. When the output voltage increases another 0.6 V, this D10 conducts current through R12, decreasing the circuit gain more, and reducing the output slope more. With even higher output voltage, D12 starts to conduct around R12, decreasing the gain some more and flattening out the output slope.

Thus, the output signal is a pseudo sinewave made up of a total of 12 straight-line segments per full cycle.

Please post larger images of each page. If possible, a link to the original article.

ak

 

Certainly the non-linear portion of the diodes going into conduction is used to shape the triangle wave closer to a sine wave. But proof in a math manner would be a challenge because it will not be a pure sine wave, only "fairly close". Generating a math proof for an incorrect statement is really difficult.

 

Certainly the non-linear portion of the diodes going into conduction is used to shape the triangle wave closer to a sine wave. But proof in a math manner would be a challenge because it will not be a pure sine wave, only "fairly close". Generating a math proof for an incorrect statement is really difficult.

There are 10 diodes in the sine shaping circuit, five for each half-cycle (polarity) of the waveform. For each polarity, the five diodes interact with the three resistors such that as the output voltage gets higher and higher, the diodes conduct one at a time, effectively adding the resistors in parallel to reduce the gain of the circuit.

The image is so small that the resistor values are difficult to read. At low voltages, R14 and R10 set the gain at 1. When the output voltage is above three diode-drops (approx. 1.8 V), the left three diodes conduct and pull R13 into being in parallel with R14. This reduces the gain of the circuit for this portion of the triangle wave, decreasing the slope seen at the output. When the output voltage increases another 0.6 V, this D10 conducts current through R12, decreasing the circuit gain more, and reducing the output slope more. With even higher output voltage, D12 starts to conduct around R12, decreasing the gain some more and flattening out the output slope.

Thus, the output signal is a pseudo sinewave made up of a total of 12 straight-line segments per full cycle.

Please post larger images of each page. If possible, a link to the original article.

ak

This is the link to the article https://magzdb.org/file/599413/dl

 

This is the link to the article https://magzdb.org/file/599413/dl

I there an English version of the site?

ak

 

I there an English version of the site?

ak

When I go to the link then Google pops up and asks if I want Russian or English. But you must login to the site to see or download stuff.

 

When I go to the link then Google pops up and asks if I want Russian or English. But you must login to the site to see or download stuff.

When I click on the URL it bring up the English version of the magazine to download.

 

Nothing is wrong with the site

 

Nothing is wrong with the site

Nothing wrong with the site? If I understood Russian text that would be true. Th fact at is that I do not understand it one bit.
I look at the screen capture and it is a very simple circuit and it is a simple square wave generator with shaping circuits following. I also see that the rest is not shown. explanation for the very specific amplitude requirement is that the distortion level depends on the diodes operating in the very non-linear portion of their forward conduction curve. What we do not get to see is that the forward current/voltage relationship is also dependent on the temperature. of the diodes. Thus the adjustment for the best approximation of a sine wave is closest at one specific temperature. In addition, that precisely set amplitude feeding the diode sine-shaper depends on the power supply voltage, and we do not see the power supply portion of the circuit at all.
So, given that a correctly adjusted WEIN BRIDGE oscillator is able to produce a quite accurate sine wave, I suggest that if a low distortion sine wave is required, the circuit presented is not the best choice. Consider that the best case sine wave distortion is only given for one frequency, not a spread of frequencies.

 

In addition, that precisely set amplitude feeding the diode sine-shaper depends on the power supply voltage

Don't think so.

The triangle wave amplitude is set by a pair of clamping diodes, a gain-of-5 amplifier, and hysteresis resistors R2 and R3. This is an improvement over the standard low-cost function generator circuit. The article discusses hand-selecting diodes with matched Vf to improve symmetry between the positive and negative half-cycles.

The power supply is a 7815 and 7915, with decoupling caps at each IC. Not great, but good enough for this.

ak

 

I see nothing at all controlling the output of either section of IC2. The square wave amplitude may be stable, but as the supply voltage varies the outputs of the IC2 . And without seeing the power supply we do not know if the positive and negative outputs track. Nor can we consider if te filtering is adequate

 

Thanks guys for your inputs. My pain point is not yet understand. I am trying to gain understanding about where the pi came about in my first post. I am studying underlying theory.

 

During a half-cycle, the voltage at the IC1a output is +/- 5 x diode Vf, about 2.5 V according to a Vishay datasheet. These go through the frequency adjument and drive the IC2b integrator. These drive the hysteresis loop around the IC1b-IC1a comparator. Nothing in the circuit is referenced to either of the supply rails. Everything is referenced either to GND or diodes D1 and D2.

The output of IC1b bangs up and down between its rails, and that voltage will change with regulator performance. But that is not the voltage going to the integrator. It is hard-clipped at +/-Vf about GND. Current through the diodes is around 1 mA, well away from the curvy part of its conduction curve. If one of the IC1b rail voltages wanders around 0.1 V, that is a change in diode current of only 10 uAand it will have a very small effect on the clipping diode's Vf. Again, for a basic circuit like this it seems fine to me.

ak

 

Thanks guys for your inputs. My pain point is not yet understand. I am trying to gain understanding about where the pi came about in my first post. I am studying underlying theory.

For a sine wave with a positive slope going through its zero-crossing at t=0, the slope of the sine wave at that point is 1 This is one of the defining characteristics of a sine wave. A tangent line at this point also has a slope of 1, and its y value increases steadily going to the right.

As shown in fig. 2, the slope of the triangle wave matches the slope of the sine wave at the zero crossing. Because it has a slope of 1, then at t = pi, the y value of this line is pi. Thus, at 90 degrees, the peak of the sine wave, t = pi/2 so the y value of the tangent line, and thus the value of the triangle wave, is pi/2.

ak

 

For a sine wave with a positive slope going through its zero-crossing at t=0, the slope of the sine wave at that point is 1 This is one of the defining characteristics of a sine wave. A tangent line at this point also has a slope of 1, and its y value increases steadily going to the right. Because it has a slope of 1, then at t = pi, the y value of this line is pi. Thus, at 90 degrees, the peak of the sine wave, t = pi/2 so the y value of the tangent line, and thus the value of the triangle wave, is pi/2.

ak

I will look at this tomorrow. Thanks