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Recently, i see such a circuit in a board schematic, but can't figure out how it works, why the first Op Amp has positive feedback ????

enter image description here Thanks.


Question, again, now i complete the circuit with part number and values, just as below.

Triangular wave generator with frequency synchronization????

By simulation with PSpice, i know the "AC coupled" negative input of the comparator is to synchronize the oscillation frequency of the generator. But it seems there is one 'working window', if the pulse rate is two high, the output wave form will be oscillation on the amplitude.

I want to know if this a 'classic' or a widely used technique. And i want to know if there are some accurate mathematical method to predicated the behavior, particularly the synchronizable frequency range.

Thanks.

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  • \$\begingroup\$ Without resistor values you can't tell if the feedback is positive or negative or whether it changes sign, the latter depending on the value of \$C_1\$ (AND the resistors) \$\endgroup\$ – Vladimir Cravero Apr 19 '14 at 9:19
  • \$\begingroup\$ this thing might also be a Vco but then what is \$C_2\$ there for? \$\endgroup\$ – Vladimir Cravero Apr 19 '14 at 9:24
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Probably:

It's a fixed (approximately) frequency PWM or pulse width ratio modulator with high/low ratio of IC2_out controlled by Vin_AC voltage level. [See below for IC2 defn].

Certain conditions need to be met. See below.


Because:

lh (=left hand) opamp = IC1
rh opamp = IC2

Set Vin = 0 initially. Assume IC1 output high initially and V_C1 = 0V.

IC2 is a classic integrator.
As +ve current enters inverting node IC2_-, the opamp will supply -ve current via C1 to hold IC2_- at zero so IC2 output will ramp negative.

IC1_+ is at V caused bt IC1_out and IC2_out weighted by R5 and R4. Make R4 such that IC2_out will dominate IC1_out before IC2_out reaches max negative. For identical opamps with symmetrical vsupply and symmetrical V+ and V- swing then R4 is < R5 for this to occur.

As IC2_out ramps negative a stage is reached where IC1_- goes for >0 to <0 and IC1 toggles. R5 provides hysteresis (+ve feedback so the system starts back in the other direction until other extreme is reached and system toggles back.
So far we have a fixed frequency oscillator with square wave at OIC1_out and triangle wave at IC2_out.

Now apply an offset voltage to Vin.

Sav Vin is 0.1Vdc at left end of R6 to start.
The switching point of IC1 is now offset from ground. so the relative influences of the ramp and feedback voltage are asymmetric relative to the switching point. The relative charge and discharge times vary and the mark-space ratio varies.

Now instead of a fixed DC input via R6 apply AC at Vin proper and the mark-space ratio of the "oscillator" will vary with Vin. Ifg the period of the oscillator is short relative to the mean period of the Vin signal you have a PWM signal that tracks Vin amplitude.

Vout1 is a square wave.
Vout2 is a triangular wave.
Usually Vout1 would be the main output.

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\$IC_1\$ is a configured as a comparator, \$IC_2\$ is configured as an inverting integrator.

The output of \$IC_1\$, \$v_{OUT1}\$ is either \$V+\$ or \$V-\$.

When \$v_{OUT1} = V+\$, the output of \$IC_2\$, \$v_{OUT2}\$ is a decreasing ramp.

The voltage at the non-inverting input of \$IC_1\$ is given by

$$v_{+,1} = v_{OUT1}\frac{R_4}{R_4 + R_5} + v_{OUT2}\frac{R_5}{R_4 + R_5}$$

With \$v_{OUT2}\$ ramping down, \$v_{+,1}\$ is ramping down and, after some time, becomes less than the inverting input \$v_{-,1}\$ at which time \$v_{OUT1}\$ rapidly changes to \$V-\$ making \$v_{+,1}\$ more negative.

Now, \$v_{OUT2}\$ begins ramping up and so, \$v_{+,1}\$ begins ramping up and, after some time, becomes more positive than \$v_{-,1}\$ at which time \$v_{OUT1}\$ rapidly changes to \$V+\$ making \$v_{+,1}\$ more positive and the cycle repeats.

The ratio of the time \$v_{OUT1}\$ spends at \$V+\$ to the time spent at \$V-\$ depends on the voltage at the inverting input.

So, this is a PWM modulator circuit.

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