Hallo Raghunathan,

the simplified schematic I posted earlier, had two disadvanatges: Feeback loop was too fast, and after turning on power supply voltage a voltage of 2.5V appears at output.

Meanwhile, I have improved the circuit a bit and I want to explain it in detail:




First, TL052 is much too fast. That's not necessary and it presents a disadvantage. Why? If the comparator toggles at output a sharp edge is fabricated, which can make trouble at the integrator: If a fast change (glitch) reaches the integrator, Opamp can not suppress it by the help of its local feedback loop and a certain portion of this glitch will again appear at the comparator. This means, that oscillation can occur! Because of this it's very important, that the feedback loop from comparator to integrator does not allow glitches from traveling here!!

The first measure against these glitches is to use a slow comparator. The TL031 shows a slew rate of about 4V/µsec. The TL051 is with 22V/µsec much faster! So, for the comparator TL031 is the better choice.
I have choosen such a TL031 for integrator, too, but here you can also use a TL051.

Another measure against glitches is the use of two separate OPamp for integrator and comparator, so that internal stray coupling does not play a role.

And third, a simple low pass filter behind comparator will also help to prevent glitches from travelling arround.

By the way, the same is valid of course for the connection of DAC to comparator! A DAC can produce relevant glitches, too, so even here a simple low pass filter is inserted. A low pass filter of 10kOhm and 100nF will show a time constant of 1msec. That's well below your DAC actualization rate of 50msec, so no negative effect will occur.

How does the circuit work now?
Assume, that output of integrator and output of DAC show the same voltage. Now, let's have the DAC make a change of +50mV. Then, comparator's output will toggle to -13.9V. This voltage is divided by the 8.2KOhm and 1.8kOhm resistors down to about -2.5V. This causes a current to flow into 2.2µF capacitor of 2.2µA. This current is driven by the output of integrator OPamp. As this current is constant, voltage drop across 2.2µF capacitor, and by this output voltage of integrator, will linearily increase, according to:

U = I x t / C

This formula can simply be found from C = Q / U = I x t / U.

If I = 2.2µA and C = 2.2µF, then U = t, means within a period of 50msec, voltage will increase by 50mV. So, this fits to your application.

Unfortunately, 2.2µF capacitor comes with some manufacturing tolerances of about 10%. So, a trim poti of 250kOhm is needed to adjust for this tolerance.

If the output voltage at integrator has reached the DAC output voltage and becomes greater than this, then the comparator will toggle back to +14.0V. This causes the 2.2µF capacitor now to become discharged by a current of 2.2µA. This will last up to the moment, when the comparator output toggles back to -13.9V again, and so forth and so forth.

As the comparator has got some hysteresis by the help of 10MOhm resistor, you will notice a very little sawtooth with amplitude of a few millivolts at output of comparator. But this only, when DAC output voltage does not change for more than 50msec. If the DAC output changes every 50msec, then, integrator will only show linear ramps, either positive ones or negative ones, assumed that trimpoti was adjusted properly.

Why this voltage divider at output of comparator?
Without this divider, which decreases output voltage to +-2.5V, part values of integrator had to be increased. But to work with bigger capacitor and/or resistor at integrator is not a good idea.

TL031 has a certain advantage over other OPamps: When having a load resistance of about 10kOhm at its output, output voltage is symmetrical! Means, comparator output is either +14.0V or -13.9V. So, both values are nearly identical. This is important, because rise time and fall time at output of integrator must be identical of course. With such an OPamp it's possible to use its output directly for drivening the integrator, without the need to insert some sort of bipolar switch, as shown with my earlier schematic.

At the output of this circuit you will notice another low pass filter. Time constant is 4.7msec. Its purpose is to round a bit eventually remaining edges of sawtooth. As 5 x 4.7msec = 23.5msec is less than 50msec (actualization rate of DAC), this filter will not have any effect on overall curve, but will only provide some 'allowed' additional smoothing.


Kai