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I've made triangular wave oscilator as shown in Fig. 1 and it works perfectly at frequancy 10 kHz with peak voltage +3 Volt to -3 Volt.

Figure 1. Oscilator circuit

I intend to generate SPWM. But when I feed this circuit to comparator, the output drops to 0 Volt. How can this happen?

enter image description here

I assume that output current of op-amp IC TL-082 is very small, not enough to feed comparator. If my assumption is correct, can I use transistor amplifier to produce enough current?

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    \$\begingroup\$ Th comparator should have very high impedance inputs, so the problem must be something else. Did you measured the waveform on the second opamp with comparator connected? how is the - input of the comparator connected? \$\endgroup\$ – Bruno Ferreira Oct 28 '14 at 9:45
  • \$\begingroup\$ The oscilator output is +3 to -3 Volt. When connected to comparator, the oscilator output drops 0 Volt. \$\endgroup\$ – smsidki Oct 28 '14 at 9:53
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    \$\begingroup\$ Have you tested the comparator alone? The TL082 should be abe to source/sink at least 1.4mA, that should be more than enough for the comparator. \$\endgroup\$ – Bruno Ferreira Oct 28 '14 at 10:49
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    \$\begingroup\$ You don't mention what comparator you're using, but feeding negative input signals into a comparator with only a positive supply is typically not acceptable. \$\endgroup\$ – Spehro Pefhany Oct 28 '14 at 11:07
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    \$\begingroup\$ What Spehro said! Does this oscillator work by railing the opamp? (What does it look like when it leaves the rail?) I wonder if a comparator would be better than the first opamp? \$\endgroup\$ – George Herold Oct 28 '14 at 11:44
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You say:

I use TL082 as comparator with negative supply (-Vcc) connected to ground directly.

In the datasheet for the TL082, under APPLICATION HINTS, it clearly states:

However, neither of the input voltages should be allowed to exceed the negative supply as this will cause large currents to flow which can result in a destroyed unit.

Your integrating opamp, IC(2/2), is using a bipolar supply, and the output signal swings both positive and negative. However, when you connect this signal to the third opamp, which has VEE connected to ground, the signal gets clamped at ground. This is because when it tries to go below ground, the third opamp allows a "large current to flow" (i.e., its input impedance drops dramatically). This stops the oscillation, because now the first opamp, IC(1/2), can't switch at -3V like it normally does.

The solution is to connect VEE of the third opamp to the same negative supply as the other opamps. Why didn't you do this to begin with?

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  • \$\begingroup\$ Already did. I got SPWM output with peak values +Vsat and -Vsat. Thank you. \$\endgroup\$ – smsidki Nov 4 '14 at 8:23

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