Kevin Weddle
- Feb 23, 2004
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audioguru2
- Apr 6, 2004
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Hi Kevin,
Your circuit won't oscillate because it doesn't have any positive feedback. It also doesn't have a bandpass filter.
If you changed parts around and made it into a similar Wien Bridge oscillator, then it won't produce a sine-wave because its uncontrolled output level will clip.
Attached is a Wien Bridge oscillator. It is similar to your circuit except its RC networks are a bandpass filter and are in the positive feedback loop so that it oscillates at the peak frequency of the filter, which is also the frequency that the filter has no phase shift. Since the filter has a loss of 3 due to its equal-value resistors and capacitors, the gain of the circuit as determined by its negative feedback voltage divider must be a little more than 3.
The negative feedback can be controlled for a fixed output level below clipping level by making its resistor to ground a positive-temperature-co-efficient resistor (PTC) such as a small lightbulb.
Frequently Wien Bridge oscillators use a junction FET in the negative feedback loop for output level control. ;D
View attachment 36685
Your circuit won't oscillate because it doesn't have any positive feedback. It also doesn't have a bandpass filter.
If you changed parts around and made it into a similar Wien Bridge oscillator, then it won't produce a sine-wave because its uncontrolled output level will clip.
Attached is a Wien Bridge oscillator. It is similar to your circuit except its RC networks are a bandpass filter and are in the positive feedback loop so that it oscillates at the peak frequency of the filter, which is also the frequency that the filter has no phase shift. Since the filter has a loss of 3 due to its equal-value resistors and capacitors, the gain of the circuit as determined by its negative feedback voltage divider must be a little more than 3.
The negative feedback can be controlled for a fixed output level below clipping level by making its resistor to ground a positive-temperature-co-efficient resistor (PTC) such as a small lightbulb.
Frequently Wien Bridge oscillators use a junction FET in the negative feedback loop for output level control. ;D
View attachment 36685
Kevin Weddle
- Feb 23, 2004
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Audioguru, there is a good sinewave generated from my oscillator. There is a low pass and a high pass filter. I just altered the wein bridge to show how it could be done. My oscillator has 360 degrees feedback or it must. Remember that positive feedback can cause oscillation before 360 degrees. Mine oscillates somewhere close to 360 degrees as the capacitors and the low value resistors make the source close to 90 degrees away. That is a total of 270 degrees oscillation. That is close enough to 360 degrees for oscillation.
I even have an update to my circuit. I bumped up the amplitude with a transistor to 10Vpp.
I even have an update to my circuit. I bumped up the amplitude with a transistor to 10Vpp.
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Kevin Weddle
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Here is a circuit which takes the apparent sine wave to 10Vpp. The transistor is a TIP 31 NPN type.
View attachment 36687
View attachment 36687
Kevin Weddle
- Feb 23, 2004
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There is something interesting to point out here. Notice the output capacitor to separate bias. It is often desireable in a circuit to avoid the bias seperation. Sometimes it is enevitablely necessary.
Incidentally, you might want to change the 1kohm into a 10kohm if you experience clipping.
Incidentally, you might want to change the 1kohm into a 10kohm if you experience clipping.
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audioguru2
- Apr 6, 2004
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Kevin,
Aren't your "104" capacitors actually 0.1uF? At 60Hz, a 0.1uF cap has a capacitive reactance of 26.7K ohms. Therefore the 10M resistor across one doesn't change the phase, nor does the 500 ohm resistor across the other.
Your circuit is nothing more than a single-stage integrator circuit with a DC gain of 20,000. Single-stage integrator circuits don't oscillate. Maybe your "oscillator" is just amplifying the ripple voltage of its power supply.
The opamp's DC gain of 20,000 is interesting. If the opamp has an input offset voltage of 6mV, its limit, then the DC output of the opamp will try to be 120V! So the opamp's ouput will be saturated against a supply voltage, even if you used a "typical" opamp with an input offset voltage of only 2mV.
Your new transistor amplifier stage is also interesting. I wonder why you added it since if your opamp circuit did oscillate, it doesn't have anything to limit its output so it certainly wouldn't need an amplifier.
Without any negative feedback, the transistor's output will be very distorted. Also, since the current gain of transistors vary about 8 times from one batch to another, you can't predict the DC voltage of its output without having any negative feedback:
The 10K bias resistor has 11.3V across it providing 1.13mA base current.
1) A transistor having its minimum current gain of 38 will have a DC output voltage of +9.85V. It is almost cutoff.
2) A transistor having its maximum current gain of 300 will have a DC output voltage of -4.95V. It is almost saturated.
Aren't your "104" capacitors actually 0.1uF? At 60Hz, a 0.1uF cap has a capacitive reactance of 26.7K ohms. Therefore the 10M resistor across one doesn't change the phase, nor does the 500 ohm resistor across the other.
Your circuit is nothing more than a single-stage integrator circuit with a DC gain of 20,000. Single-stage integrator circuits don't oscillate. Maybe your "oscillator" is just amplifying the ripple voltage of its power supply.
The opamp's DC gain of 20,000 is interesting. If the opamp has an input offset voltage of 6mV, its limit, then the DC output of the opamp will try to be 120V! So the opamp's ouput will be saturated against a supply voltage, even if you used a "typical" opamp with an input offset voltage of only 2mV.
Your new transistor amplifier stage is also interesting. I wonder why you added it since if your opamp circuit did oscillate, it doesn't have anything to limit its output so it certainly wouldn't need an amplifier.
Without any negative feedback, the transistor's output will be very distorted. Also, since the current gain of transistors vary about 8 times from one batch to another, you can't predict the DC voltage of its output without having any negative feedback:
The 10K bias resistor has 11.3V across it providing 1.13mA base current.
1) A transistor having its minimum current gain of 38 will have a DC output voltage of +9.85V. It is almost cutoff.
2) A transistor having its maximum current gain of 300 will have a DC output voltage of -4.95V. It is almost saturated.
