Oscillators such as a Wein-bridge sinusoid or a square wave Schmitt trigger oscillator. Is it possible to bias these so they oscillate above 0V?
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\$\begingroup\$ Of course they can. \$\endgroup\$– Andy akaCommented Jul 3, 2014 at 11:42
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\$\begingroup\$ @tgun: Don't you mean below 0V? If you do, the answer is "yes". \$\endgroup\$– EM FieldsCommented Jul 3, 2014 at 11:47
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\$\begingroup\$ @EMFields No..as in if I only have a positive rail, can I make a waveform which oscillates around a positive voltage. \$\endgroup\$– tgun926Commented Jul 3, 2014 at 11:54
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\$\begingroup\$ @Andyaka could you give an example of such an oscillator, or how to modify a +ve and -ve rail one? Google revealed nothing. \$\endgroup\$– tgun926Commented Jul 3, 2014 at 11:55
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\$\begingroup\$ @tgun926 isn't that what most oscillators do? \$\endgroup\$– Phil FrostCommented Jul 3, 2014 at 11:55
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2 Answers
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Creating a single-supply version of a dual-supply circuit is fairly trivial - the key is to generate a pseudo-ground. Take a dual-supply oscillator of your choice. Replace the positive supply with one with twice the amplitude. Instead of a negative supply, ground these points. Now, create a pseudo ground by making a voltage divider with 2 equal resistors of, let's say, 1k to 10k. Buffer the output of the divider with an opamp follower. All of the ground connections of the original circuit get connected to this opamp, and you should be in business.
The buffer amp should be faster than the other opamps, and it should have a fairly high current capability, but other than that it's a straightforward proposition.
Note that the new oscillator output is referenced to pseudo-ground, rather than ground, and the amount of noise and distortion present if you reference it to ground will depend entirely on your ability to generate a clean, stable pseudoground. With this in mind, the lower leg of the voltage divider may profitably be replaced with a zener, or the entire divider with a good voltage reference. The reference does not have to be exactly half the positive supply voltage, but doing this will maximize the voltage swing you can get for the AC value of the oscillator.
answered Jul 3, 2014 at 12:08
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\$\begingroup\$ Silly me for only biasing the feedback path and not the remaining ground points of the circuit. \$\endgroup\$– tgun926Commented Jul 3, 2014 at 12:13
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\$\begingroup\$ (+1) wonderful and clear explanation. It would be perfect if you had time to add a schematic to show a beginner exactly what you mean. I wish I could upvote twice! \$\endgroup\$ Commented Jan 7, 2017 at 13:08
