# RC oscillator amplitude?

I'm trying to make this sine wave oscillator, but I want the amplitude of the voltage sine wave to be 4.8V and the output current amplitude to be 45A. I am not sure how to go about calculating for the amplitude for either the current or the voltage. Has anyone had some experience with these?

• What frequency should be achieved with this? Why do you need such a high peak current? – Grebu Dec 28 '16 at 10:33
• Hum, do you really mean 45 ampere? – pipe Mar 29 '17 at 10:05
• @meagh: If you specify the output voltage (BTW it's not clear if you mean $V_{pp}$ or $V_{max}$ or $V_{RMS}$ or...) the output current is given by your load resistance (Ohm's law) not by other parameters of the oscillator. ...and as pipe noted 45A is quite a high current; did you mean mA? – Curd Mar 29 '17 at 10:49

You are really asking two quite separate questions: 1) how do I make a 4.8 volt phase-shift oscillator (and I notice that you do not specify pk-pk, rms, or mean amplitude), and 2) how do I drive a 0.1 ohm load at this voltage?

1) The simplest way to regulate the amplitude is to make Rf slightly greater than necessary, and let the op amp saturation limit the amplitude by clipping. For this sort of circuit, having the gain slightly too high will not cause the output to develop into a square wave. Instead, it will stabilize with slightly flattened peaks. The actual amplitude will depend on power supply voltages and op amp saturation characteristics, so you'll need to do some research on that. You would follow such an amplifier with a voltage divider to give about 1 volt, which is a standard voltage level for the next stage, which is

2) Making high-current output will be a challenge. Assuming your voltage is rms, your output power will be (4.8 x 45) or a bit more than 200 watts. Your best bet might be to build a high-power audio amplifier capable of, say 300 watts, then feed a step-down transformer to lower the voltage. Assuming a nominal 8-ohm load for the amplifier, the output voltage is about 42 volts rms, so you'd need about an 8.6:1 ratio. If your signal frequency is around 60 Hz, you could use a standard power transformer designed to take 120 VAC and give 14 volts, which is a standard process control voltage. The transformer, of course, would have to be rated for more than your nominal 216 watts. If you're using a different frequency, you'd need to either design and build your own transformer, or find a source of specialty units. It's entirely possible that the audio amplifier connoisseur community might be a source of such a transformer.

On the other hand, if your frequency isn't too high, you might have a go at making an amplifier using an op amp followed by some high-current MOSFETs, although that discipline has some gotchas which you would need to learn about.

Correction: Only now I have discovered that the shown circuit is not identical with the classical phase shift topology with 3 C-R units. The most right resistor R of the C-R highpass chain must be in series to the most right capacitor - and connected directly to the inverting opamp input. (As shown, this resistor is between the differential input which - for ideal opamps - has zero voltage. Hence, R has no meaning and no influence).

However, the circuit (without the mentioned resistor R) can work - but for another dimensioning if compared with the "classical" topology. Here, the third C-R highpass block is replaced by a differentiating active block.

(A) As mentioned already by WhatRoughBeast, the value of the feedback resistor should be (no: MUST be) slightly larger than the theoretical value. This is really required because otherwise the circuit cannot start safely. Moreover, due to unavoidable tolerances you have no other chance to be on the safe side (loop gain >0 dB at the oscillation frequency).

(B) In case you want to avoid hard-limiting of the oscillation amplitude you can introduce a "soft-limiting" device. As the most simple solution you could replace the feedback resistor Rf by a parallel combination Rf=[Rf1||(Rf2+rd)]. Here, rd is the differential resistance of a diode. However, because the diode must open for both half waves (both polarities) you must use two antiparallel diodes in series to Rf2.

Now it is your task to select proper values for Rf1 and Rf2 so that an opening of the diode allows a value of the resulting Rf value which fulfills the oscillation condition for the desired output voltage.

(C) In a simplified version of this principle you could place the two diodes directly across Rf (only one resistor). However, in this case, the signal quality will be degraded (if compared with version B) because the limting action is "less soft" as in case B.

(D) An alternative solution (with diodes) can be found here: http://electronicsarea.com/wp-content/uploads/phase_shift_oscillator_amplitude_limiter.gif