5
\$\begingroup\$

I was handed the following circuit in order to determine what it does and what is its operating conditions. The circuit was reverse engineered from an old equipment.

schematic diagram

I have simulated with NI Multisim v.11 and the oscilloscope shows only a constant voltage of 2.7[V], this are measured at \$C_3\$.

I need help to analyze this circuit in order to determine what it does, what is its output waveform and its operating conditions.

The op amp stage seems to me like a comparator/oscillator, it is comparing \$V_0\$ which varies over time because of the capacitors against the 2.5[V] that is a square wave generator with 5V amplitud. So is this an square wave oscillator or what? What is its output?

\$\endgroup\$
  • \$\begingroup\$ your problem with simulation arises because SPICE calculates the operating point and brings up the circuit in an ideal state. You can add some asymmetry into the circuit via a large 10Mohm to ground or rail, OR you can bump the non-inverting input with a voltage source 100 ns into the sim. Starrt it at say 5 v and then kick it to 2.5V and it should light up if you have it connected properly. \$\endgroup\$ – placeholder Jan 28 '13 at 2:25
  • \$\begingroup\$ Since each RC link entails a phase delay, at the frequency where the total phase shift is 180°, there ought to be some positive feedback. What happens in your simulation if you start by force-feeding a 3 Hz square wave at the inverting input for a few seconds and then take it away? \$\endgroup\$ – hmakholm left over Monica Jan 28 '13 at 2:41
  • 1
    \$\begingroup\$ The circuit is probably wrong, even though you have checked :-). You have 2 DC paths from U1A _out tp inverting input - one is 1.5M and the other is 86.2K (so precise). If it was positive feedback it would be interesting. Time constants are about 0.1s. Overall it was either a rubbish circuit or intended to mislead or it has been reconstructed wrongly. | What is it part of? \$\endgroup\$ – Russell McMahon Jan 28 '13 at 2:43
  • \$\begingroup\$ @Russell McMahon: It is not 86.2k but 56.2k it was a measure straight from the Ohmeter. Ok, let's suppose it is wrong and it is as you say but let's concentrate on the given circuit arrangement (I'm open to suggestions and guesses of course). Can't tell if rubbish or misleading, maybe the inputs of op-amp are inverted, but then again let's concentrate on the current circuit shown. Thanks. \$\endgroup\$ – mongoose85 Jan 28 '13 at 2:55
  • \$\begingroup\$ @Russell - it's a "typical" phase shift oscillator (topology - don't know about the exact values, what they are now, what they were originally, etc) I have built quite a few of these during my audio tinkerings. I do wonder about the frequency of ~4.5Hz though here (is it part of an guitar FX pedal or something? e.g. phaser, flanger, chorus, etc) \$\endgroup\$ – Oli Glaser Jan 28 '13 at 4:07
9
\$\begingroup\$

Looks like a Phase Shift oscillator to me. Here is almost the exact same circuit (from here):

enter image description here

It does as it's name suggests, uses a series of RC filters to shift the phase and create positive feedback.

Here's an LTspice (crude, no gain control) circuit and simulation:

Phase Shift

Simulation:

Phase Shift Simulation

With values shown in question

I could not get the circuit to oscillate using the exact values, a change to the gain resistor was needed, from 56.2kΩ to 5.62kΩ (less than ~40kΩ):

Phase Shift 2

Simulation:

Phase Shift 2 Simulation

I have a feeling that somewhere in the reverse engineering an order of magnitude type mistake has been made with one of the values. For a frequency this low I would use buffer amplifiers in between phase shift stages (to get rid of the issues caused by the loading effects of the cascaded filters)
I'd be interested to know what the circuit came from, might help give a clue or two to the desired frequency/amplitude.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ could you simulate it with the C values above? the pole will shift substantially, and if it doesn't start "bump" it like I mentioned in the comments above. \$\endgroup\$ – placeholder Jan 28 '13 at 3:17
  • \$\begingroup\$ @rawbrawb - Okay, I added the results - I used the original circuit but I had to increase the gain by 10 (by dropping the 56.2k resistor to 5.62k) Using the larger value dies out quickly, and as you can see the 5.62k is borderline, with only a 600mV amplitude. \$\endgroup\$ – Oli Glaser Jan 28 '13 at 3:36
  • \$\begingroup\$ @Oli, the output (1.2Vp-p) is probably all that can be expected from a 5V supply and a non- rail-to-rail opamp. Output at the opamp itself is healthier! I would guess the lower gain version stabilises at nearly the same amplitude given a longer simulation. It's common to reduce the excess gain to minimise harmonic distortion, or even vary the gain by some means to eliminate the excess (as in a Wien bridge osc) \$\endgroup\$ – Brian Drummond Jan 28 '13 at 10:09
  • \$\begingroup\$ @BrianDrummond - I was referring to my dual +-15V SPICE example circuit, although at the low frequency it seems to settle around this level anyway, which from what I recall in the real versions is not inaccurate (no gain control as in Wien bridge is necessary if the correct values are chosen). Time permitting, I might make the circuit up later and add some real data to the answer. I agree the output is fine as it is, the point was if I use the 56k value the circuit does not maintain oscillation (in SPICE). \$\endgroup\$ – Oli Glaser Jan 28 '13 at 10:22
  • \$\begingroup\$ @Oli : my apologies, I didn't study your Spice cct in enough detail. In that case I'm curious about the amplitude at the opamp output. \$\endgroup\$ – Brian Drummond Jan 28 '13 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.