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I'm a total beginner hobbyist trying to understand the concept of an audio synthesizer, with a view to building one. Rather than just build a kit I'm keen to understand the concepts.

Goal: build a simple sawtooth oscillator using an integrator and Schmitt trigger

I'm working through the VCO module and just wanted to strip it down an see if I could build a simple sawtooth oscillator, using op-amps; an integrator circuit feeding to a Schmitt Trigger that would short the capacitor on the integrator when a certain voltage is reached.

My circuit so far is below:

Simple Voltage Controlled Oscillator Circuit - so far

R4, R5 provide voltage to the RC circuit R6, C2; I feel I need R4, R5, R6, C2 to supply a change of voltage to the integrator in order for to start to integrate C1 with the Integrator U1 - although this seems messy. Please see the waveform outputs below. The is the linear change of voltage from the integrator I'd expect V(TP1), however, this voltage is negative (I'm not sure if this should be positive). This then feeds R3 and the Schmitt Trigger and the next step would be for V(out) of U2 to turn on a JFET across C1 to short it at certain voltage point along V(tp1).

Should the voltage to the JFET just be a very short single pulse at the point the threshold voltage is reached, say of 20us? Currently, V(out) just seems to be flatlining.

When working I'd expect to see a sawtooth wave of 2v peak to peak centred on 0v from v(tp1)

Any help for me to gain further understanding into these fundamental concepts and my goal in creating this simple sawtooth oscillator would be greatly appreciated.

Many thanks
Alex

Waveforms so far

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  • \$\begingroup\$ Just as an initial nudge, analog synthesizer VCOs need to have a logarithmic response to the control voltage (e.g. 1V/octave), not a linear response as you're building here. That way, a given increase in input voltage always produces the same increase in pitch. \$\endgroup\$ – Finbarr Mar 10 '17 at 23:25
  • \$\begingroup\$ @Finbarr Yes this is abit of an aside to the practical point you make about 1V/octave. Presumably for what I'm trying to achieve the Schmitt Trigger isn't setup correctly? \$\endgroup\$ – Alex2134 Mar 10 '17 at 23:28
  • \$\begingroup\$ U1 will try to hold its -ve input at 0V, so with a positive voltage at TP2 the highest voltage you will ever see at TP1 will be when C1 is fully discharged and will therefore be 0V. As C1 charges, TP1 will go further and further negative, exactly as per your output trace. You'll need to design the Schmitt so it flips at a large negative voltage and flips back as TP1 gets almost back to 0V as C1 discharges. \$\endgroup\$ – Finbarr Mar 11 '17 at 0:27
  • \$\begingroup\$ No positive feedback = no oscillation. \$\endgroup\$ – Andy aka Mar 11 '17 at 11:46
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You need no fet. You can add another resistor to pin 2 of U1. It makes the integrator summing. If the new resistor is small, the time constant will be low. Draw that way the capacitor empty fast. In practice you can connect a diode from U2's output to 2/U1, anode to 2/U1. The diode prevents the loading of the cap.

It generally is not a good idea to have high pulse current reset. It will not be fast enough for linear Volt/Hz dependence. You will do better by having a triangle oscillator and a switching circuit that inverts the other ramp of the triangle voltage to another qrowing sawtooth.

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