I'm currently designing a saw-tooth oscillator that should be frequency adjustable, I don't want to use potentiometers, so it has to be voltage controlled. The frequency is controllable in the circuit shown, although it has not a very wide range, but has problems when the control voltage (VC) exceeds a perticular value.

At the moment I have trouble understanding this Sawtooth Oscillator fully. These kind of oscillator are always a bit tricky to me to understand, most of the time there is too much going down. Which gives some problems at certain voltages. Saw-tooth oscillator schematic

My current understanding:

  • The first part concerning Q1 is a current source for C1, which integrates voltage VC, which is 11V in this example. We choose VC to be adjustable for frequency control of the oscillator
  • The last part is a voltage divider for static voltage VA. It ensures a static voltage at Vbase if it isn't pulled to ground.
  • The middle part: If the integrated voltage of C1 (Vout) gets higher than Vbase + Vbe of Q2, current can flow through Q1 from emitter to collector, which saturates Q3 and pulls Vbase and C1 to ground. C1 gets discharged, Q3 isn't saturated anymore, Vbase isn't pulled down and C1 (Vout) can't make the Vbe jump of Q2 and VC gets again integrated by the current source. And the cycle continues.

Seems to work: Resulting transient analysis for VC of 11V Bigger Green: Vout, Blue: Vbase

As expected the output voltage rises to Vbase + ~0.6V (Vbe), then gets quickly discharged. The blue line also shows that the divider voltage gets quickly pulled to ground.

There is only one problem. What if I increase the voltage VC to 12V? Resulting transient analysis for VC of 12V Bigger Green: Vout, Blue: Vbase

The first cycle everything is going right, but the cycle doesn't continue. Q3 stays saturated and pulls Vbase to ground. Vout stays at +Vbe.

I'm fairly certain I'm missing something in understanding this circuit. Can someone explain me why the cycle doesn't continue, although it does for lower voltages, and maybe a way to fix this to increase the frequency range.

Here is a Pastebin link to the schematic file for LTSpice, if anyone needs a different file/output please give me a comment! http://pastebin.com/VMVLEJkv



What if you connect a resistor from Q3-base to ground to drain "excess" current? You want the Q3-base voltage to drop below 0.6V at discharge. Otherwise if Q1 current source delivers too much current, to "discrete" thyristor will not shut off.

  • \$\begingroup\$ A 1K Resistor: Voltage caps at 3.6V, Q3 never achieves full saturation, A 100K Resistor seems to work, 10meg produces the same results for 12V. \$\endgroup\$ – Rick van Loo Jul 20 '15 at 19:16
  • \$\begingroup\$ The problem is that it seems to shift the working control voltage range, but I might be wrong. The end result should be an VCO from 20Hz-20kHz, any guess if that's actually possible with this circuit with some value tweaking? \$\endgroup\$ – Rick van Loo Jul 20 '15 at 19:25
  • \$\begingroup\$ @RickvanLoo I think you have to find a balance between the programmed Q1 collector current (which depends on VC) and the resistor. Actually, it took me a minute to realize what your problem really is. You'll have to find a way to turn off Q3 which is probably less trivial than just adding a resistor. Not sure, I'd have to pull out a bread board, but I think I would experiment with a forth transistor or an entirely different architecture. \$\endgroup\$ – jippie Jul 20 '15 at 19:47

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