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I have figured out how to make a sawtooth VCO using a 555 IC , by using a current mirror.

The issues I am having is I can't make it stable at all. Its having random issues like it randomly changes frequency output or if I turn it off for awhile then back on it changes frequency output even if the input voltage on the current mirror stays the same.

So what components can I add to this circuit to make it stable enough? I need this circuit to be able to be able to keep the same frequency output with the same voltage input.

enter image description here

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  • \$\begingroup\$ What frequency you want? Why use a 555 instead of an actual VCO? \$\endgroup\$
    – Justme
    Nov 14, 2021 at 22:38
  • \$\begingroup\$ @justme , I only need frequencies between 50 hz to 200 hz. Why not use a actual VCO: I am trying to make a really simple DIY VCO. \$\endgroup\$ Nov 14, 2021 at 22:44
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    \$\begingroup\$ Define "stable enough". Engineers like to work with real specifications, not vague descriptions. Also, keep in mind that the 555 was never intended as a precision oscillator. \$\endgroup\$
    – Barry
    Nov 14, 2021 at 23:07
  • \$\begingroup\$ The circuitry looks ok. Current mirror works well, and 555 is a good proven device. What is the connection between out & ctl for? That will make it hard to hit trig, and will be noisy. Remove that and try. \$\endgroup\$
    – jay
    Nov 14, 2021 at 23:22
  • \$\begingroup\$ @jay That makes sure, the sawtooth output goes to ground, if the control input and the input isn't connected using a diode, the entire sawtooth output will float above ground. \$\endgroup\$ Nov 14, 2021 at 23:26

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As shown the current mirror will be very sensitive to temperature variations. Even the self-heating due to the dissipation in the transistors will cause their temperature to change enough to cause unbalance between the two transistors. That is probably why you notice the change when it has been switched off for a while.

They should be matched and both be at exactly the same temperature, for example clamped together into a common heatsink.

The imbalance can be improved by putting a resistor in the emitter of each of the transistors. The value should be selected so that it drops a few hundred millivolts or more. In that circuit there is about 300uA flowing in the first transistor so a 1 kilohm resistor would be a good place to start. An example is shown below.

What type of capacitor are you using for the timing? A ceramic capacitor such as X7R is probably not good enough for temperature stability. A film capacitor would be better.

enter image description here

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  • \$\begingroup\$ The addition of emitter degeneration, right at the level you mention, is good to do. That's still got the Early Effect to worry over. The addition of a cascoded device will help that. (There are a few other ideas to add, but that's the larger problem.) \$\endgroup\$
    – jonk
    Nov 14, 2021 at 23:46
  • \$\begingroup\$ @jonk - emitter degeneration will help with Early effect as well because it will reduce the effect of the base-emitter voltage changes. \$\endgroup\$ Nov 14, 2021 at 23:50
  • \$\begingroup\$ Agreed. I haven't yet calculated how much is removed by what allowed degeneration overhead voltage. Have you done that, yet? \$\endgroup\$
    – jonk
    Nov 14, 2021 at 23:54
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A current mirror is great on paper, but it rarely works well with real-world components. Every parameter about a transistor, including its gain and base-emitter junction voltage, is either temperature-dependent, collector-current-dependent, or both.

Commercially available VCO IC's have a lot going on internally, and a lot of that is compensating and offsetting circuits hovering around the main VCO structure.

Linear Technology and Texas Instruments have many app notes on opamp circuit design, with sections on VCO's. IIRC these are not "simple" circuits.

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Add some emitter degenration to the current mirror to improve its stability. (perhaps 3.3K in series with each emitter - you have plenty of headroom)

The trick you're pulling on the control-voltage pin is defeating the 555s internal ripple rejection strategy. you may want to reconsider that. perhaps restore the ripple rejection capacitor on pin 5 and use a comparitor to drive the trigger pin if you don't like a trigger level of VCC/3

Use a high stability capacitor such as polystyrene film for the timing capacitor.

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When the DIS pin is driven low it is forced to sink a very high current for a very short time as it discharges the timing capacitor. You could try putting a small resistor, say 100R, in series with the DIS pin to limit the discharge current. Note that adding the extra resistor will slightly increase the slope of the vertical drops of the output waveform.

If that doesn't improve matters then you could try the circuit below.

The op amp and its associated resistors form a level shifter. It also doubles the amplitude of the output waveform compared to your circuit. If the amplitude is too big for you then you can reduce it by adding a resistive potential divider at the output. The frequency is about 125 Hz, as you can see.

Saw tooth waveform generator

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  • \$\begingroup\$ weirdly if I try this in spice it doesn't go to ground for me, for some reason it slightly floats above ground if the op amp is grounded , if the op amp has both positive and negative rails , the sawtooth goes positive and negative. \$\endgroup\$ Nov 15, 2021 at 16:26
  • \$\begingroup\$ @MichaelWeaser The LM358's output will not quite reach ground when its negative supply rail is ground, but it should get within a few 10s of millivolts of ground. If you use plus and minus supplies for the LM358 make sure that the bottom end of R4 (and the rest of the circuit) is still connected to ground (0V). \$\endgroup\$
    – James
    Nov 15, 2021 at 16:50
  • \$\begingroup\$ @MichaelWeaser Accuracy of my circuit's output depends on the 555's internal thresholds being 1/3 and 2/3 of Vcc. The thresholds will probably not be spot on those values which will lead to a dc offset at the output. You could adjust any dc error out, bringing the lower extreme of the waveform to almost exactly 0V, by using a + and - dual supply for the LM358 and then replacing R3 with a 4k7 resistor in series with a 10k 22-turn cermet trimmer. Adjusting the trimmer would vary the output's dc level. You could test this in spice by varying the value of R3. Dual supply needed for LM358. \$\endgroup\$
    – James
    Nov 15, 2021 at 17:33

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