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Q:Is my work right, finding the frequency of the sawtooth (voltage accross the capacitor)?:

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My work:

  • Vdd(t) and Vss(t) are DC voltages and: $$\text{V}_{\text{ss}}(t)=-\text{V}_{\text{dd}}(t)$$
  • Assume at t=0, Vc(0)=0V and Vout(0)=Vdd(t)
  • Charging the capacitor: $$\text{V}_\text{C}(t)=\text{V}_{\text{dd}}(t)\left(1-\exp\left[-\frac{t}{\text{C}\text{R}_1}\right]\right)$$
  • When: $$\text{V}_\text{C}(t)>\frac{\text{R}_3}{\text{R}_3+\text{R}_2}\cdot\text{V}_{\text{out}}(t)=\text{V}_-(t)$$ then Vout(t)=-Vdd(t)=Vss(t)
  • Discharging the capacitor: $$\text{V}_\text{C}(t)=\text{V}_{\text{C}}(0)+\left(\text{V}_{\text{dd}}(t)-\text{V}_{\text{C}}(0)\right)\left(1-\exp\left[-\frac{t}{\text{C}\text{R}_1}\right]\right)$$
  • When: $$\text{V}_\text{C}(t)<-\frac{\text{R}_3}{\text{R}_3+\text{R}_2}\cdot\text{V}_{\text{out}}(t)=-\text{V}_+(t)$$ then Vout(t)=Vdd(t)

So, when we want to find the frequency we can say that, solving for 't':

  • $$\text{V}_-(t)=\text{V}_{\text{C}}(0)+\left(\text{V}_{\text{dd}}(t)-\text{V}_{\text{C}}(0)\right)\left(1-\exp\left[-\frac{t_1}{\text{C}\text{R}_1}\right]\right)\Longleftrightarrow$$ $$t_1=\text{C}\text{R}_1\ln\left(\frac{\text{V}_{\text{C}}(0)-\text{V}_{\text{dd}}(t)}{\text{V}_-(t)-\text{V}_{\text{dd}}(t)}\right)$$
  • $$-\text{V}_+(t)=-\left(\text{V}_{\text{C}}(0)+\left(\text{V}_{\text{dd}}(t)-\text{V}_{\text{C}}(0)\right)\left(1-\exp\left[-\frac{t_2}{\text{C}\text{R}_1}\right]\right)\right)\Longleftrightarrow$$ $$t_2=\text{C}\text{R}_1\ln\left(\frac{\text{V}_{\text{C}}(0)-\text{V}_{\text{dd}}(t)}{\text{V}_-(t)-\text{V}_{\text{dd}}(t)}\right)$$

Then, the time of one period is given by:

$$\text{T}_{\left[\text{s}\right]}=t_1+t_2=2\text{C}\text{R}_1\ln\left(\frac{\text{V}_{\text{C}}(0)-\text{V}_{\text{dd}}(t)}{\text{V}_-(t)-\text{V}_{\text{dd}}(t)}\right)$$

So, the frequency is given by:

$$\text{f}_{\left[\text{H}z\right]}=\frac{1}{\text{T}_{\left[\text{s}\right]}}=\frac{1}{2\text{C}\text{R}_1\ln\left(\frac{\text{V}_{\text{C}}(0)-\text{V}_{\text{dd}}(t)}{\text{V}_-(t)-\text{V}_{\text{dd}}(t)}\right)}$$

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    \$\begingroup\$ Do you have a question? \$\endgroup\$
    – Eugene Sh.
    May 5, 2016 at 17:30
  • \$\begingroup\$ @EugeneSh. yes, Am I right, about my function for the frequency? \$\endgroup\$
    – Jeans Boss
    May 5, 2016 at 17:31
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    \$\begingroup\$ You have a page of math here, but you need a frequency counter to get exact value. Are you asking if your math is correct? If so this might be a question for the mathematics forum. \$\endgroup\$
    – user105652
    May 5, 2016 at 18:14
  • \$\begingroup\$ @Sparky256 I know that my math is correct, but are my equations right for the opamp? So, is the function I've got for the frequency right? \$\endgroup\$
    – Jeans Boss
    May 5, 2016 at 18:34

1 Answer 1

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Yes. I had to read your entire equation to see how Vss(t) was accounted for. To get the frequency of course you take the reciprocal of the time period, and your using the negative point of the waveform for a ref.

[s] Is just a local reference for the waveshape. For these simple circuits, R and C1 dominate the total time period (and thus the frequency) while R2 and R3 adjust the trip point.

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  • \$\begingroup\$ @JeansBoss. Welcome. Normally R2 = R3 for center stability so R and C1 dominate the time period, but each can vary by 50%. This will change the frequency, as you are changing the trip point. \$\endgroup\$
    – user105652
    May 5, 2016 at 19:33
  • \$\begingroup\$ Yes I undestand that \$\endgroup\$
    – Jeans Boss
    May 5, 2016 at 19:35

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