# Find the frequency of the sawtooth (voltage accross the capacitor)

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

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:

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

• Do you have a question? May 5, 2016 at 17:30
• @EugeneSh. yes, Am I right, about my function for the frequency? May 5, 2016 at 17:31
• 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.
– user105652
May 5, 2016 at 18:14
• @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? May 5, 2016 at 18:34