Question 10.5: (a) Find the peak-to-peak amplitude and frequency of the saw...

(a) First, find the gate voltage in order to establish the approximate voltage at which the PUT turns on.

V_{\mathrm{G}}=\frac{R_4}{R_3+R_4}(+V)=\frac{10   \mathrm{k} \Omega}{20  \mathrm{k} \Omega}(+15 \mathrm{~V})=7.5 \mathrm{~V}

This voltage sets the approximate maximum peak value of the sawtooth output (neglecting the 0.7 V).

V_{\mathrm{P}} \cong 7.5 \mathrm{~V}

The minimum peak value (low point) is

V_{\mathrm{F}} \cong 1 \mathrm{~V}

So the peak-to-peak amplitude is

V_{p p}=V_{\mathrm{P}}-V_{\mathrm{F}}=7.5 \mathrm{~V}-1 \mathrm{~V}=6.5  \mathrm{~V}

The frequency is determined as follows:

V_{\mathrm{IN}}=\frac{R_2}{R_1+R_2}(-V)=\frac{10   \mathrm{k} \Omega}{78  \mathrm{k} \Omega}(-15 \mathrm{~V})=-1.92 \mathrm{~V}

f=\frac{\left|V_{\mathrm{IN}}\right|}{R_i C}\left(\frac{1}{V_{\mathrm{P}}-V_{\mathrm{F}}}\right)=\left(\frac{1.92 \mathrm{~V}}{(100  \mathrm{k} \Omega)(0.005  \mu \mathrm{F})}\right)\left(\frac{1}{7.5 \mathrm{~V}-1 \mathrm{~V}}\right) \cong \mathbf{5 9 1 ~ H z}

(b) The output waveform is shown in Figure 10–33 . The period is

T=\frac{1}{f}=\frac{1}{591 \mathrm{~Hz}}=1.69 \mathrm{~ms}

P R A C T I C E EXERCISE

If R_i  is changed to 56 kΩ in Figure 10–32 , what is the frequency?