The relaxation oscillator of Example 36-8 is used in Fig. 36-25 to drive the integrator. Assume that the peak voltage out of the relaxation oscillator is 13.5 V. If the integrator has R4 = 10 kΩ and C2 = 10 μF, what is the peak-to-peak value of the triangular output wave?

Question

The relaxation oscillator of Example 36-8 is used in Fig. 36-25 to drive the integrator. Assume that the peak voltage out of the relaxation oscillator is 13.5 V. If the integrator has [latex]R_4[/latex] = 10 k[latex]\Omega[/latex] and [latex]C_2 = 10 \mu F[/latex], what is the peak-to-peak value of the triangular output wave?

Accepted Answer

With the equations shown in Fig. 36-25, we can analyze the circuit. In Example 36-8,(What is the frequency of the output signal in Fig. 36-26?)

we calculated a feedback fraction of 0.9 and a period of 589 [latex]\mu [/latex]s. Now, we can calculate the peak-to-peak value of the triangular output:

[latex]V_{out(p-p)}=\frac{589 \mu s}{2(10 K\Omega)(10 \mu F )} (13.5 V)=39.8 mV p-p[/latex]

The circuit generates a square wave with a peak-to-peak value of approximately 27 V and a triangular wave with a peak-to-peak value of 39.8 mV.

Question 36.9: The relaxation oscillator of Example 36-8 is used in Fig. 36...

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