Explanation Given: The saturation output voltage are +10 and -5V. R 1 = R 2 = 20 k Ω R x = 50 k Ω C x = 0.01 μ F Referring to the circuit for the Schmitt trigger oscillator: Writing the expression for the voltage across the capacitor in the RC network ( V x ) for first cycle of the Schmitt trigger. V x = V H + ( V L 2 − V H ) e t 1 t x V x = V H − ( − V L 2 + V H ) e t t x .......... ( 1 ) Where, V H = high saturation output voltage V L = low saturation output voltage t 1 =time for first cycle τ x =time constant . The output of Schmitt trigger switches for first part of the cycle at time t 1 with the value of voltage V H 2 across the capacitor in the RC network . Hence, V x = V H 2 Substituting 10V FOR V H : V x = 10 V 2 = 5 V Re-arranging the expression 1 to evaluate the time t 1 . ( − V L 2 + V H ) e t t x = V H − V X e t 1 t x = V H − V x ( − V t 2 ) + V H Applying natural log on the both sides of the expressions. ln ( e t 1 τ x ) = ln [ V H − V x ( − V t 2 ) + V H ] − t 1 τ x = ln [ V H − V x ( − V t 2 ) + V H ] t 1 = − τ x ln [ V H − V x ( − V t 2 ) + V H ] { ∵ ln ( e x ) = x } Substituting the known values in the above equation: t 1 = τ x ln [ 10 V − 5 V { − ( − 5 V ) 2 } + 10 V ] = τ x ln ( 5 V 2.5 V + 10 V ) = − τ x ln ( 5 V 12.5 V ) = − τ x ln ( 0.4 ) t 1 = − ( − 0.916 ) τ x t 1 = 0.916 τ x The expression for the voltage across the capacitor in the RC network ( V X ) for the next cycle of the Schmitt trigger. V x = V L + ( V H 2 − V L ) e t 2 τ x .......... ( 2 ) Here, t 2 =time for next cycle. The output of the Schmitt trigger switches for next part of the cycle at time t 2 , and the value of voltage V L 2 across the capacitor in the RC network . Hence, V x = V L 2 Substitute -5v for V L . V x = − 5 V 2 = − 2.5 V Re-arranging the expression 2 to evaluate the time t 2 . ( V H 2 − V L ) e t 2 τ x = V L − V x e t 2 τ x = V L − V x ( V H 2 ) − V L Apply natural log on both sides of expression

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

4th Edition

ISBN: 9781266368622

Author: NEAMEN

Publisher: MCG

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Chapter 15, Problem 15.8EP

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The frequency of the oscillation and the duty cycle also sketch the v_o and v_x versus time over the two periods of the oscillation.

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MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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The frequency of the oscillation and the duty cycle also sketch the v o and v x versus time over the two periods of the oscillation.

Solution Summary: The author illustrates the frequency of the oscillation and the duty cycle. The saturation output voltage is +10 and -5V.

Concept explainers

The frequency of the oscillation and the duty cycle also sketch the v_o and v_x versus time over the two periods of the oscillation.

Want to see the full answer?

Chapter 15 Solutions

The frequency of the oscillation and the duty cycle also sketch the v o and v x versus time over the two periods of the oscillation.

Solution Summary: The author illustrates the frequency of the oscillation and the duty cycle. The saturation output voltage is +10 and -5V.

Concept explainers

The frequency of the oscillation and the duty cycle also sketch the vo and vx versus time over the two periods of the oscillation.

Want to see the full answer?

Chapter 15 Solutions

The frequency of the oscillation and the duty cycle also sketch the v o and v x versus time over the two periods of the oscillation.

The frequency of the oscillation and the duty cycle also sketch the v_o and v_x versus time over the two periods of the oscillation.