Absolute-Value Output Circuit MCQ [Free PDF] – Objective Question Answer for Absolute-Value Output Circuit Quiz

An absolute value output circuit produces an output signal that swings positively only, regardless of the polarity of the input signal; because of the nature of its output waveform, the circuit is used as a full-wave rectifier.

The voltage at the terminal

V₁ = (V_p -V_d1) /2

V₁ = (12-0.7) /2 = 5.65 v (V_d1= voltage drop across diode=0.7)

Similarly, the voltage at the negative terminal

V₂ = (V_o -V_d3 ) /2 = (V_o – 0.7) /2

Since V_id ≅ 0v , ∴ V₁ = V₂

V_o = (5.65 *2 ) + 0.7 = 12v.

A conventional ac voltmeter is designed to measure the RMS value of the pure sine wave whereas, the peak value of the non-sinusoidal waveforms can be a peak detector.

For proper operation of the circuit, the charging and discharging time constant must satisfy the following: CR_d ≤ T/10 and CR_L ≥ to 10T.

The resistor is used to protect the op-amp against the excessive discharge current, especially when the power supply is switched off.

The diode-connected at the output of op-amp conducts during the negative half cycle of input voltage. Hence, prevent the op-amp from going into negative saturation. This in turn helps to reduce the recovery time of the op-amp.

The negative peaks of the input signal V_in can be detected by reversing diodes D₁ and D₂.

The time periods of the sample and hold control voltage during which the voltage across the capacitor is equal to the input voltage are called the sample period.

The hold period is the period during which the voltage across the capacitor is constant and the output of the op-amp is processed or observed during hold periods.

LF398 has a significant reduction in size and improved performance and requires only an external storage capacitor.

All types of modulation involve taking samples of an input signal and hold on to its last sampled value until the input is sampled.

In an op-amp clipper circuit, a rectifier diode is used to clip off certain portions of the input signal to obtain a designed output waveform.

The clipping level is determined by the reference voltage which should be less than the input voltage range of an op-amp.

In a positive clipper, the diode conducts until V_in = V_ref (during the positive half cycle of the input), because when V_inref, the voltage (V_ref) at the negative input is higher than that at the positive input.

If the potentiometer R_p is connected to negative supply -V_EE instead of +V_CC, the reference voltage V_ref will be negative. As a result, the entire output waveform above -V_ref is to be clipped off.

When the input voltage is higher than the reference voltage, the op-amp operates in an open loop, and the diode becomes reverse biased. Thus, the output voltage will be equal to a reference voltage.

A positive small-signal halfwave rectifier can rectify signals with peak values down to a few millivolts because the high open-loop gain of the op-amp automatically adjusts the voltage drive to the diode so that the rectified output peak is the same as the input.

The diode acts as an ideal diode since the voltage across the ON diode is divided by the open-loop gain of the op-amp. As the input voltage starts increasing in the positive direction, the output of the op-amp also increases positively till the diode become forward biased.