Applications of Operational Amplifiers: Rectifiers

Active Half Wave Rectifier:

The Active half wave rectifier is shown in fig. 8.

Fig. 8	Fig. 9

If v_in is positive then output of the OPAMP becomes negative (the non inverting terminal is grounded). Thus diode D₂ conducts and provides a negative feedback. Because of the feedback through D₂a virtual ground exists at the input. Thus diode D₁ acts as open circuit. The output voltage under this condition is given by

v_o = v - = 0.

If v_in goes negative, then output of the OPAMP becomes positive. Thus D₁ is conducting and D₂ is off. Thus, the circuit behaves as an inverting amplifier. The output of the circuit is given by

The resultant output voltage will be positive. If v in is a sinusoid, the circuit per­forms half wave rectification. The transfer characteristic of the half wave active rectifier is shown in fig. 9. The output does not depend upon the diode forward voltage (v_d). Thus, because of the high open loop gain of the OPAMP, the feedback acts to cancel the diode turn-on (forward) voltage. This leads to improved performance since the diode more closely approximates the ideal device.

Axis Shifting of the Half Wave Rrectifier:

The half wave rectified output waveform can be shifted along the v_in axis. This is done by using a reference voltage added to the input voltage of the rectifier as shown in fig. 10. This termed axis shifting. It adds or subtracts a fixed dc voltage to the input signal. This process shifts the diode turn-on voltage point. If a negative reference voltage, V_REF, is applied to the circuit, the diode turns on when the input voltage is still positive. This shifts the v_out/ v_in transfer characteristic to the right. If a positive reference voltage is applied, the v_out/ v_in transfer characteristic shifts to the left. These shifted characteristics are shown in fig. 10.

Fig. 10	Fig. 11

The input-output voltage characteristics can also be shifted up or down. This is termed level shifting and is accomplished by adding a second OPAMP with a reference voltage added to the negative input terminal as shown in fig. 11.

Active Full Wave Rectifier:

Method 1:

A full wave rectifier, or magnitude operator, produces an output which is the absolute value, or magnitude, of the input signal waveform. One method of accomplishing full wave rectification is to use two half wave rectifiers. One of these operates on the positive portion of the input and the second operates on the negative portion. The outputs are summed with proper polarites. Fig. 1 illustrates one such configuration. Note that the resistive network attached to the ouput summing opamp is composed of resistors of higher value than those attached to the opamp that generates v₁. This is necessary since for negative v_in, v₂ follows the curve shown above the node labled v₂. That is, as the input increases in a negative direction, v₂ increases in a positive direction. Since the input impedance to the non-inverting terminal of the summing opamp is high, the voltage, v+ is simply one half of v₂ (i.e., the two 100KΩ resistors form a voltage divider). The voltage at the negative summing terminal, v-, is the same as v+, and therefore is equal to v₂ / 2. Now when v_in is negative, D₂ is open, and the node v₁ is connected to the inverting input of the first opamp through a 5 KΩ resistor. The inverting input is a virtual ground since the non-inverting input is tied to ground through a resistor. The result is that the voltage divider formed by the 100 KΩ and 5KΩ resistors. In order to achive a characteristic resembling that shown in the figure, this voltage divider must have a small ratio, on the order of 1 to 20.

Fig. 1

Method 2:

The method of full wave rectification discussed above requires three separate amplifiers. One simpler circuit or active full wave rectifier, which makes use of only two OPAMPs, is shown in fig. 2. It rectifies the input with a gain of R / R₁, controllable by one resistor R₁.

Fig. 2

When v in is positive then v' = negative, D₁ is ON and D₂ is virtual ground at the input to (l). Because D₂ is non-conducting, and since there is no current in the R which is connected to the non-inverting input to (2), therefore, V₁ =0.

Hence, the system consists of two OPAMP in cascade with the gain of A₁ equal to (-R / R₁) and the gain of A₂ equal to (-R / R) = -1.

The resultant at voltage output is

v_o = (R / R₁ ) v_in > 0 (for v_in > 0 voltage output of (1) )

Consider now next half cycle when v in is negative. The v' is positive D₁ is OFF and D₂ is ON. Because of the virtually ground at the input to (2) V₂ = V₁ = V

Since the input terminals of (2) are at the same (ground) poten­tial, the current coming to the inverting terminal of (1) is as indicated in fig. 2.

The output voltage is v_o = i R + v where i = v / 2R (because input impedance of OPAMP is very high).

The sign of v_o is again positive because v_in is negative in this half cycle. Therefore, outputs during two half cycles are same; and full wave rectified output voltage is obtained also shown in fig. 2.