Applications of Operational Amplifier: Precision Diode

Precision Diodes:

If a sinusoid whose peak value is less than the threshold or cut in voltage V_d(-0.6V) is applied to the conventional half-wave rectifier circuit, output will remain zero. In order to be able to rectify small signals (mV), it is necessary to reduce V_d. By placing a diode in the feedback loop of an OPAMP, the cut in voltage is divided by the open loop gain A of the amplifier. Fig. 5, shows an active diode circuit.

Fig. 5

Hence V_D is virtually eliminated and the diode approaches the ideal rectifying element. If the input V_in goes positive by at least V_D/A, then the output voltage (=A v_d ) exceeds V_D and D conducts and thus, provides a negative feedback. Because of the virtual connection between the two inputs v_O= v_in-v_d=v_in- v D / A » v_in. Therefore, the circuit acts as voltage follower for positive signals (above 60mV=0.6 / 1*10⁵) when V_in swing negatively, D is OFF and no current is delivered to the external load.

By reversing the diode, the active negative diode can be made.

Active Clippers:

By slightly modifying the circuit, an active diode ideal clipper circuit is obtained. Fig. 6, shows an active clipper which clips the input voltage below v_R.

Fig. 6

When v_in < V_R , then v' is positive and D conducts. Under these conditions, the OPAMP works as a buffer and the output voltage equals the voltage at non-inverting terminal

V_out = V_R.

If v_in > V_R, then v' is negative and D is OFF and v_O = v_in R_L / (R_L + R) » V_i if R << R_L Thus, output follows input for v_in > V_R and v_O is clamped to V_Rif v_in < V_R by about 60 mV. Fig. 7, shows the output waveform of clipper circuit.When D is reverse biased a large differential voltage may appear between inputs and the OPAMP must be capable to withstand this voltage.

Fig. 7