Design of small signal precision rectifier circuit

灞波儿奔

Design of small signal precision rectifier circuit [Copy link]

Circuit Principle The small AC signal first passes through the half-wave rectifier to generate a half-wave signal, which is then sent to the next stage to be superimposed and reversed with the input signal, and the output waveform is a full-wave rectified signal. This signal can obtain a relatively stable DC signal after passing through the first-order filter circuit. In the circuit diagram, U1, D1, D2, R3, and R2 constitute the half-wave rectifier part; U2, R4, R6, and R5 constitute the superposition and reverse part; and R1 and C1 constitute the first-order filter part. The diode conduction voltage in the circuit is about 0.6V, and the integrated operational amplifier's magnification is generally in the ten thousandth level. At this time, the operational amplifier input only needs a net input of microvolts to turn on the diode. Therefore, a slight change in the voltage at the input of the operational amplifier can cause the output to change with it. The small signal precision rectifier circuit uses this feature to realize the rectification of small AC signals. The model of the integrated operational amplifier in the circuit is mainly selected according to the voltage amplitude and frequency of the input signal. Here, OPA6951D is selected, and the maximum supported bandwidth can reach 500M. Half-wave rectification 1. When the input AC small signal is the positive half cycle of voltage: Because ui>0, the output voltage uo1 of U1 is <0, so D2 is turned on and D1 is turned off. At this time, R3, R2, and U1 form a reverse proportional amplifier circuit, and its output voltage uo1=-(R2/R3)ui. In the circuit, R3=R2, so uo1=-ui, and the circuit is a reverse amplifier circuit with an amplification factor of -1. 2. When the input AC small signal is the negative half cycle of voltage: Because ui<0, the output voltage uo1 of U1 is >0, so D2 is turned off and D1 is turned on. Since D2 is cut off, the signal uo1 at the output end of U1 is not sent to the next stage (i.e., the input end of U2); because the same-direction end is grounded, according to the virtual short principle, the voltage at the reverse end is zero, and at this time D1 is turned on, so the output voltage uo1 of U1 is clamped at 0V (i.e., uo1=0). Superposition Reverse According to the schematic diagram, R6, R4, R5 and U2 together form a reverse addition circuit, which performs reverse superposition operation on the input signal ui and the output signal uo1 of U1. uo2=-(R6/R4)uo1-(R6/R5)ui, because 2R4≈R6, R6=R5, so uo2=-2uo1-ui. When the input AC small signal is the positive half cycle of voltage: ①Because the voltage output by U1 is uo1=-ui, the output of the signal after passing through U2 is uo2=2ui; ②The output of the signal is uo2〞 =-ui; ③The output of U2 is uo2 =uo2 +uo2〞 = ui. That is, when the input is the positive half cycle, the output is equal and in the same direction. When the input AC small signal is the negative half cycle of voltage: ① uo1 is clamped at 0V, that is, the voltage on the left side of R3 is 0, and the voltage on the right side of R3 is also 0 according to the virtual short, so ideally no current flows into U2 at this time, that is, uo2=0; ②The output of the signal is uo2〞 =-ui; ③The output of U2 is uo2= uo2+ uo2〞 =-ui. That is, when the input is the negative half cycle , it is an equal reverse output. Summary: During the entire cycle, the output of U2 is the positive half cycle voltage plus the negative half cycle reverse voltage, thereby realizing AC rectification

alan000345

The explanation is good, and there are simulation circuit diagrams.