Circuit principle
The AC small signal first passes through the half-wave rectification part to generate a half-wave signal , which is then sent to the latter stage for superposition and inversion with the input signal , and the output waveform is a full- wave rectified signal . This signal can get a relatively stable DC signal after the first-order filter circuit . In the circuit diagram, U1 , D1 , D2 , R3 , R2 form the half-wave rectification part; U2 , R4 , R6 , R5 form the superposition reverse part; R1 , C1 form the first-order filtering part . In the circuit, the diode conduction voltage is about 0.6V , and the open magnification of the integrated op amp is generally 10,000 times . Therefore, a small change in the voltage at the input of the op amp can cause the output to follow it . The small-signal precision rectifier circuit takes advantage of this feature to realize the rectification of AC small 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 the voltage :
because ui>0 , the output voltage of U1 is uo1<0 , so that D2 is turned on and D1 is turned off . At this time , R3 , R2 and U1 constitute an inverse proportional amplifier circuit, and its output voltage uo1=-(R2/R3)ui . Take R3=R2 in the circuit , so uo1=-ui , the circuit is an inverse amplifier circuit with a magnification of -1 . 2. When the input AC small signal is the negative half cycle of the voltage : because ui<0 , the output voltage of U1 is uo1>0 , so that D2 is turned off and D1 is turned on.
. Since D2 is turned off, the signal uo1 at the output end of U1 is not sent to the next stage (that is , the input end of U2 ); because the same-direction end is grounded, according to the virtual short principle, the reverse end voltage is zero, and at this time D1 is turned on, so The output voltage uo1 of U1 is clamped at 0V (ie 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
the voltage: ①Because the voltage output by U1 is uo1 = -ui , the output of the signal after U2 is uo2=2ui ; ②The output of the signal is uo2 " =-ui ; ③The output of U2 uo2 =uo2 +uo2 " = ui . That is, when the input is a positive half cycle, it is the same output in the same direction . When the input AC small signal is the negative half cycle of the 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 ; ② the output of this signal is uo2
" =-ui ;
③ U2 's output uo2= uo2+ uo2 " =-ui . That is, when the input is a 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, thus realizing AC rectification
The AC small signal first passes through the half-wave rectification part to generate a half-wave signal , which is then sent to the latter stage for superposition and inversion with the input signal , and the output waveform is a full- wave rectified signal . This signal can get a relatively stable DC signal after the first-order filter circuit . In the circuit diagram, U1 , D1 , D2 , R3 , R2 form the half-wave rectification part; U2 , R4 , R6 , R5 form the superposition reverse part; R1 , C1 form the first-order filtering part . In the circuit, the diode conduction voltage is about 0.6V , and the open magnification of the integrated op amp is generally 10,000 times . Therefore, a small change in the voltage at the input of the op amp can cause the output to follow it . The small-signal precision rectifier circuit takes advantage of this feature to realize the rectification of AC small 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 the voltage :
because ui>0 , the output voltage of U1 is uo1<0 , so that D2 is turned on and D1 is turned off . At this time , R3 , R2 and U1 constitute an inverse proportional amplifier circuit, and its output voltage uo1=-(R2/R3)ui . Take R3=R2 in the circuit , so uo1=-ui , the circuit is an inverse amplifier circuit with a magnification of -1 . 2. When the input AC small signal is the negative half cycle of the voltage : because ui<0 , the output voltage of U1 is uo1>0 , so that D2 is turned off and D1 is turned on.
. Since D2 is turned off, the signal uo1 at the output end of U1 is not sent to the next stage (that is , the input end of U2 ); because the same-direction end is grounded, according to the virtual short principle, the reverse end voltage is zero, and at this time D1 is turned on, so The output voltage uo1 of U1 is clamped at 0V (ie 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
the voltage: ①Because the voltage output by U1 is uo1 = -ui , the output of the signal after U2 is uo2=2ui ; ②The output of the signal is uo2 " =-ui ; ③The output of U2 uo2 =uo2 +uo2 " = ui . That is, when the input is a positive half cycle, it is the same output in the same direction . When the input AC small signal is the negative half cycle of the 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 ; ② the output of this signal is uo2
" =-ui ;
③ U2 's output uo2= uo2+ uo2 " =-ui . That is, when the input is a 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, thus realizing AC rectification