AD8512 Half-Wave and Full-Wave Rectifiers

It was produced in the p16 of the data sheet "Half-Wave and Full-Wave Rectifiers" circuit uses the AD8512 our customers.

It is the output level of the output signal is different.
I will attach the output waveform.

Why level or probably would be different?
In addition, the What is the cause?

Regards,
Yuji

AD8512_Waveform.pdf
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  • Hi Yuri,

    In theory, it should work with ideal amplifiers but due to the inherent offset of the amplifiers and the gain of the circuit the output will not be perfect. The device doesn't really rail to the ground during the negative half cycle. So, base on this equation during the negative half cycle: Vout= (1+R7/R2)Vo -(R7/R2)Vin, where Vo is the output of the first stage and supposedly zero/ground but is not and is being amplified by gain of (1+R7/R2) which is not present during the positive half cycle of your input causes that difference in the out level of the circuit.

    On the other hand, may I ask how your costumer measures the output based on the screenshot, the output voltage is 200mv while the input is 2V. do use something to attenuate the signal?

    Regards,

    Phil

Reply
  • Hi Yuri,

    In theory, it should work with ideal amplifiers but due to the inherent offset of the amplifiers and the gain of the circuit the output will not be perfect. The device doesn't really rail to the ground during the negative half cycle. So, base on this equation during the negative half cycle: Vout= (1+R7/R2)Vo -(R7/R2)Vin, where Vo is the output of the first stage and supposedly zero/ground but is not and is being amplified by gain of (1+R7/R2) which is not present during the positive half cycle of your input causes that difference in the out level of the circuit.

    On the other hand, may I ask how your costumer measures the output based on the screenshot, the output voltage is 200mv while the input is 2V. do use something to attenuate the signal?

    Regards,

    Phil

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