Remove DC Bias in Precision Rectifier

Thread Starter

bonchenko

Joined Apr 11, 2013
2
Hello, my name is Bontor. I am trying to measure current using Hall effect sensor ACS758. Its output have DC offset 0.5*Vcc (I use 5Vcc). The resulting signal is AC signal with 2.5V DC bias.
I am trying to convert the AC signal to DC using precision rectifier circuit in here (http://sound.westhost.com/appnotes/an001-f8.gif). The problem is, with DC bias, the output will not rectified. If I am adding capacitor in series with the input, the bias is still exist (only reduced)

Do any of you have suggestion as how can I get DC output (with voltage = Vpeak of the AC signal) from the Hall effect sensor? I am going to read the value using Xbee, which is a radio module that cannot be programmed. Therefore, I am planning to read just the DC value and calculate RMS by dividing it with 1.4 (square root of 2)
 

#12

Joined Nov 30, 2010
18,224
The first answer is: Use a power supply that has both a positive supply and a negative supply, and rectify from 0V DC input. Because nobody wants to do that, rail-to-rail operational amplifiers were invented so you can input to a 0 volt level at your precision rectifier.
 

Thread Starter

bonchenko

Joined Apr 11, 2013
2
I have already done that, if I generate 22 mVpeak signal using signal generator, 50 Hz without DC bias, the output is rectified (tried on two circuit, one using op amp with positive and negative supply, second circuit use single supply). It works.

The real problem is that the sensor (ACS 758) automatically add DC bias in its output. How can you remove the DC bias without cutting down the output voltage?
 

#12

Joined Nov 30, 2010
18,224
I did not read carefully enough. I will post a circuit that can be set for any DC offset in the positive and have independent AC gain. I have not customized it for you. Is it enough for you to finish the design for your needs?

and..another idea where the input DC creates it's own compensation voltage. This amplifier chip has a rather large input bias current compared to modern chips. The filter on the negative input could be a lot higher impedance if you change the amplifier number.
 

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#12

Joined Nov 30, 2010
18,224
It doesn't matter. X number of peaks per second or 2X number of peaks per second will read the same on a peak detector.
 

Ron H

Joined Apr 14, 2005
7,057
It doesn't matter. X number of peaks per second or 2X number of peaks per second will read the same on a peak detector.
I know that, but how is that relevant to my question? The implementation for a peak detector is different than for a rectified but unfiltered output. I thought that if he wants DC out, we could maybe design something for that. Of course, you can put a LPF on the rectified output, which might be OK.
 

#12

Joined Nov 30, 2010
18,224
how can I get DC output (with voltage = Vpeak of the AC signal) Therefore, I am planning to read just the DC value and calculate RMS by dividing it with 1.4 (square root of 2)
The OP covered your question in post #1.
I am not trying to be relevant to your question. I am trying to be relevant to the OP's question.
 

Ron H

Joined Apr 14, 2005
7,057
The OP covered your question in post #1.
I am not trying to be relevant to your question. I am trying to be relevant to the OP's question.
I don't mean to nitpick or be argumentative. I was just trying to resolve the difference between his statement:
how can I get DC output (with voltage = Vpeak of the AC signal)
and his link to a schematic.
I still am not sure which he wants, but I would guess it is his statement, and not the circuit in the link.
 
I have found that the circuit shown in Figure 6 on Rod Elliot's page about precision rectifiers to have very low DC offset and it uses very low cost op amps:
http://sound.westhost.com/appnotes/an001.htm

If you add a capacitor to the circuit before R1 it will block the DC portion of the input signal. 1u will give you a 15 Hz cutoff frequency, 10u 1.5 Hz, etc.

-Charlie
 
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