Precision rectifier with OPA330

The diodes are 1N4148, by the way.

Hi Eric,

Would you please also include the simulation file results that you're referring to so we can see what type of input you're providing to the circuit?

We will work the circuit out in TINA-TI, a free simulator that we use here.  I would suggest you download the free copy so you can view the circuit simulation file and results that we'll send back to you.

Thanks!
Collin Wells

Forgot the link to TINA-TI:

http://www.ti.com/tool/tina-ti

Collin,

I'm not using a simulator.  I used a function generator to apply the input voltage, and read the output on a scope.  But let me download the simulator you linked and I'll post what I get from that.

Thanks,

Eric

Collin,

Attached is the simulation file.  I was unable to upload the results file, a .tdr file - "file type not allowed".  The results reflect the full-wave rectification I'm seeing, but not the different amplitudes.  I've attached a picture of what I'm seeing on the oscilloscope I'm using.

Thanks,

Eric

1447.Precision_rectifier.TSC

Hi Eric,

The input signal you've provided to the circuit is GND referenced and the circuit is operating with a single supply.  The circuit won't function properly if the output of the op-amp can't swing negative to make up for the diode drop (~0.7V) that allows the circuit behave as a "Precision Rectifier".  Basically, the reason the circuit doesn't have a diode drop is that the op-amp decreases its output to make up the diode drop such that the output node where the resistive feedback is taken from has the correct voltage.  Please see the results below with a single supply power-supply first and a dual power-supply second.  In the dual supply circuit it becomes apparent that the OPA output must swing negative to achieve the proper transfer function.  Then see the results of how to fix the circuit to operate on a single supply using a second op-amp and a few resistors to provide a bias.  If you can not tolerate the DC offset in the output then you will have to high-pass filter the output of the final circuit.

Here's a solution using an OPA2330 that uses the second channel to provide a 1V bias to the rectifier circuit.  This will ensure that the OPA output can swing low enough to make up for the diode drop associated with the forward voltage for the selected diode. 

4834.OPA330_Dual_Supply.TSC

Thanks Collin.  I can't believe you were able to come up with that at midnight!  Can you explain what's happening in my original circuit that produces the quasi-full-wave rectification?  I understand that the amp can't swing negative, but why is it going positive again in the "off" cycle?  I'm still learning about an op amp's response in this type of circuit.

Collin,

I got these interesting results while playing with the circuit:

It appears this does the same thing as the 2-diode, dual supply version.  Is there any drawback to using this simplified version?

Thanks,

Eric

Got it.  Thanks again!

Collin,

I'm coming back to this design after a couple of months working on other stuff.  In the last design proposal, could I use a standard resistor voltage divider to provide the 1V bias to the precision rectifier amp, instead of the unity gain buffer?  This is for an AC project, so current draw isn't a major concern.

Thanks,

Eric

Hi Eric,

I don't remember where we left off.  Could you please attach an image of the circuit in question?  The buffered version will probably be better but depending on the impedance of the nodes it must drive and the impedance of the divider then it may be fine.  Please confirm the circuit and we'll get it working.

Sorry about that.  I was just looking at the thread, but you probably got my post in an email.  Here's a picture of the 2-amp solution you posted, where the bottom amp is simply providing a 1V bias for the top amp.

Hi Eric,

Thanks for clarifying things.  I thought this was the circuit you were referring to, but I wanted to double-check before answering.

You can replace the OPA330 buffer in the circuit with a resistor divider.  Just keep the resistor divider impedance much smaller than the 499k resistor that biases the input and you shouldn't experience any issues.  If you don't are about power, keep the divider impedance lower than 499 Ohms.

Thanks Collin.  What's the purpose of the 499k biasing resistor, and how did you arrive at its value?

-Eric

Hi Eric,

I didn't look too close when I first put the circuit together and that resistor is actually not needed here.  If the impedance of the resistor was lower than the impedance of the op-amp input resistor then it would dominate the high-pass filter cutoff frequency.  

Ok, just checking.  It looked like it was serving a high-impedance role, and I wondered why an open circuit wouldn't be better than a 499K resistor.

One more thing - for some reason, the output voltage in my simulation is a compressed version of yours, even though I wired the circuit the same as your example.  I've attached the image below.  Do you see where the error might be?

Hi Eric,

Are you sure it's getting compressed?  It's a little hard to tell from your image but the "Vout" signal looks to be about the same magnitude as the "Vin" signal.  If you don't like how the results were framed you can change the y-axis by double-clicking it.

If you attach your simulation file I'll check it out.

Collin,

You're right, the Vin and Vout signals are about the same magnitude.  But I thought the Vout signal should be offset by 1V since I'm using the voltage divider in the non-inverting side of the OP330.

I've attached the simulation file for your reference.

Thanks,

Eric

6557.SingleSupplyBias.TSC

Thanks Collin.  I was getting confused by the default y-axis ranges.  After studying the graphs some more, I see the 1V offset.  I think I'm ready to prototype this circuit now.  Again, I really appreciate all your help with this.

-Eric