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Ideal rectifier with single supply opamp

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hash

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Hi

Please tell me if this would work as a half-wave rectifier for a 1Vpp bipolar square wave. Would the capacitance of the schottky diode interfere with the reading? I need to find out the peak value of the input signal.

Sorry for the mspaint drawing but i wasn't able to find schottky diodes in orcad (I'm using a 1N5819 or 1N5711)

Thanks.
 

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Hero999

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hash

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Thanks for the respose but the problem is protecting the opamp input from reverse voltages in excess of -0.3V (the maximum for single supply according to the datasheet), that's why that diode is there, and I think I've put it in the wrong direction (I want to clip the negative side of the waveform)...
I need to know if it can hold out indefinitely, given that I'm driving the opamp with some reverse voltage.
 

crutschow

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Yes, to clip negative voltages you need the diode anode to ground. Also add a resistor in series with the input to limit the current, 10kΩ or so.

The diode capacitance should not be a problem unless the frequency is high. What is the maximum operating frequency?
 

hash

New Member
1KHz but the signal is a square wave and I need to have a somewhat precise measurement of the peak voltage. The filtering cap lowers it a bit but I can compensate for that.
 
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Hero999

Banned
Does the squarewave ever go below 0V?

Oh I didn't think about it enough.

You need to treat it as a dual supply circuit and AC couple the input. If you want the output to go down to 0V you'll need to level shift it back to 0V.
 

hash

New Member
Yes, it's a 1Vpp bipolar square wave. That means it's going from -0.5 to 0.5 volts at its maximum. It is changing its amplitude based on a sensor's output, and I need to find its (positive) peak value.
I don't have a negative power supply in my circuit, so a single supply opamp is all I can afford :)
I need to know if this is a good design, that is, if you'd be willing to place it in a real world application/product (it's not that, just a project, but the teacher will sure ask me).
 

Hero999

Banned
I've just thought about it more and what I've said above only applies if it's a fullwave rectifier not halfwave.

The design in Wikipedia is a good design but you'll need to add a 10k resistor and diode to protect the op-amp.

It's also very easy to integrate gain into this circuit if you need it.
 

hash

New Member
Is there a reason to include the diode (or diodes) in the feedback loop given the fact that the opamp is limited to 0V by its power supply?
 
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Hero999

Banned
No there isn't, the original circuit will work, as you said the output is limited to 0V by the supply.
 

dougy83

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The output of the opamp will be a series of 500us pulses. Is that what you want? Or do you want a constant voltage that represents the peak of the input voltage? (this would be called a peak detector)

Also, is the negative part of the signal as accurate as the positive part? You could use a full-wave precision rectifier following by a precision peak detector (2 opamps in total). Or just use an inverting precision peak detector (1 opamp). Neither of these have to worry about that protection diode you mentioned.
 

hash

New Member
The output of the opamp will be a series of 500us pulses. Is that what you want? Or do you want a constant voltage that represents the peak of the input voltage? (this would be called a peak detector)
Yes, a peak detector is the final goal of the circuit. I was going to accomplish this with a filter cap after the opamp-rectifier stage.
I thought about a full wave rectifier but I only have at my disposal a single power supply (+5V). The designs that I've found that incorporate a single supply opamp are far more complex that the one I drew in my first post.
The output goes into an A/D converter so with some calibration I'll be able to compensate for the loss of voltage from the half wave rectifier... at least in theory :)

Thanks for the replies!
 

ericgibbs

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Yes, a peak detector is the final goal of the circuit. I was going to accomplish this with a filter cap after the opamp-rectifier stage.
I thought about a full wave rectifier but I only have at my disposal a single power supply (+5V). The designs that I've found that incorporate a single supply opamp are far more complex that the one I drew in my first post.
The output goes into an A/D converter so with some calibration I'll be able to compensate for the loss of voltage from the half wave rectifier... at least in theory :)

Thanks for the replies!
hi,
Have a look at this option.:)
 

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ericgibbs

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hi,
Have a look at this option.:)
For a 5V supply you should get about +3Vout,, use a Vref of +3V on the PICs adc.


EDIT: cannot explain this extra post, selected Edit.?
 
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dougy83

Well-Known Member
I thought about a full wave rectifier but I only have at my disposal a single power supply (+5V)
If you use the op amp as an inverting amp, the negative input becomes positive at the opamp output. Also, a positive input signal stays positive (due to the diode being reverse-biased). So no negative supply is needed. Also, neither input will be dragged below zero volts, so the opamp is happy:)


Have a look at the attached cct. The single opamp inverts a negative signal and charges the cap if the input signal magnitude is greater than the voltage on the cap (just like a perfect diode). When the input is positive, it has little effect on the cap (200k charging/discharging the cap). Therefore it's basically a half-wave rectifier.

The 2 opamp cct has a similar rectifier cct but the output is buffered by a precision diode. So now both +ve & -ve parts of the input are used to update the value on the capacitor.

Enjoy!
 

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dougy83

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@ericgibbs: Why do you use that precision diode configuration with the 6dB attenuation?
 
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ericgibbs

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@ericgibbs: Why do you use that precision diode configuration with the 6dB attenuation?
hi,
We get a number of OP's asking for an absolute precision low signal rectifier, mainly to suit mains current transformers.
The peak filter on the second stage is a simple addon.

The first stage rect circuit is from the Intersil applications data.

If you test/simulate the peak circuits you have posted you will find that the dont perform well at low mV ac levels.

Most times, its a guessing game trying to find out what an OP really wants.:)
 
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hash

New Member
If you use the op amp as an inverting amp, the negative input becomes positive at the opamp output. Also, a positive input signal stays positive (due to the diode being reverse-biased). So no negative supply is needed. Also, neither input will be dragged below zero volts, so the opamp is happy:)


Have a look at the attached cct. The single opamp inverts a negative signal and charges the cap if the input signal magnitude is greater than the voltage on the cap (just like a perfect diode). When the input is positive, it has little effect on the cap (200k charging/discharging the cap). Therefore it's basically a half-wave rectifier.

The 2 opamp cct has a similar rectifier cct but the output is buffered by a precision diode. So now both +ve & -ve parts of the input are used to update the value on the capacitor.

Enjoy!
Well... that seems to do the job but I can't quite grasp the fact that the negative input of the opamp becomes positive... From the simulation I've run it seems that way, with a spike of negative voltage in front of it (perhaps just before the feedback loop works its magic). Will this damage the opamp in any way?

@ericgibbs: Thanks, but I think that design is a little too complicated for me :) I need something simple which I can fully understand, if the teacher decides to ask questions.
 

dougy83

Well-Known Member
Well... that seems to do the job but I can't quite grasp the fact that the negative input of the opamp becomes positive...
The opamp is configured as an inverting amplifier. So the opamp output voltage, for a negative input will be positive. The output voltage will be such that the voltage at the negative input - the junstion of the resistor divider - will be 0V (i.e. the same as the +ve input).

When the input goes positive, the output of the opamp tries to go negative, but it's limited by the negative supply of the opamp - which is 0 volts. The diode is also reverse biased in this scenario.

If you were to use a negative supply rail for the opamp, when the input went positive, the opamp output would go to the negative rail. This is because the diode is reverse biased, and therefore there is no negative feedback for the opamp; the gain becomes huge. Try simulating the circuit with a negative rail on the opamp; you'll see the large swing just mentioned.

From the simulation I've run it seems that way, with a spike of negative voltage in front of it (perhaps just before the feedback loop works its magic).
Any negative voltage will be an artifact of the simulation software I would think. There will be a delay for the peak detector due to the opamp having to jump from 0V to the peak voltage+0.7V (diode drop), especially for low slew rate opamps. You can reduce this with a extra diode and resistor (see attached). The extra diode stops the output from going more the 0.7V below the positive input. The extra resistor in the feedback path does degrade performance a little.

@ericgibbs: Ok, thanks. The rectifier is just a copy.
 

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