Load needed for op-amp used as an ideal rectifier?

[brett]

Hi
I have a circuit rectifying an ac signal of about 1 kHz. It's an inverting op-amp (TL071) with diodes to make it an ideal rectifier. It has a single sided power supply (0, 9V) and 4.5 V for Vref ( fed into the +ve input, as is usual). The input signal goes to the op-amp -ve input and is DC isolated by a film cap.

I will put a DC voltmeter across the output to read the rectified voltage.

My questions are:
1. will this work if I don't put a load (e.g. 10k resistor) on the output?
2. The load should be between the output pin and Vref. Yes?
3. Would the meter (eg. A 10 MegOhm DMM) be enough of a load to make the op-amp 'behave'?

I am novice as far as DC op-amp applications are concerned. AC amplification seems so much easier.
Thanks for your help.

[Keppy]

Hi
I have a circuit rectifying an ac signal of about 1 kHz.

Half-wave or full-wave rectifier? I'm assuming half, since you're using a single opamp.

It's an inverting op-amp (TL071) with diodes to make it an ideal rectifier. It has a single sided power supply (0, 9V) and 4.5 V for Vref ( fed into the +ve input, as is usual). The input signal goes to the op-amp -ve input and is DC isolated by a film cap.

Are you feeding it through a resistance? That would be the usual method when using the inverting input.

1. will this work if I don't put a load (e.g. 10k resistor) on the output?

It depends. If you mean the output pin of the opamp isn't connected to anything except the feedback loop, then yes, it should work fine. An open connection is essentially an infinite resistance load, which is fine for an opamp. Opamps respond poorly to very small resistance loads, such as a short to ground/Vref.

2. The load should be between the output pin and Vref. Yes?

It depends. If the load resistor is attached directly to the output, then Vref. If it follows a coupling cap, then it depends on the next stage of the circuit (if there is one), since in that case the load resistor for the rectifier would likely double as a biasing resistor for the next stage.

3. Would the meter (eg. A 10 MegOhm DMM) be enough of a load to make the op-amp 'behave'?

Yes, easily.

I am novice as far as DC op-amp applications are concerned. AC amplification seems so much easier.
Thanks for your help.

You keep mentioning DC. If you are trying to create DC from 1kHz AC, then you will need additional smoothing. Otherwise, your rectified signal sounds like 1k through a fuzz pedal (half-wave) or like distorted 2k (full-wave). Pure DC, by contrast, is silence. The usual smoothing method is to send the output of the rectifier through a small series resistance (100R-1k) into a large shunt capacitor (1-5 uF is common, YMMV). These components form a lowpass filter to reduce ripple.

I hope this helps. If you need any more input, you'll probably have to provide a schematic or a description of what you're trying to do with the rectifier.

[brett]

Thanks Keppy.
It's a full wave rectifier. An op amp with a diode in the neg feedback loop, as well as a diode on the output. There are dozens of them on the web as 'ideal rectifiers'. But they never seem to show the loads.

Yes, it's got 10 k of input resistance (and 47 k of feedback resistance for a gain of about 4.7).

I am concerned that the 'load' is too small (in current terms. i.e. too much resistance). Most application notes indicate loads of 10 to 100k and some capacitance (about 100pF, which increases stability, despite what I assume is a reduction in high frequency feedback).

Yes, I'm after DC, so I'm not using a DC blocking cap, but I am using a smoothing cap. The TL071 has some output impedance (128 ohms as protection against short-circuit) and I'm adding 1k to that, and using a 22uF cap.

Thanks for helping me think this through.

Cheers

[knutolai]

Its nice to add a picture

[merlinb]

It's a full wave rectifier.

More likely half-wave.

In the inverting configuration, the feedback resistor is a load, so your smoothing cap will discharge through that (although there's nothing to stop you adding another resistor in parallel with the smoothing cap).
 In the non-inverting configuration you would indeed need a extra load resistor to allow the cap to discharge, otherwise it would charge up to the peak voltage and stay there for ages; it wouldn't "follow" the input signal.

[brett]

Thanks everyone.
Apologies for not posting a picture : my internet is broken except for poor phone link (lightning strike!)
Doh! Feedback resistor is a load. Of course.
Building it tmrw.
Cheers

[PRR]

There are at least 345 different ways to do "precision rectifier", and one-opamp versions were a VERY popular sub-set in days when an opamp was two tubes and $70.

If you can't post a picture, you need to point-to one, or we will all be talking about different things.

> The load should be between the output pin and Vref.

Only IF the load "floats". A battery DMM (or a passive meter) floats. A good solid bench VTVM doesn't. "Other parts of the box", like control inputs to compressors or bar-graphs generally have one side hard-grounded.

It is very awkward to develop your DC referenced to V/2, then try to get it re-referenced to ground. (I actually built a whole 2-stage stereo dynamic control and monitor amp once, and didn't see this until the end.)

Many of the one-opamp plans, especially full-wave, are critically sensitive to loading. I worked with one recently, if I find the sketch I'll post.

1KHz?? Most $3 DMMs will do 1KHz pretty good and for signals as small as a simple "precision rectifier" can do. (The AC/DC converter inside the DMM may be as sophisticated as your plan, or even a penny fancier.)

If you are really reaching into the depths of hiss-noise, use an *AC* amplifier to bring it up to part-Volt, then rectify (perhaps with your meter's AC mode).

[PRR]

1 opamp rectifier with pretty good precision and ripple filtering.



This was to read my house current consumption, from 100A down to 1A, with good precision in the 10A area. A passive coupling (not shown) converted 0A-100A AC to 0.1V-10V AC from 100 ohm resistance.

The 1,520 ohm string in the middle is *critical*, because the conversion is half-active. When input goes negative, opamp inverts and drives the string strongly with R2/R1 of gain. But when input goes positive, the opamp slams to zero, and the string is driven through the opamp's feedback string R1+R2. Just this far, the equation for symmetry is non-obvious.

But the output is all ripple and I wanted fair filtering before the DC DVM. Low ripple but a rise-time shorter than my motor loads. Active filtering was out because I am lazy. Large resistors would skew the accuracy into a 10Meg meter load, small resistors skew the symmetry because of the large difference in rectifier output impedance for each polarity of input.

The solution shown is inexact, it is a compromise of several ~~1% errors different ways.

It would have done the job, for about $90 of parts (current transformer, reliable 5V power, and etc etc, for two phases); and then I found a Chinese module with everything pre-built and back-lit AND with voltage monitor, for $20.