I stumbled across the circuit shown in Fig. 36 on PDF page 21 of Jim William´s "High Speed comparator technieques" Application note (see Link)and also the attached schematic which shows a fast precision recitifier.

enter image description here

Can someone explain how this circuit works? I don´t get it based on the description in the appnote, especially the "biasing" of the diode bridge.

P.S.: I also have done an LTSpice simulation (luckily, besides of the special schottky diodes, LTSpice has all the models of the used parts ;-) ).

When looking at the "output" of the bridge the simulation only shows "spikes" (positive and negative) at the moments where the comparator outputs switch, but no rectified input signal

[EDIT]: Added schematic to this question [EDIT2]: Added hyperlink to LTSpice simulation file LTSpiceSimulation

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    \$\begingroup\$ Show the circuit here. Not gonna chase a link. Make sure the schematic has component designators, has logical layout, etc. Fix this or be closed. Starting the process. \$\endgroup\$ – Olin Lathrop Jan 3 '18 at 12:02
  • \$\begingroup\$ Just out of curiosity: What´s your problem with a link to the LT Website? Sure, I could now redraw the schematic shown in the appnote but what would be the point of that? (I would have to redraw since even the pretty schematic in the appnote does not have reference designators...) \$\endgroup\$ – Junius Jan 3 '18 at 12:11
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    \$\begingroup\$ Stop arguing and start fixing. The clock is ticking. 4 more close votes to go. \$\endgroup\$ – Olin Lathrop Jan 3 '18 at 12:14
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    \$\begingroup\$ @Three: It's still the OP's job to produce a proper schematic here. It doesn't matter whether someone else created some or all of it for him. That only determines how much work the OP must do, but none of that is our business. \$\endgroup\$ – Olin Lathrop Jan 4 '18 at 12:01
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    \$\begingroup\$ @Three: Schematics without part designators are not acceptable here. They make the circuit very difficult to talk about. The OP's solution is acceptable. He added his own component designators on top of the existing schematic. Again, we don't care how it is accomplished, only that it is. \$\endgroup\$ – Olin Lathrop Jan 4 '18 at 13:20

That's a lot more complicated than the typical precision rectifier I've seen.

Let's get the easy stuff out of the way. R18 and C6 low pass filter the result out of the rectifier, then IC2 amplifies it. Depending on the time constant relative to the intended AC frequency, this yields the DC "average".

At first glance D5-D8 might look like a full wave bridge, but it's really not. It seems it's really two diode pass gates. Note that the raw input is connected to the left corner. Think of D5-D6 separately, and D7-D8 separately.

Look just at D5 and D6. They block the incoming signal when the bottom of R15 is high, and pass it when low. Each of the two diodes ideally has the same voltage drop, so when R15 is driven high, the input voltage appears at the output.

D7 and D8 work similarly, except that a positive voltage (on R16) is needed to pass the signal from input to output.

The mess of stuff between each of IC1's outputs and the diode gate control signals seems to be drive circuitry with a bunch of fancy feed-forward speedup compensation, although I'm not really sure about that. I don't have time to analyze that carefully right now.

It can be a lot easier to come up with a circuit from specs, than to look at someone else's circuit to infer the specs it was designed to. At least in this case you can assume competent design. Linear app notes and sample circuits are usually quite carefully done.

The input circuit to IC1 is essentially a differentiator. The comparator output will go high on rising slope of the input, and low on falling slope. I haven't worked out how switching the pass gates on the slope of the input achieves the desired result, but I have to go now and get some real work done.

The first thing to do is look to see if Linear offers any explanation of how this circuit is supposed to work. Sometimes they explain carefully, sometimes not. Maybe someone else can jump in here with a more detailed analysis.

  • \$\begingroup\$ Thanks for the reply! I cannot follow your thoughts about D5/D6. When the bottom of R15 is high, the incoming signal is blocked. But when bottom of R15 is low (i.e. -5V), isn´t the raw signal "shorted" to -5V via D5 and R15? One additional comment: I want to analyse this circuit to find out whether it can be used for very small input signals (i.e. below 15mV down to about 1mV). All the "typical" rectifier circuits I know have a problem below that, even with fast opamps. You see, It has a reason that I have a look into more complex circuits". \$\endgroup\$ – Junius Jan 3 '18 at 14:29

This is actually pretty simple, but it makes use of a diode bridge, which is not commonly done nowadays.

Take this circuit


simulate this circuit – Schematic created using CircuitLab

Start by assuming V1 = V2 and R1 = R2, and R3 is infinite. Then for Vin = 0, so does Vout. Call the current through R1 or R2 the reference current. Note that, for a fairly large range of input voltages, the output voltage will follow the input voltage quite closely. Although the input source will provide extra current through D2, since the diodes are biased on with about half the reference current, this will have little effect on the diode voltage. So in general, the top and bottom of the bridge will follow the input, and so will the output.

Similarly, making R3 non-infinite will cause inaccuracies in the output voltage, but the effect will be small so long as R3 does not become too small. Note that in the ap note, R17 is about twice R15 and R16, and at these levels you can expect errors of about 1%.

Of course, this only applies with the voltages shown. If the polarities of V1 and V2 are reversed, the bridge becomes reverse-biased, no current flows, and Vout becomes zero. In addition, from an AC standpoint, Vin and Vout are decoupled by the capacitance of the diodes, and at the frequencies in the ap note this provides a high impedance. So for one polarity the bridge acts as something like a short circuit, and for the other it acts as an open circuit. In other words, a rectifier. Admittedly, it's not as good as a standard op-amp precision rectifier at low frequencies, but it works well for frequencies one to two orders of magnitude higher than such rectifiers.

The rest of the circuitry simply senses the polarity of the input and switches the bridge voltage as appropriate. Of course, as with all Jim Williams' circuits this involves clever handling of various important (but not always obvious) effects. In this case, the switching of the gate voltage must be very fast and precisely synchronized. If not, you'll get switching transients feeding the output.

  • \$\begingroup\$ Thanks, with the "bridge" redrawn and your explanation, I think I understand that part of the circuit. So that means from an AC standpoint the input source is more or less loaded by R1||R2 + R3 ? Also, can you maybe say some words about the "biasing" circuitry? For what I see do far, the circuit also has troubles with small amplitude levels of the input voltages.. \$\endgroup\$ – Junius Jan 3 '18 at 15:55
  • \$\begingroup\$ @Junius - You're correct about loading. Not so much about signal amplitude. For the circuit shown, anything below a couple of volts AC will show identical performance. \$\endgroup\$ – WhatRoughBeast Jan 3 '18 at 19:44
  • \$\begingroup\$ I will try to rethink the limitations of the circuit for small signal amplitudes, however I fear that without a good simulation, I am lost. Unfortunately (as I said in my post), I haven´t managed to get useful results (also not for totally suitable input signal amplitudes) from an LTSpice simulation, though I have checked the schematics. Is it possible to somehow upload my sim file here? \$\endgroup\$ – Junius Jan 4 '18 at 12:17
  • \$\begingroup\$ I added a link to my LTSpice simulation file in my post \$\endgroup\$ – Junius Jan 4 '18 at 12:29

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