Precision detector Question

[N9NEO]

I am looking at the Precision detector and it looks like there may be errors with wiring as presented in the link below.  Has anyone built this circuit and didja hafta rewire where the diodes connect to?

http://www.amwindow.org/tech/htm/alowdisdet.htm

73
NEO

[WA1GFZ]

Bob,
Rob built the circuit and sent me a pile of test data. he also built a board and put it in his R4.
He reads the posts and will respond when he sees this.
It looked quite good. fc

[N9NEO]

Hey Frank.

Yea, to get specific it looks like the pin 5 of U6B should be connected to the cathode of CR801 while pin 3 of U6A should be connected to the anode of CR800. The way it is wired now will give you a negative voltage output.  No big deal.

I should hear tomorrow about that job I was looking for.

73
Bob

[WA1GFZ]

Bob,
I think it will work either way. I don't have his scope pictures at home. Maybe he needed to reverse phase in the R4. He will check in by Monday. I can tell you it was very clean.

[N9NEO]

Ok, good. I look forward to hearing Rob's comments. 

 I think the full wave rectification allows using an RC avaraging function on the output.  There is of couese a linear scaling factor between avarage and peak when in full-wave mode. This is lost with simple diode rectification.  I didn't read the whole paper - the math scares me.

I'm going to stick it on a pcb express board with some kind of elliptic filter and a headphone driver for my portable radio. I think LT makes a very sharp switched capacitor IC that takes up hardly any room and you don't need giant inductors.

Sort of an in-between exercise to help me get up to speed one day with the Sync Detector Yahoo Group.

I got a Pic Microcontroller seminar to go to on Tuesday up in Westboro which should be fun.

73
Bob

[Rob K2CU]

HI Bob,

Thanks for your interest in the precision detector!  You are correct that the rectified signal will drive the output negative from the 6 V bias point. The two halves of U6 make a simple differential amplifier. You can exchange the diode outputs, as you suggest, and it then the signal will provide a positive drive from the resting bias. In the board I designed for the R4, I actually created dual supplies with the modified power supply board and had +12 VDC as well as -12 VDC to work with. Instead of the two 18K resitors making +6 Volts to bias the circuit, it is referenced to gnd as the op amps have +/- 12 VDC for their supplies. There was no need for recovering the DC (carrier) component form the incoming signal in my application, but it could certainly be used in some receiver to develope an AGC voltage. Please note that the R4 has a 50 KHz I.F. frequency, and easy to work with.

This circuit is simply two precision half wave rectifiers that are combined to produce a full wave rectified output.  A simpler single op-amp stage differential amplifier could have been used, but then there would have been loading on the output of the rectifying stage and errors in the feedback in the LM318 stage. I wanted the diode outputs working into the high impedance non inverting inputs of the next two opamps.  Normally, such a Diff amp would have yet another op-amp to do the combining, and with more complex resistor matching. In this circuit, for the ultimate in precision, you only need match the two 3.3 K's used with the diodes, and then only the actual feedback-to-input-resitor ratios for the two LM660 stages, and not the absolute valuses of the four 18K resitors. Only the ratios to be as close to 1:1 as possible.

Each LM660 Op-amp stage has a gain of 1 + Rf/Rin which is very close to 2. BY matching the rations of Rf/Rin, you can make them the same. The output stage also inverts with a gain of 1 and adds the one to the other.


I have experimented with this circuit at a 455 KHz I.F. and there, it works better if you use three LM318's. Of course, the error in combining the two signals becomes more apparent since one goes through an additional stage and is delayed slightly. This delay only slightly increases the I.F. ripple on the output, something readily taken out by whatever additional low pass filtering is placed on the output.  Ideally, the ripple is at twice the I.F. frequency. The LM660 may be a bit slow to use at 455 KHz.


You can always just bag the full wave bit and just use one of the diode outputs. But be sure to build the LM318 circuit with both halves.


rob - K2CU

[WA1GFZ]

INA163 would be a good amplifier to use as the output summer.

[AB2EZ]

Bob, Rob, et. al

This (see link below) is a variation of a circuit published by K2CU (Rob) that employs a precision rectifier and an averaging circuit to create a precision detector for AM demodulation. The limitations from ideal performance are imposed by the maximum slew rate of the first operational amplifier, the amount of peak output voltage, from the first operational amplifier, that is required to drive the peak current that must flow through each diode to develop the necessary feedback signal, and the peak output voltage that each operational amplifier can produce. This circuit works well for me, with a 455 MHz i.f. as input, for both listening, and for monitoring the modulated envelops of the received signals. It is DC coupled. The feedback circuit of the second op amp filters the r.f. from the output signal. Additional filtering can, of course, be used. The 20k ohm feedback resistors on the first op amp are selected to minimize the current that must flow through the diodes to produce the required peak output voltages... and thus to minimize the peak diode drops. In the ideal case, this wouldn't matter... but the slew rate limitations of the first op amp make it important to minimize the diode drop.

http://mysite.verizon.net/sdp2/id17.html

[K1JJ]

A very nice, well laid out website, Stu!


Very clear and spoon fed: 

http://mysite.verizon.net/sdp2/id11.html

I may build up your modified AM detector later on...

T

[Rob K2CU]

Interesting point Stu. But...The LM833 has only a 7 V/uS slew rate compared to the 70 V/uS of the LM318....and that's at unity gain. My original circuit was set at unity gain for this reason. The LM833 circuit has a gain of 10. The peak current for a 1 V I.F. signal would be less than a milliamp. The slew rate is important as the output has to "jump" past the diode drop as it attempts to follow either the positive or negative going input as it passes the reference point. The difference in diode voltage drop is relatively minor yet has to be enough to make a big difference between the forward current and reverse leakage.

[AB2EZ]

Rob

Yes... I'm going to switch to the LM318 because it has a higher slew rate than the LM833.

Best regards
Stu

[AB2EZ]

Rob, et. al.

I switched to an  NTE918M op amp (available from Mouser), which is a direct substitute for an LM318 (rated at a 50V/usec slew rate, same pinout). It works great.

See my updated web site: http://mysite.verizon.net/sdp2/id17.html

Happy holidays to all

Stu

[WA1GFZ]

Stu,
How about trying some 1N5711 or 1N5712 with a lower junction voltage to slew across.
  gfz
Happy Holidays

[Rob K2CU]

Frank,

What especially makes a the 5711/5712 Schottky better is the very fast turn off time (100ps). Sure, the forward drop vs slew rate oF the op-amp are important when going from Off state to ON state, but equally important is the time to turn off the diode when the voltage swings on the opposite direction, combined with the low capacitance (2 Pf).

Stu, I'll have to order some of those amp my next order. Look good for this and other projects. Wonder if there is a quad pack version.

ras

[WA1GFZ]

Rob,
We had problems at work with quad amps due to substrate cross talk...I avoid them now.

[AB2EZ]

Rob
Frank

I've been thinking about the impact of the diode characteristics that you have both, correctly, pointed out...

For the previous circuit that was shown in the diagram on my web site... the peak (of the rf cycle) current through either of the 20 kohm resistors (now changed to 2.7 kohm resistors), for 5 volt peak output from the first stage (i.e., 2 volts peak at carrier, 5 volts peak at 150% positive modulation) is 250 uA. For 95% negative peak modulation, the peak of the rf cycle produces an output current of:  .1 volts/20,000 ohms = 5 uA.

The reverse capacitance of the diode I'm using is specified at 4pF.

On 150% positive modulation peaks, when the current switches to the other diode, the reverse capacitance must charge up to around 5.5 volts [i.e., the difference between the output of op amp (approximately -5.7 volts) and the negative input of the op-amp ( -5.0/21 ~ -0.24 volts]. The time constant for charging the reverse capacitance is ~ 4pF x 20,000 ohms ~ .08 microseconds = ~80 nanoseconds. This corresponds to ~8% of the duration of a half cycle of the 455kHz input signal. So that doesn't seem to create a significant problem on positive modulation peaks. I.e., the output of the detected AM envelope will be about 8% low.

On 95% negative modulation peaks, when the current switches to the other diode, the reverse capacitance must charge up to around 0.8 volts (the difference between the output of the op-amp (~ -0.8 volts) and the negative input of the op-amp (-.1 /21 ~ -.005 volts). Again, the time constant is ~8% of the duration of a half cycle of the 455 kHz input signal... but the peak voltage (of the wrong polarity) is ~0.8 volts / 0.1 volts ~ 8 x as large as the peak voltage during the desired half of the cycle. Thus the area of the undesired (wrong polarity) half cycle signal is ~ 8 x 8% ~ 64% of the area of the desired half cycle (correct polarity) of signal. Thus, at 95% negative modualtion peak, the detected AM envolope will be about 64% low.

The reverse current of the diode I'm using is specified at around 25 nA at room temperature (typically a little less at this reverse voltage), which is about .01% of the forward current at the peak of an rf cycle, at 150% positive peak modulation (5 volts output)

So... it would appear the the time constant of the 4 pF diode reverse capacitance, in combination with the 20 kohm feedback resistors I was using, is too long (8% of a half cycle of the rf input signal) to get good fidelity on negative peaks. It would be better to use a time constant that is ~1% of a half cycle of the rf input signal. Using a smaller value of feedback resistor will also increase the peak current... which will also be helpful (but not as critical an issue in this instance).

I changed the two feedback resistors from 20 kohms to 2.7 kohms, and I changed the gain of the first stage from 21 to 10 (i.e., I now use a 330 ohm resistor from the negative input to ground)

Stu

[WA1GFZ]

Stu,
Sounds like you also want to use fast op amps with a low output Z. A.D. Makes some cool parts worth checking out (810 maybe). I think some of them have an output Z of 7 ohms. This would support the extra load on the output driving the low C diodes. I know we  are getting crazy here but what the heck.