# N-stage rectifier as AM demodulator?

I want to downconvert a simple 100kbit/s NRZ OOK/ASK signal from 900 MHz. In a previous design I found two options:

Option 1: simulate this circuit – Schematic created using CircuitLab

Option 2: simulate this circuit

L1 and C1 are an L-match (input is 50 Ohm). The diodes are Schottky diodes (HSMS-285C). I recognize both options as an instance of an N-stage rectifier (option 1: one stage, option 2: two stages).

I simulate both options in LTspice and the output voltage of option 2 is twice as high as option 1 which sounds definitely better to me. I assume this would not increase the SNR (?) but it would make threshold detection of Vout easier when the voltage is larger.

Questions:

1. Why would I pick option 1 over option 2? Is there any advantage of using option 1?
2. Why stop at 2 stages? Why not multiple stages? (I know that the voltage multiplication will be limited by the parasitics and the diode drops). But could there be a deeper reason why the original author used only one or two stages?
• What's the peak to peak voltage at RFIN? What's the forward voltage of your diodes at the current you expect to draw from the demodulator? – JRE Jul 20 at 15:07
• And, finally, what will the total impedance of your multiplier be? – JRE Jul 20 at 15:09
• ^ Questions for you to consider in trying to figure out the reasoning behind the circuit. – JRE Jul 20 at 15:10
• @divB: The Schottky symbol is hidden down at the bottom of the regular diode properties under "DISP". Just note it in your post rather than edit the schematics. – Transistor Jul 20 at 15:27
• @MarcusMüller this ia a great question ... this is from a passive RFID tag and I think these tags don't have a high selectivity in general. I assume the reason is that the typical sensitivity of a tag is far higher than everything else (> -20dBm) ... – divB Jul 20 at 17:03

You might achieve more sensitivity by operating a diode with 1uA or 10uA of bias current. A 1 MegOhm resistor from +9 volts will do this. For reduced capacity, use a small leaded resistor??

You need a R+C load: 100Kohm and 10pF is 1uS time constant, giving you excellent response to your 10uS bit times.

Your ZERO RF output will be +9v * 100,000/1,000,000 = 0.9 volts.

You can drop the Grounded resister (the 100,000) to 10K if you want.

The primary load on the RF input, given the diode is biased ON, will be the 10pF capacitor.

Given 1pF = -J 159 ohms at 1GHz, the 10pF will be -j 16 ohms, a heavy load on an antenna. You might use just a few picoFarads, then into 10Kohm series into 100pF shunt, into opamp buffer. This light load should give 5X larger rectified voltage.

Given the diode is biased ON, a fast diode should handle the 900MHz.

Given 0dBm across 50 ohms is 0.632 volts PP, and -20 dBm is 0.0632 volts PP, and -40dBm is 0.00632 volt PP (a mere 6 milliVolts), you have interesting possibilities here.

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To recover the OOK/AM modulation, you need to remove the RF (1.1 nanosecond period) yet preserve the 10uS bit times.

A 3_component cascade: 220,000 ohm to +1v, into the diode anode, with the cathode to a 10,000 ohm resistor (in parallel with 100 pFcap to remove the 900MHz), should do this.

GIven 100pF has only 1.6 ohms reactance, compared to the 10,000 ohm resistor, the RF will be well suppressed limited by your Ground Plane and your lead inductance.

• This is nice additional info but unfortunately does not address the actual two questions at all. Apart from this: 1.) There is only 1-3V supply. You would connect the 1MOhm from there to Vout node? 2.) Why do you need explicit R load? The diodes have parasitic resistance – divB Jul 20 at 17:08

The diode pump will give more DC volts when the RF input is well above the voltage drops of the unbiased diodes.As you add more stages the Diode drops also go up in proportion.This means that weak signal detection will not be improved by using a large number of stages .Capacitive loading will also become a problem at your frequencies.I did consider this for an AM detecter at 455KHz.