Help on envelope detector circuit

[faisal78]

Hi
I had found a reference design from ST application notes on inductive proximity switch circuit.
**broken link removed**



The circuit is basically a simple resonant circuit in which 'inductance' is changed due to metal proximity, and then a simple envelope detector circuit to convert it to DC to be measured by a ADC.
Circled in green is the oscillator circuit, operating ~800kHz.
Circled in red is the 'resonant' circuit...L1 & C12 with L1 being ~68uH.
Circled in black is probably the envelope detector circuit.

Some googling on envelope detector does not include the diode D5 in parallel.
Can somebody shed some light on what is the purpose of D5.
It is however needed based on some simulation that I did but I cannot explain why.

Also, what is the criteria to choose values for C10 and the RC circuit C13 and R7. How do I choose this RC value?

Thanks in advance.

 

[crutschow]

D5's purpose is to recharge the capacitor on the negative portion of the signal. Otherwise as the capacitor passes unidirectional current (charge) through D4 its output would become negatively charged until no more current could flow through D4. A capacitor can not pass DC current.

C10 just needs to be large enough so it has a relatively low impedance at the carrier frequency as compared to its load impedance.

The RC value of C13 and R7 acts as a low-pass filter that roughly determines the high frequency modulation rolloff frequency so it should be set somewhat above the maximum modulation frequency to be detected.

 

[faisal78]

Crutschow.,
So if I understand you correctly, D4 would pass thru the positive +ve swing of the AC signal from the resonant circuit, whilst, D4 would pass thru the negative swing of the AC, kinda like a full wave rectifier?
How is this then different from a full bridge rectifier using 4 diodes?
;Why wouldn't a AM diode detector need a parallel diode ?

 

[FvM]

Explanation of the voltage doubling rectifier (Greinacher circuit)
https://en.wikipedia.org/wiki/Voltage_doubler

You can turn the circuit in a simple half-wave rectifier by omitting D5 and C10. In any case, there must be a DC path for the output current of the rectifier.

 

[faisal78]

FvM - The Greinacher circuit tip really helped.
However I would think the capacitance value of C10 (47pF) and C13 (1.5nF) would need to be the same value if it was to 'double' equally right?

- - - Updated - - -

Actully i think I get it now. C10 & D5 would for a Villard doubler.
D4, C13 & R7 would then for the envelope rectifier.
Did i get it right?

- - - Updated - - -

Villard doubler + Rectifier = Grinacher doubler