# How does a Schottky diode rectify RF signals

In low frequencies, the diode rectifies the input signal depending of the bias polarity. This is because (the simples case) the diode is consider either an open or closed circuit.

In high frequencies (> 300MHz) the diode is consider as a equivalent circuit (Fig.1) but I do not understand how the rectifying process occurs.

My assumption is that, when talking about RF, the source must be adapted to the input impedance of the diode. When the impedances are adapted the signal is rectified.

• Where did fig 1 come from - you may be misrepresenting what it means. – Andy aka Feb 28 '18 at 17:41
• Have you considered using two resistors to improve the rectification? Have 100 ohm to GND from the cathode (the bar on the symbol), and perhaps a capacitor to remove RF half-wave; this is the output. Have 1MegaOhm to +9volts, attached to anode (base of the arrow), along with some antenna; the 9uA ensures the diode is "on", and the hundreds of milliVolts of DEAD BAND becomes much smaller. – analogsystemsrf Mar 1 '18 at 4:44

A fatal misconception: The equivalent circuit is for small signals, so low amplitudes added to DC that the polarity of the signal does not change. The DC value affects Cj and Rj.

=> The equivalent is useless for large signals. They are rectified like diodes generally rectify AC. Only the Schottky diode materials are different than in normal diodes. Schottky diode turns faster to non-conductive than normal PN diode, when the signal polarity reverses.

BTW. the same equivalent circuit is valid for large signals when the diode is not Schottky, but a PIN diode. That's ultra slow. It doesn't rectify at all UHF signals, it can be used as DC controlled resistor. DC current decrease the effective Rs and increase Cj. With forward DC current Cj is so high that UHF signals get through without rectification. With reverse DC voltage Rs is high and Cj is very low. PIN diodes are used in voltace controlled attenuators and rf signal switches.

A shottky diode has a voltage/current transfer function that follows the standard Shockley equation $$I=I_S(e^{\frac{V_D}{nV_T}}-1)$$

The I and the V are instantaneous voltage and current.

If you take this equation, apply it to a circuit consisting of an RF source, a capacitor to a grounded diode carrying a small DC bias current and integrate it over one cycle of applied RF, then for large signals you have the normal 'big' signal' result of the output DC voltage changing with the applied voltage, less a relatively constant diode drop.

Interestingly for small signals, you get the output DC voltage changing with the applied power, the so-called 'power law region'.

Any real diode will have residual resistance, and a junction capacitance, which will reduce the output as the frequency rises.