Have no idea from your limited informationWhen the left port is left unconnected I measure about -1v DC at the left port and about 2vpp at the right port.
When a DC is applied to the left port, I measure a lower voltage (~1vpp) on the right port.
What is going on?
This is quite expectable.
What worries you? What did you expect?
That's just the diode working as a rectifier.When the left port is left unconnected I measure about -1v DC at the left port and about 2vpp at the right port.
If you apply a negative voltage, the diode junction capacitance will drop and you might get marginally more RF at the right.When a DC is applied to the left port, I measure a lower voltage (~1vpp) on the right port.
If you apply a positive voltage, the diode will conduct and start to look more like a resistor so the input capacitor will start to shunt the signal.
I think what you are trying to do is make a simple diode RF switch rather than in inverter gate. The trick is to pass the signal through two back to back diodes in series. At the input and output use resistors or ideally chokes to ground to provide a DC path. At the junction of the two diodes feed the switching DC voltage, again through a resistor or chokes. The idea is that if you apply a voltage to make the diodes conduct, they become like low value resistors and the signal passes through. If you reverse the voltage the diodes look like low value capacitors and the signal is blocked.
Diodes are quite commonly used for signal steering, many receivers use them to direct signal paths through different filter circuits for example. They use the two diode approach I described earlier though but add the filter between the diodes.My idea for this weird inverter is to switch the RF to different parts of the circuit
I'm confused when you say logic '0' or logic '1' at the output. By logic state are you referring to the presence or absence of signal?
If so, be careful with 'shunt switches' using diodes to steer signals into ground because they have no isolation, they literally short the signal out. You will also get clipping of the signal unless you provide a reverse bias voltage to keep the switch 'on', just removing the DC turns the circuit into a rectifier.
I don't call this "leaking". It' the normal forward biased diode operation, where the anode is more positive than the cathode.I think that my problems with this schematic are related to the RF leaking through the input diode (negative cycle) ans back to the input (left port). This negative rectified voltage is combined to the preceding stages positive DC and pulls this down.
Hi,
I don't call this "leaking". It' the normal forward biased diode operation, where the anode is more positive than the cathode.
The diode simply is conductive.
Klaus
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