I am trying to prove/disprove the concept. I hear that a diode could be used to switch low power analog signals. The idea is this:

In this case the diode is reverse biased, so it is off and the signal does not pass thru it.


In this case the diode is forward biased, so the analog signal will pass thru.


So, I built the circuit. I use a 300 Hz, 1 V sine as source and oscilloscope to view the output signal. What happens in reality is the signal does pass in both cases, since the GND-VCC circuit conducts too -- whatever it is: a microcontroller, a PSU or anything else. Even if I use a bare battery with a 4.7k resistor to apply forward/reverse voltage directly to the diode, the circuit does pass signal even when the diode IS reverse-biased.

Does that mean this circuit is a hoax? Or, is there a correct way to do it?

p.s. It sort of does work if I DC offset the input signal relative to the diode without applying voltage to the diode. But this has to be done with an isolated current source which sort of renders the whole idea pointless.

  • \$\begingroup\$ Your schematics are missing the component values and the input impedance of your oscilloscope or whatever you're using to measure. \$\endgroup\$
    – Transistor
    Feb 8, 2022 at 22:24
  • \$\begingroup\$ The resistors are 4.7k each and the capacitors are 0.68uF each. Oscilloscope is using a probe with a 1:10 divider so the impedance should be infinite. \$\endgroup\$ Feb 8, 2022 at 22:27
  • \$\begingroup\$ But what is the amplitude of the signal before and after it goes through the diode? Do you notice any phase shifts? \$\endgroup\$
    – Voltage Spike
    Feb 8, 2022 at 22:28
  • \$\begingroup\$ No, it looks the same. The concept sort of does work. When the diode is forward biased, the sine passes thru unchanged. Same amplitude and shape as input. However, there is no way to reverse bias the diode without leaking the signal. That's the problem. \$\endgroup\$ Feb 8, 2022 at 22:31
  • \$\begingroup\$ As a diagnostic (to check for trouble with the setup), try removing the diode altogether and measure how much signal still leaks through. If you're using solderless breadboard, that may have significant parasitic capacitive coupling between rows. \$\endgroup\$
    – MarkU
    Feb 8, 2022 at 22:38

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Figure 1. Top - biased off. Bottom - biased on.

enter image description here

Figure 2. The CircuitLab simulator results. Blue = Node1. Orange = Node2. Brown = Node3.

It looks like it works. The NODE3 voltage will decay over time given by C4 × R8.


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