2
\$\begingroup\$

In my research for this question, I found this by a new user that hasn't been seen before or since. His low investment here and the "say goodbye to traditional bad stuff" language makes his answer a bit dubious, but just in case he's onto something, I thought I'd fish for some clarification.

His original answer is the classic unformatted block of text with no diagrams, but I think he's describing this:

schematic

simulate this circuit – Schematic created using CircuitLab

I'm pretty sure that won't work as described. Or did I read the description wrong?

\$\endgroup\$
5
\$\begingroup\$

He's talking about biasing a diode with a voltage source.

schematic

simulate this circuit – Schematic created using CircuitLab

I've taken the liberty of adding C2 since the output will have some offset, and left out his "tweaks". The threshold will depend on the ratio of currents between the two diodes. The current through D2 will obviously be less so there will be a non-zero threshold, but much less than without bias.

\$\endgroup\$
  • 1
    \$\begingroup\$ Ah! Okay, so I did read it wrong. Yours will work. This is why blocks of text are bad. Thanks Spehro! \$\endgroup\$ – AaronD Apr 14 '15 at 16:26
  • \$\begingroup\$ How exactly does this work? \$\endgroup\$ – Golaž Apr 14 '15 at 20:58
  • 1
    \$\begingroup\$ @Golaž R1 and D1 establish a bias voltage that turns D2 just on a bit. So a small AC voltage at In gets rectified. Without the bias, a much larger AC voltage (hundreds of mV) would be required to get much rectified signal (one reason why germanium diodes were preferred for crystal radios). It's not as good as an op-amp precision rectifier for low frequency applications, but it will work at RF frequencies, which most op-amps will not. Since the two diodes are likely closely thermally coupled and will change with temperature similarly, it's better to use a second diode than a voltage divider. \$\endgroup\$ – Spehro Pefhany Apr 14 '15 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.