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I have been reading about the Wikipedia article on negative resistance and in particular, I'm interested in this circuit (also linked at Wikipedia):

I was wondering if we could convert this into DC and feedback some of it into the circuit so we don't need Vb. Is this possible or not?

So im looking for something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ what specifically do you mean with "external source"? It's really not clear what you want. But preemptively: the fact that you can not generate energy out of nothing is fundamental; no way around that. \$\endgroup\$ Commented Aug 11 at 17:16
  • \$\begingroup\$ well you need Vb to set the bias of the NDR device initially but I was wondering if we took the output converted it into DC then set the bias of the NDR by feeding back the output. \$\endgroup\$ Commented Aug 11 at 17:21
  • \$\begingroup\$ it's really not clear what you mean. there's no "output" or "back" here, without you defining what these were, and you cannot "send" something in electronics. Please be much, much precise in your wording! Giving us the full context of what you mean is really necessary here. My best guess is that your "negative resistance device" is underdefined here, and the question is how that behaves in different situations, which are not described by the fact that it's got a negative effective resistance at all. \$\endgroup\$ Commented Aug 11 at 17:24
  • \$\begingroup\$ @MarcusMüller sorry I wasnt clear enough I will edit. \$\endgroup\$ Commented Aug 11 at 22:11
  • \$\begingroup\$ Like you said, you got that first picture from Wikipedia. You should read the text that goes with the picture. The AC source in the diagram, \$V_i\$, represents a weak signal, and the output, \$V_o\$, is supposed to be a stronger (more powerful) version of that same signal. The DC source, \$V_b\$, does not merely bias the tunnel diode. It also is the source of the extra output power. It would be impossible for the circuit (or any circuit) to amplify an input signal without some source of "extra" power. \$\endgroup\$ Commented Aug 13 at 0:10

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