2
\$\begingroup\$

I'm not sure it entirely counts. It certainly isn't useful in the way that the current-controlled voltage source is useful as the original transcitor paper describes, but more importantly, I am unsure how much its voltage output depends on current characteristics.

The gate-controlled diode is essentially an ordinary diode, but with a MOS gate fabricated on top of it. The electric field present on the MOS gate can increase the width of the depletion region, thereby controlling the breakdown voltage. I thought that this would make it a voltage-controlled voltage source (VCVS) in the same way that a MOSFET can be configured as a voltage-controlled current source (VCCS). But on the other hand, the configuration of the current-controlled voltage source transcitor in the original transcitor article appears to be able to develop voltage without needing a specific loading. (Then again, the hall-effect device described in the original paper appears to be a (passive, albeit nonreciprocal) hall-effect gyrator serving to convert current into voltage.)

\$\endgroup\$
4
  • 3
    \$\begingroup\$ Be careful with your language. It is not true that a MOSFET can be configured as a "voltage-controlled current source" because it can not source current by itself. In some circuits the MOSFET can be modeled as a voltage-controlled current source, but that model is only useful in certain situations. It's a big difference. \$\endgroup\$ Jan 12, 2021 at 19:31
  • \$\begingroup\$ @Elliot -- "mosfet as VCCS" is the high level paradigm by the first paper that is referenced in the question, in its Figure 1 introducing their objectives \$\endgroup\$
    – Pete W
    Jan 12, 2021 at 20:29
  • 1
    \$\begingroup\$ The I-V curves look different. Doesn't that mean they are not the same? \$\endgroup\$
    – Aaron
    Jan 12, 2021 at 20:52
  • \$\begingroup\$ @PeteW I take your point, but I wasn't particularly impressed with the paper either. \$\endgroup\$ Jan 12, 2021 at 23:28

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.