0
\$\begingroup\$

I am creating a current sink that can draw a step current from the Vnode of the circuitry in blue.

The reason to draw a step current from Vnode is that I need to measure the characteristic impedance of the TLIN by drawing a sudden current by applying a VCCS in red (voltage controlled current source) with a step voltage.

However I tried to combine these 2 circuitry as below, red and blue. The Vg of the NMOS is around 15V positive, but since there is no voltage difference between source and drain, the current read at I_probe1 and I_probe5 are nearly zero.

How should I modify this circuit so can get a step current from Vnode so I can measure the characteristic impedance of TLIN when a sudden current is drawn from Vnode with a VCCS?

enter image description here

\$\endgroup\$
9
  • 3
    \$\begingroup\$ You have made the same mistake you made in your question about current mirrors not working. Your current sink only sinks current if there is some supply (which you are missing). Your "VCCS" is not a supply, as you seem to think. \$\endgroup\$ Commented Aug 22, 2022 at 23:42
  • \$\begingroup\$ why not use some voltage instead of ground in the blue part of the circuit, so there is something to push current through? \$\endgroup\$ Commented Aug 22, 2022 at 23:46
  • \$\begingroup\$ If you just said you wanted to measure transmission line impedance, and asked how to measure the impedance of a transmission line, maybe someone would already know a circuit to do it. \$\endgroup\$ Commented Aug 23, 2022 at 0:01
  • 1
    \$\begingroup\$ I will repeat my other comment: try learning basic electricity, voltage and current \$\endgroup\$ Commented Aug 23, 2022 at 0:07
  • \$\begingroup\$ Your transmission line has 180 degree electrical length at 1 GHz. It will have 0.18 degree electrical length at 1 MHz (the bandwidth of your op-amp), and the overall circuit has less bandwidth than that. You need to make the transmission line maybe 250 times longer (or your source's bandwidth 250 times higher) if you want to be able to measure the line characteristics with good accuracy. \$\endgroup\$
    – The Photon
    Commented Aug 23, 2022 at 2:31

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.