0
\$\begingroup\$

I am reading a chapter in a thesis about using HEMT (high electron mobility transistor) as a cryogenic amplifier. The measurement scheme is to use the gate voltage fluctuation's influence on the drain source current.

In order to use a HEMT as an amplifier, it is necessary to find out what the gain factor is going to be, for a given circuit. To that end, the paper shows plots of \$I_D\$ as a function of \$V_{DS}\$(image a) and as a function of \$V_{GS}\$(image b) for fixed power dissipation values \$100\mu \$w, \$10\mu \$w, and \$1\mu \$w.

My question is about plot (b). The gain factor from the fitting of plot (b) is only valid, to my knowledge, if you can somehow adjust the drain voltage bias such that, throughout the operation of the HEMT, \$100\mu \$w \$= I_D V_{DS}\$.

Based on the plot, it appears that if you can somehow adjust the drain voltage bias to keep the power dissipation of the HEMT constant, you have a nice well defined gain factor over the entire range of the gate source voltage.

It is difficult for me to imagine a feedback circuit where the drain voltage bias does change to keep the power dissipation constant though. Based on the circuit shown below, it doesn't seem to have such a feature.

My question is, given the fact that you are measuring the drain voltage and the drain current has to come from the source of drain voltage bias, is there a scheme to change the drain bias voltage to keep the power dissipation constant?

enter image description here. enter image description here

\$\endgroup\$
1
\$\begingroup\$

Presumably the signals to be measured are very small with respect the gate bias voltage. So there is a single operating point for the HEMT transistor (and the input and output signals represent small-signal variations about that point).

So you pick an operating point (in this case you pick Id * Vds) and a load resistor. Say you pick 100uW and 510\$\Omega\$.

For every Vds from 50mV to 450mV there is a corresponding Id. So you can pick, say 200mV Vds. So Id must be 500uA. So the gate bias voltage is approximately -0.575V. You set the drain bias voltage to exactly 200mV + 255mV + 255mV (2x 510 ohms), and then trim the gate bias voltage so that the DC drain voltage (VD) reads 200mV.


P.S. If that thesis is available online it would be interesting to read.

\$\endgroup\$
  • \$\begingroup\$ @Sphero Pefhany that makes perfect sense, thank you. yes, the gate voltage fluctuation is on the order of \$ \mu V\$, very tiny compared to \$ V_{GS}\$. I found the thesis here repository.tudelft.nl/islandora/object/… \$\endgroup\$ – Blackwidow Sep 5 '18 at 1:16
  • \$\begingroup\$ I would appreciate it if you get a chance to read it and leave your thoughts about their HEMT circuit. (it's chapters 6 and 7) \$\endgroup\$ – Blackwidow Sep 5 '18 at 1:18
  • \$\begingroup\$ Actually, could you explain a bit about how you got 200mV + 255mV = 455mV for the drain bias voltage? I understand that you want 200mV for Vds. I am struggling to see how you knew that you had to set the drain bias voltage at 455mV, knowing 255mV would get distributed across the resistive elements up until the drain gate. \$\endgroup\$ – Blackwidow Sep 5 '18 at 1:27
  • \$\begingroup\$ I think you got 255mV because you 510\$\Omega \$* \$ 500 \mu A \$ but shouldn't it be 2* 510\$\Omega \$* \$ 500 \mu A \$ because there are two 510 \$ \Omega \$ resistors between the drain and the source of the drain voltage bias? \$\endgroup\$ – Blackwidow Sep 5 '18 at 1:48
  • \$\begingroup\$ One more thing, in the caption below the image, it says Vout/VG, but I am not sure which voltage they mean by Vout. I thought Vout is really just Vds since the Vout is being measured at the drain gate (based on the circuit diagram above) \$\endgroup\$ – Blackwidow Sep 5 '18 at 2:03

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.