How to determine the DC operating point of a GaAs RF HEMT

Question

I've been reading datasheets of RF GaAs HEMTs, such as this one: ATF-33143

I understand the theory behind HEMTs, they are basically JFETs with higher electron mobility, thus, I thought that Schokley's equation to determine the DC drain current operation point of the transistor would also apply. The equation im refering to is this:

$$I_D=I_{DSS}\bigg( 1-\frac{V_{GS}}{V_P}\bigg)^2$$

Here is an excerpt of the ATF-33143 transistor datasheet:

There is something that doesnt make any sense to me, if you look at the specs it says that Idss is around 237 mA, and the pinch off voltage is around -0.5 V, so how is it possible that the quiescent bias current is 80 mA with Vgs=-0.5V? I would like to think that if -0.5V is the pinch off voltage, then in order to achieve an 80 mA current, Vgs would have to be between -0.5 and 0? using the before mentioned formula and solving for Vgs, I get that for an 80 mA current I would need a Vgs of -0.209V.

The transistor data sheet provides S-parameters for 3 different bias points ID=40 mA, 60 mA and 80 mA. So which formula should I use to calcule the drain current?

I'm aware that Idss is a poorly defined parameter that varies from transistor to transistor, however I would like to be able to analytically find the ball park figure for the DC operating point of the transistor.

Second question: If the pinch off voltage is -0.5V does it mean that in order to bias the transistor in Class A, then I would have to set Vgs to -0.25V? does that mean that the input signal cant be larger than 0.25V peak?

Answer 1

Pinchoff is defined in that spec as Ids=0.1*Idss, i.e. 24mA. But that's at Vds=1.5V.

As it's a transconductance device, you could expect Ids to be roughly proportional to Vds, or about 4/1.5*24mA or about 64mA at the higher Vds.

So you're looking at the difference between 64 and 80mA between those two specifications, and I'd be surprised to find much closer agreement than that from a simple linear analysis of a not very linear device.

Why that disagrees with the Shockley equation I don't know, perhaps that relied on a different definition of the pinchoff voltage?