Biasing criteria for pHEMT switch

For HEMT FET switch, its terminal bias criteria never seems to be clearly stated in any standard text.

For example, is it okay to set the desired Vgs for switch ON and OFF, but leaves Vd open (or floating, or no bias)?

Most people tend to set the desired Vgs, and then leave both drain and source biased with equal potential (so Vds=0). Does it really matter if Vds is not equal to 0V set to any other voltage?

Set Vds = 0V and drive the gate with a pulse voltage around the Vgate treshold doesn't make sense.

It is a common way to use this device. Vgs is used to switch on/off the device.

I answered your question.

Don't agree.. but thanks for your participation.

Please, could you be more explicit...

And I wrote : Set Vds = 0V and drive the gate with a pulse voltage around the Vgate treshold doesn't make sense.
That means you need a voltage on Vd to switch something. If you have a look on some MOSFET datasheets, the switching time test circuit uses a large resistor (and "bias" voltage) on drain to pull the drain down.

MOSFET could be different. But pHEMT is the one we are talking about, bro.

All transistors need some output bias voltage and current to operate. Heterojunction devices aren't any different. Where did you hear otherwise?..

Where did u hear what u just said above?
This is switch application, dude.

Everybody knows that, you are not the only one working on pHEMT switch application. You should read some books explaining how transistors work. A pHEMT is a field effect transistor based like MOSFET. Are you a student?

Yes..
If you apply Vgs voltage either positive for E-Mode pHEMTs or negative for D-Mode pHEMTs, a depletion layer will be appeared just underneath of the gate and channel resistance will drop very sharply and the FET will act as a switch.Vds voltage is not necessary because FET became a simple resistance and deplation layer touches both Drain and Source.

Regardless of the answer, I always treat myself a student. :)

---------- Post added at 12:14 ---------- Previous post was at 12:11 ----------

Thank you, BigBoss. This is the exact answer I am looking for. I joked around long enough. Now some one finally loses the patience and cannot hold the answer.

Some finite amount of Vds is necessary, even if it is functioning as a switch and even if it is just a "resistance". If there is no bias voltage, then there is no current, and the device simply does nothing.

Or are you seriously claiming that I could short the drain and source together directly and it would still work somehow?

For HEMT switch, no operation current is present. It is a voltage control device anyhow. This is a good thing- don't you like it? Having Vd=Vs does not mean they are RF shorted.

Case#1: one can bias up both drain and source terminals, or even set Vd=Vs (so Vds=0), In this case, all 3 terminals have DC bias voltages. But having both drain and source bias voltages is really not a requirement for switch operation.

Case#2, one can bias up only the gate and the source terminals, and leave the drain floating without any DC voltage.

Case #3: I suspect, one can even use the relative voltage difference between gate and drain to switch the device on/off, and leave source floating without any DC bias. In other words, drain and source are interchangeable when used as switch. Either drain or source needs to be biased. Biasing both drain and source is not required. But this needs to be confirmed.

BigBoss (and other experts), please comment on the above cases. Thank you all.

If a pHEMT will be used as switch, there will compulsorily be a potential between drain and source because of signal, RF signal etc. This potential is sufficient to create a Electrical Field between drain source.