- Jan 1, 1970
It seems to me that as the transconductance goes down with colder
temp, that there will be less current and less voltage drop across
the resistor, and so a higher drain voltage.
I don't see the value in that reasoning, because gm = Ic/VT,
where VT = kT/q, for a negative transconductance tempco with
constant current. [To eliminate gm tempco you need to use a
PTAT (proportional to absolute temperature) current source.
To get a PTAT CS, use a BJT with Vb = Vg0 = 1.23V (e.g. an
LM385-1.2 or LM4041 reference) and an emitter resistor.
Vg0 is the bandgap of silicon extrapolated to 0K, where 0K
means absolute zero... alphabet soup.]
I suppose my question is really this:
Will the ouput power of a MMIC vary less over temp
with a series dropping resistor (more like a current source)
than with a straight voltage source.
The question is complicated by the fact that the
MMIC will have built in RF chokes on the die, so we are not
actually tied to the drain directly.
You may find saving a bunch of regulators appealing for a
personal one-off, but I wouldn't do it for any serious use
or for production. A low-current part would experience
more voltage than intended, perhaps even damaging it, and
a high-current part would have inadequate voltage, and be
unable to reach the specified output power. Even if you
trim the resistors for your one-off, if you ever needed to
change a component, you'd be back to square one. Not good.
That's not really a concern because even if
the device draws zero current, the voltage from the regulator
will still be less than the absolute maximum drain voltage rating on
the data sheets.
And as the other poster mentions, the MMIC supplier
recommends a dropping resistor in the data sheet for proper
operation. It would have to be extreme lot to lot and wafer variations
for us to need to adjust these during production (once we dial
them in during prototyping).
Check the app notes, many MMIC RF amps REQUIRE a sereis resistor to
stabilize the current vs temperature...
That's the reason we are using dropping resistors in
the first place. It seems you agree with me then that one of the
for using the dropping resistor is for temp comp.
what you reaaly want is to construct the V/I load line of the amp at
various temperatures and then see what the voltage and current will be
for various supply and resistor values.
Well, if you look at an AC load line (assume
linear class A for simplicity) over your DC I/V curves, you
can see how lowering your drain voltage will begin to compress your
amplifier sooner, at a lower output power. So if we lower
the transconductance at high temp, the entire AC load line is shifted
and the average drain current drops. But if we increase the
drain voltage, then we shift the load line to the right, and can
get the voltage swing back to the room temp values.
A bit simplistic of a model, i admit, but seems to