I
have several questions about the
voltage regulator presented in Sec. 24.5.

1.
Why estimate the location of the *f*_{2}
output pole, which will be
our dominate pole, using the closed loop output resistance and not the
open
loop output resistance?

Since
the
load capacitance is used to compensate the op-amp we can Thevenize the
op-amp
and reference voltage so that the result is a single-pole *RC* circuit with a BW of 1/(2p*R _{out,CL}C_{L}*).
The points are increasing

When
we think about loop stability, we
think about reducing the total open loop phase lag appearing at the
v(-)
input.

Yes,
however, you always have a -90 degree phase shift due to a dominant
pole. In
this case you want this phase shift to be due to the load and not the
internal
high-impedance node. In other words, you want *f*_{2}
(the pole on the output of the op-amp) to be much
smaller than *f*_{1}
(the pole
associated with the output of the diff-amp).

With
op-amps, the dominant pole of the
system is the 3-dB frequency of the open loop gain (which
comes from *R*_{1}
and *C*_{1}). But
here we get that dominant pole frequency using
the closed loop resistance. I wish I could justify that.

You
don't
want the design to have a dominant pole due to *R*_{1}
and *C*_{1}
but rather from *R*_{2}
and *C _{L}*
(which is approximately
equal to

Further,
once this choice is made, the
open loop gain appears in the numerator of the expression for *f*_{2}. One can then
make the
argument that there is a gain versus stability trade, since any
increased gain
pushes out *f*_{2}
(the
dominant pole). But I can't justify that claim if I can't explain using
the
closed loop expression for *R*_{2}
in *f*_{2}.

Yes,
increasing the open-loop gain does push out *f*_{2}
which is bad and why we can't have too high open-loop gain (as
discussed in the
book). If the gain is increased by increasing *R*_{1}
then *f*_{1}
moves down towards *f*_{2}
which is also bad (again, discussed in the book). The bottom line is
that you
have to have a big decoupling capacitor on the output of the regulator
to
supply charge for fast current needs so to minimize the power drawn by
the
regulator you have to compensate with the pole at *f*_{2}.
Of course, you can still design a regulator to be
compensated with *f*_{1}
(the
pole on the output of the diff-amp) but you will need a very large
compensation
capacitor and the design will burn more power (and be much, much,
larger).

2.
At interviews everyone says that
decreasing *R _{L}*
decreases
stability, and I find that as well in my simulations. However, we
wouldn’t
suspect that by looking at the equation for

Yes,
the *R*_{2}s
cancel in Eq.
(24.71). But if *R*_{2}
drops
this means M7 is supplying more current and thus *g _{m}*

If
smaller *R*_{2}
reduces stability why reduce it further with *R _{L}*
(

So if
we make *R _{L}*
larger than the
open-loop gain of the op-amp grows and

Well,
if you remember Miller's theorem, as the gain of the second stage of
the op-amp
grows the capacitance on node 1 (the diff-amp's output) grows too, (1
+ *g _{m}*

I
can agree with that if the idea is
to trade stability for a smaller *C _{L}*.
To have a smaller

But
then you seem to say that adding *R _{L}*
won’t reduce stability. The
reason is that as current increases in M7, the decrease in

My
problem with that is that the
product *R*_{2}(*g _{m}*

We
need the *R _{Lmax}*
to keep the second stage gain from getting too big so that

The
expression for *f*_{2}
given in
Eq. (24.71) does have both *R*_{2}
and *g _{m}*

Yes, *R _{L}* keeps the
second stage
gain down to keep