On Thu, 27 Jul 2006 17:17:01 -0400, "Michael" noreply@nospam put
finger to keyboard and composed:

With full wave rectification, a good sized filter capacitor,
virtually no load, and a nominal RMS AC line of 120V, I would expect the
input voltage to be close to 170VDC.

Here are my measurements, with the TV running and Iin around 500mA.

AC Line input VDC output VDC

119 155 132.5
123 160 134

Also, the DC input to the regulator has about 6V p-p sawtooth ripple
(there's no RC filter as in the test circuit, only a capacitor).

Applying I = C x dV/dt, and assuming a ripple frequency of 120Hz and a
load current of 0.5A, this would imply that the main filter capacitor
is of the order of 680uF.

Allowing for a 1.2V drop through the full wave rectifier, and a drop
of 3V to account for ripple, I'd expect that your DMM should read
about 166VDC at the input to the regulator. Instead you seem to be
losing about 10V somewhere. If this loss is resistive, then something
must be dissipating around 5W.

The fact that the datasheet specifies a test range of 153-180 suggests
that the nominal expected input is 166.5V. Is it possible that your
variac is affecting the DMM reading? Are you measuring the AC at the
bridge?

There could be measuring differences: The TV's circuit and that of the data
sheet are not exactly the same. The data sheet has R3 of 12K Ohms and R4 of
220k Ohms whereas in the TV the equivalent resistances are 15k and 470k.
Also, the DC input to the regulator has about 6V p-p sawtooth ripple
(there's no RC filter as in the test circuit, only a capacitor). I don't
know how well the IC handles this.

I notice that the regulator's performance is specified in reference to
a particular test circuit. IMO a regulator's specs should not be
dependent on the choice of external support components. In fact the
test circuit produces absurd results.

For example, Io is the load current, not the regulator current. It is
the sum of the regulator current plus the current through the bypass
resistor, R5. If the entire load current of 0.5A were to be drawn by
R5 (47 ohm), then the voltage drop across it would be 23.5V. This
would give a maximum input voltage of only ...

Vin = 135V + 47 x 0.5 = 158.5V

Conversely, an input voltage of 180V would require a current of
approximately 1A through R5.

- Franc Zabkar
--
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