"robb" wrote in message ...

....snip...

yes the sine wave hop hop hop, was the simple bridge (+) output
with no typical circuit additions (caps) they were 16v DC
measured but that must be an average as the divisions were set to
10 V vert/div so then ~ 22 volts at peaks

Okay, this gives us a little more information about what you have.

Yes, a typical DC measurement for a full wave rectified sine wave
does not match the peak voltage. Ideally the meter would measure
the 'rms' voltage. 'rms' stands for 'root mean squared'.
See http://en.wikipedia.org/wiki/Root_mean_square for more info
about calculating RMS voltages. rms voltages are used instead
of peak voltages since they track the power dissipation in a resistor.
I.e. a 1 volt rms across a 1 ohm resistor will dissipate 1 watt of heat.
This is true for both DC and AC signals. For a DC signal, the rms
voltage is equal to the DC voltage. For an AC signal, the peak voltage
in 1.414 (i.e. sqrt(2)) times the rms voltage. Thus a 16 volt rms AC
signal would have a peak of 22.6 volts.


and as you pointed out at abot 5-6 volt the regulator cuts out
and causes the regulated output dips.

Yes. The regulator will stop regulating properly when the input
voltage goes too low. A 7805 regulator does not have any form
of internal energy storage. Thus its output drops out when its
input voltage gets too low.

Do you get a reasonable operation of the regulator when you
add a storage capacitor? Please note that most regulators also
require some small capacitors for stability. These small capacitors
need to placed right next to the leads of the regulator.


With the storage capacitor, one should see a DC voltage with
some
ripple voltage. The ripple voltage should peak at about the
input sine
wave peak minus two diode drops in the bridge (about 1.4
volts).
After the input voltage peaks, the ripple will show the
capacitor
discharging as the supply current is being drawn from the cap.


interestingly, though not shown, i could only get the ripple with
fairly smallish cap (470 uF 35v) and high current load like a 12v
*auto brakes* light bulb and the ripple looked like a saw wave

Yes. The ripple voltage will rise when the peaks of the input sine
waves occur. The diodes in the bridge will be forward biased
and then current will flow through the diodes to charge the storage
capacitor. After the peak passes, the diodes will be reversed
biased. The capacitor voltage will then drop as the load current
pulls charge off of the capacitor. This will look somewhat like
a saw wave.

The magnitude of the ripple is approximately given by:
I = C * dv / dt

Thus to see a 1 volt ripple voltage with a 470 uF cap at 60 hz,
you need to draw 56 mA.


The storage capacitor needs to be large enough to provide
between
the input peaks. You can estimate the required capacitor size
using:

There was a typo in my sentence. it should have read: "The storage
capacitor needs to be large enough to provide energy storage between
the input peaks."