I'm trying to understand how you get a DC signal from an AC signal being rectified. It produces a ripple waveform and the signal gets "smoothed" out using a smoothing capacitor. Is the smoothed waveform basically DC or is there more to it? Since it's not a completely straight DC signal what am I missing.
(Image source: Electronics Tutorials - Full Wave Rectifier with Smoothing Capacitor)
In case you are wondering how e.g. a phone charger outputs a constant 5 VDC or an ATX PSU outputs a constant 12 VDC: These are switching power supplies. They replenish the capacitor much more often.. 100 kHz ballpark. Therefore, their output is less ripply, but still not a straight line, when viewed at a corresponding time scale.
In practice, the ripple can be negligibly small if you use a switching power supply (incl. the diode bridge type supply in the question), followed by a post-regulator, which basically burns off the excess peaks of the ripple into heat.
If you cannot tolerate any ripple, you have to use a battery or photovoltaics as energy source.
It's a lot easier to begin by analysing a half wave rectifier and resistive load: -
Is the smoothed waveform basically DC or is there more to it?
Then if you add a smoothing capacitor, can you see that the output waveform becomes like this (my addition in red): -
So, when the input voltage is positive, it both supplies the resistor load and, charges up the capacitor. Then, when the input voltage is falling from the positive peak, the diode becomes reverse biased and no longer supplies current to the capacitor and load resistor.
So, the capacitor slightly discharges due to the resistor taking current directly from the capacitor.
If you understand this, the full wave rectifier is a relatively short leap in understanding.
Since it's not a completely straight DC signal what am I missing.
You might be missing that the diode becomes reverse biased and the resistive load discharges the capacitor a little until the AC voltage waveform becomes positive again and recharges the capacitor close to the top of the waveform.
You can increase the output capacitance for better DC coupling. If still the waveform in not good enough for your application, try using LDO or a switching regulator in between the capacitor and the load.
abs(sin(t))thing) is considered DC. \$\endgroup\$