Stability - Driving a Capacitive Load
Let's look at the 2 circuits he presented.

The first is with Rop and C







simulate this circuit &ndash; Schematic created using CircuitLab

If we write out the transfer function for this, we get
$$ H(s) = \cfrac{1}{1+sR_{op}C} $$.

The denominator, D(s), contains a pole. 
A pole by itself is not bad. It depends. As the video explains, if you cross the 0 dB with a slope of -40dB/dec then it is unstable. 

In the example above, you already have a pole from within the opamp (which is not shown). This pole from within the opamp, causes a -20 dB/dec slope. This is seen by the first dot in the bode plot. 

Now we are driving a capacitive load, and looking at the transfer function we can see that it also adds a pole. The previous pole, plus this new pole, causes the slope of the bode to decrease by an additional -20 dB/dec. This is now -40 dB/dec.  And IF it crosses the 0 dB at this slope, it will be unstable.

So let's look at what he did by adding that additional resitor. As always, lets look at the transfer function.







simulate this circuit

$$ H(s) = \cfrac{1 + sR_{add}C}{1 + s(R_{op}+R_{add})C} $$

Now things are different arent they ?

With this additional resistor in a strategic location, we now not only have a pole but we also have a zero. To make things even better, the zero is located after the pole which means, at some point, we have a -40 dB/dec slope (due to the 2 poles) and then shortly after, since the frequency of the zero is not that far ahead, we increase our slope by +20 dB/dec so that our total slope now is -20dB/dec. And when this crosses the 0dB, it's stable.