Yes, the capacitor in that position, will act as a capacitance multiplier. The polarity is also correct.
A word of caution, since this will also cause a lag network, you may want to check stability and settling time with/without the capacitor.
Why is it called a "thrifty" regulator?? It is linear, it will waste power.
Hm... you are right. Untill the capacitor get's charged the output voltage is higher. I have tested it. This might be a problem indeed if the loag cannot tollerate a higher voltage.
There really is no advantage to adding the 100uF to the circuit in that position, if you want a soft start add a cap (47uF say) and a resistor (1k) across the zener....
There really is no advantage to adding the 100uF to the circuit in that position, if you want a soft start add a cap (47uF say) and a resistor (1k) across the zener....
The slow shutdown will be provided by the reservoir in the output capacitor.....the 4.7uf is a quite small cap, you may increase it to 100 or perhaps 220 uF. It also helps in reducing the output impedance.
The slow shutdown will be provided by the reservoir in the output capacitor.....the 4.7uf is a quite small cap, you may increase it to 100 or perhaps 220 uF. It also helps in reducing the output impedance.
Thanks!
I have just noticed that the circuit proposed in #8 works, but only if the cap is fully discharged. Maybe I will use a dpdt switch to connect a 5R resistor at parallel to the cap when the circuit is off. This will discharge the cap rapidly and the cycle will begin again dusing start up.
Any circuit modification that adds a pole to the feedback loop, e.g. capacitor from 2N2222 base to ground will not only bring up a risk of instability but also affect the regulation quality. You don't want this.
Any circuit modification that adds a pole to the feedback loop, e.g. capacitor from 2N2222 base to ground will not only bring up a risk of instability but also affect the regulation quality. You don't want this.