been reading up on powersupply design, in this diagram,
vdc, .9 (avg) * secondary voltage if my transformer secondary is 12v, i can expect to get 10.8v?
ICD, am i right in thinking this is current, so if my secondary primay winding is rated at 500ma i can only get 310ma from it? (.5*.62) ( going to be very disappointed if this is corrrect )
An unloaded, filtered, full-wave bridge will provide a DC voltage of approximately 1.414 times the AC voltage. However, as you put a load on that, you will introduce ripple. The amount of ripple will depend on the load and the size of the capacitor. This ripple is basically an AC component to the output voltage and as the load increases the available DC component decreases and the available AC component increases.
I am not sure what that diagram is getting at when it states an average DC voltage. What on earth is an average DC voltage?. If it's changing then there is AC. Even still, they did not specify a capacitor size, so how can they calculate anything other than the unloaded voltage? Maybe there is more information we don't have.
Okay, I get it. They are saying that the DC component of the unfiltered rectifier output is .9 times the Vac rating. That makes sense, but they should not call it an average DC voltage. It's misleading.
In actual use you will get somewhere between the .9 and the 1.41 depending on the load and the capacitor size. As they say, it's better to arrange to have as little ripple as possible (big cap or underrated load) to increase capacitor life.
I would seek another source of information on all this. There has to be a good book out there, or even web page on power supply design. This little paper does not appear to be the best way to learn about it. Also, the end use really defines how the supply should be designed and rated.
Do you have an end use in mind or is this just to learn about it?