Designing Regulated Power supply

[D.A.(Tony)Stewart]

IF we assume the ripple dv could be as much as 10% of Vo ...
10%Vo= IL 10ms /C or ?Vo= IL /(10*C)
Is it right?

√ Yes you are correct.
Somehow, your responses did not hit my Radar screen. Perhaps close this thread as [Solved] and start a new design question on AC to DC 35V @5A

To close this thread let me summarize on ripple for
Vo=DC out unregulated
IL= Load current=Ic during discharge and results in ripple Vac (pp)
Vac =%rip * Vo = %Ripple ac voltage (pp) * V0 (Dc out)
T= Ripple period
C= Capacitance for energy storage and ripple reduction

thus Vac(pp)/T= dv/dt ( linear decay for <50%)
.....now substituting into std. equation Ic=C dv/dt or C=Ic dt/dv

C= IL * T / (Vac)
C = IL * T / (%rip * Vo) { with limits on %rip }

Conclusions:
- thus for a constant IL, Vo drops as %rip increases
- but C increases rapidly as %rip is chosen for a lower value, and increase cost
- as IL reduces to 0, %rip also reduces to 0 but Vo also rises to peak AC voltage
- as C goes to 0, %rip goes to 100% and Vo, Average .707 %rip

- another factor not mentioned is transformer secondaries have Rs, winding loss, so the rated Vac rms is always at rated VA out to account for winding loss which is often 10% in low power transformers, so no load unregulated output voltage can be more 10% more than 1.414 * rated Vo

 

[Eshal]

Great information. So must promise me that you are going to help me all the way round in the next thread. You are going to be my teacher. Agree?

 

[D.A.(Tony)Stewart]

Nice try. What do you promise?

 

[Eshal]

that you will help me on my next thread AC to DC 35V @5A .