AC to DC power supply can't get the desired value using formula from book

Thread Starter

xso111

Joined Oct 12, 2017
24
my proffessor wants me to design an AC to DC power supply with the specifications of
ripple 5%, 1V DC filtered output voltage and 220Vrms with 60hz source

i then used Multism and made a bridge rectifier and used the formulas to know what amount of capacitance etc.. that i need to get my proff's specifications.

i used this example from floyd's book as a guideline..
https://i.imgur.com/xTmz6bt.png

heres what i did.

Vpeakprimary = 1.414(220) = 311.08V
then use 10:1 step-down transformer
Vpeaksecondary = 0.1(311.08) = 31.108
the unfiltered peak full-wave rectified voltage = 31.108V - 1.4V= 29.708 V

since i need 5% ripple factor and 1volt dc output i wanted to know how much is the capacitance that i need(C)

0.05 = {[ 1 / (120hz) (1000ohm) (C) ] (29.708) } /1 V
which gave me 4951uF.

i strapped those values in the bridge rectifier i made in multism and i'm getting 29.376V when i measured my load resistor, but i need 1 volt output..

heres a picutre of the circuit i made:
https://i.imgur.com/GHnL6vW.png

is the ripple formula from floyd wrong? or did i just commit a mistake.
 

wayneh

Joined Sep 9, 2010
16,399
You haven’t put anything in your circuit to regulate the voltage at your desired 1V, and your transformer is giving you 22V AC on the secondary.
 

Thread Starter

xso111

Joined Oct 12, 2017
24
You haven’t put anything in your circuit to regulate the voltage at your desired 1V, and your transformer is giving you 22V AC on the secondary.
Why did you think you would get 1V out?
the formula from the book says

DC value of the output voltage or Vdc = (1 - 1/ (2 x frequency x load resistor x Capacitor) ) (unfiltered peak full-wave rectified voltage)

https://i.imgur.com/xTmz6bt.png

so i thought that if i want an output of 1V i'd simply put it in the formula and set only 1 unknown variable.
 

crutschow

Joined Mar 14, 2008
25,417
You obviously do not have an understanding of how the circuit works and are just blindly plugging values into the cookbook formula.
You cannot significantly vary the output voltage from a rectifier-capacitor supply by varying the capacitance, you must change the input AC voltage to do that (why did you use a 10:1 transformer, which seems arbitrary?).
Perhaps this will help you understand.
 

Thread Starter

xso111

Joined Oct 12, 2017
24
This formula allows you to calculate the filtered DC output voltage. Try putting your values in it and see what you get.


How did you derive this formula?
isn't the filtered DC output voltage the same voltage you're gonna measure in load?

the formula came from the book.

ripple factor = Vr(pp) / Vdc

where Vr(pp) = [ 1 / (frequency x load resistance x capacitor) ] (peak rectified voltage)

0.05 is the ripple factor the i want, and 1V is the voltage that i wan to appear in the load resistance
 

Thread Starter

xso111

Joined Oct 12, 2017
24
You obviously do not have an understanding of how the circuit works and are just blindly plugging values into the cookbook formula.
You cannot significantly vary the output voltage from a rectifier-capacitor supply by varying the capacitance, you must change the input AC voltage to do that (why did you use a 10:1 transformer, which seems arbitrary?).
Perhaps this will help you understand.
i'm just confused with the formula.

Vdc = [ 1 - 1/( 2 f R C)](Vrect)

i mean the way i interpret that formula it says that capacitor value has an effect on the Vdc..

the idea i'm thinking is like V=IR. i want 12 volts with 2k resistor? then i need a 12 /2k Ampere current.
 

crutschow

Joined Mar 14, 2008
25,417
i mean the way i interpret that formula it says that capacitor value has an effect on the Vdc..
Only for a very limited amount, depending upon the ripple voltage you have.
A larger ripple voltage will give a somewhat lower average DC voltage.
But that's not how you control the voltage.
You are going down a dead-end trying to control the voltage that way. :rolleyes:
 
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