Help designing a bridge rectifier

[George317]

this is my very first project, and we haven't even tackled this lesson yet, but the time period is really short so i want to start ahead as early as i can and read some reference books about it.

my proffessor wants me to design an AC to DC power supply with the specifications of
ripple 5%, DC output voltage of 1V, 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://imgur.com/xTmz6bt

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://imgur.com/GHnL6vW

can you guys tell me which parts did i do wrong?

 

[berkeman]

this is my very first project, and we haven't even tackled this lesson yet, but the time period is really short so i want to start ahead as early as i can and read some reference books about it.

my proffessor wants me to design an AC to DC power supply with the specifications of
ripple 5%, DC output voltage of 1V, 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://imgur.com/xTmz6bt

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://imgur.com/GHnL6vW

can you guys tell me which parts did i do wrong?

Welcome to the PF.

So you are not allowed to use a linear voltage regulator at the output to smooth the DC? You have to achieve that ripple at 1Vdc output with only the rectification and output smoothing capacitor?

And what load current range is this supposed to meet the specs over? That makes a big difference in how you size your output components.

 

[George317]

Welcome to the PF.

So you are not allowed to use a linear voltage regulator at the output to smooth the DC? You have to achieve that ripple at 1Vdc output with only the rectification and output smoothing capacitor?

And what load current range is this supposed to meet the specs over? That makes a big difference in how you size your output components.

thanks very much.

my proff didn't prohibit using anything as long as we can achieve the specification. i don't actually know about a voltage regulator :D all the bridge rectifiers i saw in the book were just comprised of transformer diode capacitor and the load resistor, so i simply tried to copy that in Multism hoping it'd work.

my prof said that its up to us to decide the output current.

tnx

 

[berkeman]

thanks very much.

my proff didn't prohibit using anything as long as we can achieve the specification. i don't actually know about a voltage regulator :D all the bridge rectifiers i saw in the book were just comprised of transformer diode capacitor and the load resistor, so i simply tried to copy that in Multism hoping it'd work.

my prof said that its up to us to decide the output current.

tnx

Try doing a Google Images search on LM317 Circuits, and take a look at what a typical circuit with an adjustable linear output regulator looks like. Then do some reading about that LM317 Adjustable Linear Voltage Regulator (and see is there is a model for it in your simulation package), and go from there...

Show us what you come up with next

 

[berkeman]

BTW, funny coincidence that your username is @George317

 

[George317]

Try doing a Google Images search on LM317 Circuits, and take a look at what a typical circuit with an adjustable linear output regulator looks like. Then do some reading about that LM317 Adjustable Linear Voltage Regulator (and see is there is a model for it in your simulation package), and go from there...

Show us what you come up with next

oh wow. it worked like a charm, i simply plugged it in and its giving me the output that i need. thanks so very much.

if i may ask though, I am curious to why i wasn't able to get my desired voltage when there's no voltage regulator. because i actually plugged in the values from an example in the book http://imgur.com/xTmz6bt and i was able to get his calculated output voltage even though i wasnt using any voltage regulator back then.

again thank you very much for your help.

 

[willem2]

if i may ask though, I am curious to why i wasn't able to get my desired voltage when there's no voltage regulator. because i actually plugged in the values from an example in the book http://imgur.com/xTmz6bt and i was able to get his calculated output voltage even though i wasnt using any voltage regulator back then.

You used a 10:1 transformer and a 220V supply, which produces a peak voltage of 31.1 V. The peak voltage after the bridge rectifier is 29.7 V. This will also be the maximum voltage of your supply. The capacitor can in no way reduce the voltage that the bridge rectifier outputs, it can only prop up the voltage when the output of the bridge rectifier is lower than that on the capacitor.
If you want a 1V supply with 5% ripple you want the voltage of your supply to be between 0.975 and 1.025 V, so the output of the bridge rectifier can't be more than 1.025V.

Real world power supplies will always have some kind of voltage regulation. You will still need to control the ripple before the input of the voltage regulator, because that will have a minimum input voltage.

 

[Guineafowl]

0.05 = [ 1 / (120hz) (1000ohm) (C) ] (29.708) /1 V

Out of interest, what formula is this? I can't see the link clearly.