Calculate the percent ripple voltage? Help Need ASAP

Thread Starter

TRIAD-ELECTROMOTIVE-LTD

Joined Jun 27, 2018
17
I. For the bridge rectifier, assume a 2.2kΩ load with a sinusoidal, 6.4Vrms, 60 Hz input
voltage. Use suitable approximations to;

a) Calculate the capacitor value to output ripple voltage of 0.5V P-P with a
2.2kΩ resistor. [DONE]

b) Determine the percent ripple voltage for the half and bridge rectifiers with no
load.
[NEED HELP WITH THIS] [Can't Find any Info Online regarding This]
What formulas do I need to work this out?

Thanks in advance

Sean
 

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ericgibbs

Joined Jan 29, 2010
11,941
b) Determine the percent ripple voltage for the half and bridge rectifiers with no
load.
[NEED HELP WITH THIS] [Can't Find any Info Online regarding This]
hi T

Considering there is no load, what would you say, the ripple would be.?
E
 

Thread Starter

TRIAD-ELECTROMOTIVE-LTD

Joined Jun 27, 2018
17
hi T

Considering there is no load, what would say the ripple would be.?
E
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Thanks for your fast reply friend.

I have been reading every article online regarding this since 10am this morning. And my gut instinct is this:

"In no load condition the ripple will be virtually non-existent and the circuit will effectively be an open circuit."

 

Thread Starter

TRIAD-ELECTROMOTIVE-LTD

Joined Jun 27, 2018
17
hi,
Is the rectified output voltage smoothed by a reservoir capacitor.???

E
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The question does not mention any capacitor for this particular question, but I am thinking it might be wise to assume a 1.8uF capacitor as I calculated in part a) of this 2 part question. But the question b) does not mention any capacitor.

I can't see it doing any harm by including the 1.748uF capacitor into the assignment.
What do you think?

I calculated the capacitance for part a) to be 1.748uF

Regards
Sean

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a) Calculate the capacitor value to output ripple voltage of 0.5V P-P with a
2.2kΩ resistor. [DONE]
upload_2019-1-4_18-9-52.png
 

ericgibbs

Joined Jan 29, 2010
11,941
hi,
The question does say 'no load', I agree it suggests that the capacitor calculated in part [a] is connected as smoothing capacitor and the 2k2 load resistor is removed, to calculate part .(b)
I would agree with post #3.

The cap would charge to Vpeak. [ neglecting any leakage in the cap]

E
 

Thread Starter

TRIAD-ELECTROMOTIVE-LTD

Joined Jun 27, 2018
17
hi,
The question does say 'no load', I agree it suggests that the capacitor calculated in part [a] is connected as smoothing capacitor and the 2k2 load resistor is removed, to calculate part .(b)
I would agree with post #3.

The cap would charge to Vpeak. [ neglecting any leakage in the cap]

E
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Thanks again friend for your input.

This has been of much help

regards
Sean
 

Thread Starter

TRIAD-ELECTROMOTIVE-LTD

Joined Jun 27, 2018
17
hi Sean,
Recheck your Cap calc, line #2 of your post #5 is wrong.
Also the cap value is incorrect.
E
Hint: the 6.4Vrms * 1.414 waveform is rectified.
Don't forget the diode voltage drops.

EDIT:
Sim to show the incorrect result with 1.748u
@TRIAD-ELECTROMOTIVE-LTD
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Thanks for pointing that out to me I believe this is now the correct answer
 

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ericgibbs

Joined Jan 29, 2010
11,941
hi Triad,
Thats better, using your formula giving 54uF.:)

BTW:

For ref only.
Your equation assumes a discharge period of 1/120, 8.3mS.
Its actual discharge period is a little shorter, approx 7mS. [45uF]
Your equation is often used as it errs on the side of ensuring a slightly lower ripple value.
E
 

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