Smoothing Capacitor/Vripple

[Claire Patterson]

Hi guys,

I have been stuck on this part of my paper for 2 weeks now and I'm coming up to my deadline to hand it in. I honestly cannot see the answer because I have looked at it for far too long.

The paper has been fluently focusing on half wave and full wave rectification and has built up to a full wave bridge rectifier that started in Part 1 with a 45:1 step down transformer, please ask me if you want me to dig out any more details from any other Part of the paper if you feel it may apply in the following Part (5) that I am stuck on:

"Use the formula below to calculate the minimum capacitance required to keep the remaining ac ripple to within 10% of the desired voltage (Vs) (ie 12V) Assume R1 is 2.2Kohms"

"To calculate the ac ripple (Vripple) in a capacitor smoothed circuit: Vripple=I/2fc, where f = supply frequency and C is the value of the smoothing capacitor and I is the load current."

I have come up with two answers. One my teacher has told me IS wrong the other I know is wrong because my teacher told me that I am looking for an answer of around 11 or 13 which would be within 10% of the 12v.

Answer 1:

I = 12/2200 = 5.46ma

By transposition C = (5.46 x 10^-3) / 2 x 50 x 1.2 = (5.46 x 10^-3)/120 = 4.55uF (where C = I / 2 x f x Vripple)

Vripple = (5.46 x 10^-3) / 2 x 50 x (45.5 x 10^-6) = 1.2v

Answer 2:

I = 12/2200 = 5.46ma

By transposition C = (5.46 x 10^-3) / 2 x 50 x 12 = (5.46 x 10^-3)/1200 = 4.55uF (where C = I / 2 x f x Vripple)

Vripple = (5.46 x 10^-3) / 2 x 50 x (4.55 x 10^-6) = 12v

Please tell me where I am going wrong, otherwise I fear that I may fail this part of my distinction

 

[KrisBlueNZ]

Hi Claire and welcome to Electronics Point

"Use the formula below to calculate the minimum capacitance required to keep the remaining ac ripple to within 10% of the desired voltage (Vs) (ie 12V) Assume R1 is 2.2Kohms"

So you need to provide at least enough smoothing capacitance to ensure that the "AC ripple" is kept to "within 10% of the desired voltage". I assume this means that the peak-to-peak ripple voltage must be 1.2V or less. You've assumed that too.

"To calculate the ac ripple (Vripple) in a capacitor smoothed circuit: Vripple=I/2fc, where f = supply frequency and C is the value of the smoothing capacitor and I is the load current."

Yes, that's right for full-wave rectification.

Answer 1:
C = (5.46 × 10^-3) / 2 × 50 × 1.2
= (5.46 x 10^-3) / 120
= 4.55 μF

You've just made a decimal place error in your calculation.

Vripple = (5.46 × 10^-3) / 2 × 50 × (45.5 x 10^-6) = 1.2V

 

[Claire Patterson]

Hi thanks for your reply, however, I do apologise, answer 1 should finish with 45.5uF, I mixed up my calculations a bit there. The other thing is I did show my teacher the 1st answer of vripple 1.2v and he told me it was wrong and it should be between 11 and 13 volts? Hence my 2nd answer which is my most recent but still isn't correct. Does that make sense?

 

[KrisBlueNZ]

The other thing is I did show my teacher the 1st answer of vripple 1.2v and he told me it was wrong and it should be between 11 and 13 volts?

I guess it depends on how you interpret the specification:

... minimum capacitance required to keep the remaining ac ripple to within 10% of the desired voltage (Vs) (ie 12V)

Since ripple is normally specified in volts peak-to-peak, I would interpret that as you did - the ripple must be 1.2V peak-to-peak or less.

Another interpretation is that the voltage at the peaks must be no higher than (12V +10%) and no lower than (12V - 10%) I suppose, but that gives voltages of 13.2V and 10.8V.

Did you add the "(i.e. 12V) at the end of that quote above? Is it possible there's more information in that question that you haven't read carefully enough?

If not, I think you should point out to your teacher that the ripple is specified as a percentage of 12V and there is no mention of 11V and 13V and also no way to get those voltages by calculation.

Edit: You could also calculate C for a ripple voltage of 2V p-p just to satisfy your teacher that you know how to do it, even if that's not the "right" answer for the question as written.

 

[Claire Patterson]

Your other interpretation is what my teachers looking for, it needs to be no higher than (12v +10%) and no lower than (12v-10%). The (i.e. 12V) is as its written on the paper, I swear it must be worded incorrectly because the paper is so misleading in itself. I know that from the paper the full wave bridge rectifier has a 50Hz frequency. In part 4 they asked me to consider that it was a 400vrms circuit with a transformer loss of 5% which conflicts the 12V in Part 5.... I think? My calculations are now all over the place for Part 4 as well because my teacher told me that my averages were incorrect for a full wave and half wave rectifier. The edit could be a good idea but my teacher is very straight laced and he will congratulate the edit but say "however, its not the answer, look at the question!"

 

[KrisBlueNZ]

Oh well. I guess it depends on your priorities. Do you want to be right, or do you want to pass the course? LOL

 

[Claire Patterson]

I want to pass the whole course with distinction but I have done so much research on this paper I don't know what's right and what's wrong now.

 

[Claire Patterson]

The internet is so conflicting and my resources outside of work are limited. You are the only one who's got the closest to helping me get the answer right, because at least I have been reassured that I am not going crazy and my maths makes some sense now.

 

[KrisBlueNZ]

I think you understand what you're doing. The confusion is arising because of how the question is worded. Your teacher's interpretation is different from how you and I interpret it. Just hope that the exam questions will be more clearly worded.

 

[Colin Mitchell]

"I know that from the paper the full wave bridge rectifier has a 50Hz frequency."
The input may be 50Hz but the output is 100Hz.

 

[KrisBlueNZ]

The input may be 50Hz but the output is 100Hz.

Seagulling threads again, Colin? You need to start reading from the beginning otherwise you might embarrass yourself.

That's taken care of by the factor of 2 in the formula (which is 1 for a half-wave rectifier) as Claire explained in post #1:

V_ripple = I / 2 f c where
f = supply frequency
C is the smoothing capacitor
I is the load current.

(emphasis mine).

 

[Colin Mitchell]

"
Seagulling threads again, Colin? You need to start reading from the beginning otherwise you might embarrass yourself.

That's taken care of by the factor of 2 in the formula (which is 1 for a half-wave rectifier) as Claire explained in post #1:"


And yet is post #4 she still says"I know that from the paper the full wave bridge rectifier has a 50Hz frequency." so I don'ta thinka sha realisa the pointa.