cap size?

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Philipc

New Member
Trying to determine what would be a large enough filter cap. I have a 20 volt transformer with a rectifier, and I want to filter the dc with a cap, how do I go about choosing the correct size. By the way I will be pulling about 3 amps max.
Thanks
Philip

Russlk

New Member
Assuming that you have a full wave rectifier, the ripple will be 120 Hz. The time between voltage peaks is 1/120 = 8.3 mS. The capacitor will charge to the peak voltage, then discharge for 8.3mS according to this formula: Vd = I*T/C where I= the discharging current, T = time between voltage peaks, C = filter cap in farads, and Vd is the voltage change of the capacitor.

Philipc

New Member
Russik,
Thanks for help,
You will have to forgive me for my ignorance; I got a couple more questions.
Would the voltage change of the capacitor be 2 times the secondary output, I'm thinking we might have -20v and +20v? And I'm also thinking I should multiply by 1.4 to get peak volts not RMS?
Second question, is the discharging current the same as the load on the secondary?
Thanks again
Philip

k7elp60

Active Member
Capacitor value

The rule of thumb for linear power supplies is 3000uf per amp. The peak voltage across the capacitor in a full wave rectifier will be 1.414 times
the rms, or 28.28 volts. Another thing to consider is that because of charging current in a typical full wave rectifier circuit one should only plan on getting 0.566 X the rated current of the transformer with out overheating the transformer.

stevez

Active Member
I've asked, via email, three major amateur radio equipment manufacturers to let me know what the acceptable levels of ripple would be in an AC power supply that delivers 13.8 vdc to mobile equipment to be used in base station operation. I've only heard from one so far (Kenwood) and they indicated that the industry standard was 100 mv RMS and that their supplies typically run at 20 mv RMS. The ripple is essentially the change in voltage that you'd calculate from the formulas provided - except that the RMS value has to be converted to a peak-to-peak value.

Russlk

New Member
Phillip: You are confused. The capacitor charges to peak value (1.414 times RMS) and stays there unless there is a discharge path. The load is normally the discharge path. What kind of rectifier do you have, 1/2 wave (1 diode), full wave (2 diodes and center-tapped transformer), or bridge (4 diodes using the full secondary)?

Philipc

New Member
Re: Capacitor value

k7elp60 said:
The rule of thumb for linear power supplies is 3000uf per amp. The peak voltage across the capacitor in a full wave rectifier will be 1.414 times
the rms, or 28.28 volts. Another thing to consider is that because of charging current in a typical full wave rectifier circuit one should only plan on getting 0.566 X the rated current of the transformer with out overheating the transformer.

Does this mean the rectifier is that inefficient?

Russlk,
I'm using a bridge

Another thing I noticed while playing around, without the filter cap, voltage reads something like 16v, and with cap it reads something like 21v. Numbers are for example only, but is this because the AC ripple is counteracting the DC volts. And this question might be even crazier, but if I had a DMM that could read AC in mv, could I put this on the DC side of the rectifier to measure ripple?
Philip

Nigel Goodwin

Super Moderator
Re: Capacitor value

Philipc said:

k7elp60 said:
The rule of thumb for linear power supplies is 3000uf per amp. The peak voltage across the capacitor in a full wave rectifier will be 1.414 times
the rms, or 28.28 volts. Another thing to consider is that because of charging current in a typical full wave rectifier circuit one should only plan on getting 0.566 X the rated current of the transformer with out overheating the transformer.

Does this mean the rectifier is that inefficient?

Russlk,
I'm using a bridge

Another thing I noticed while playing around, without the filter cap, voltage reads something like 16v, and with cap it reads something like 21v. Numbers are for example only, but is this because the AC ripple is counteracting the DC volts. And this question might be even crazier, but if I had a DMM that could read AC in mv, could I put this on the DC side of the rectifier to measure ripple?
Philip

For a full wave rectifier, with a centre tapped transformer, each half of the transformer is only half wave rectified - so you only get half it's rated current (roughly). With a bridge rectifer (like you have) you get complete full wave rectification, so you can get full current output.

The best way to measure ripple is with an oscilloscope, but an AC millivolt would work - make sure you use a DC blocking capacitor to feed it. Bear in mind the ripple isn't a sine wave - so unless you have a true RMS meter the reading won't be accurate, but it will serve to compare different values of capacitor.

mozikluv

New Member
computing cap capacitance & voltage

hi

here's a formula i have been using whenever i make a power supply.

Capacitance = Current out / 1.5 (8.3^-3)
to convert to microfarad multiply the result by 1million

Working voltage > 1.41Vrms

hope this would help

Philipc

New Member
Nigel Goodwin
I also have a KHz reading on my DMM, could this also be used to look at the ripple. I have yet gotten to the theory classes yet, just my maths and sciences, so my electronics knowledge is very limited, so please forgive my ignorance.

mozikluv

Capacitance = Current out / 1.5 (8.3^-3)
can this formula also be written as
C = (I/1.5)*8.3^-3

Nigel Goodwin

Super Moderator
Philipc said:
Nigel Goodwin
I also have a KHz reading on my DMM, could this also be used to look at the ripple. I have yet gotten to the theory classes yet, just my maths and sciences, so my electronics knowledge is very limited, so please forgive my ignorance.

No, you can't use a KHz reading to measure ripple - you already know the frequency, it's twice the mains frequency (for a full wave rectifier), so 100Hz in the UK, and 120Hz in the USA.

Philipc

New Member
I was thinking with the right cap, the ripple would go away, and I could use the meter to prove that? Or will the ripple always be there?
Also can you tell me more about the DC blocking capacitor?
Thanks
Philip

Nigel Goodwin

Super Moderator
Philipc said:
I was thinking with the right cap, the ripple would go away, and I could use the meter to prove that? Or will the ripple always be there?
Also can you tell me more about the DC blocking capacitor?
Thanks
Philip

The ripple will never go away, it will just get smaller the larger the capacitor you use.

For a blocking capacitor just use a 1uF non-polarised capacitor in series with one of the meter leads - this will block the DC component and allow just the riple through.

Philipc

New Member
Just wanted to thank everyone for your help, this has been a great learning experience. Here is the data I just pulled up from my test. Granted this is a cheap DMM, so numbers do no reflect accuracy but concept only.

With a bridged transformer here is the ripple I was able to read.

470uf - .6mv
2200uf - .3mv
6800uf - .1mv

I hope you don't mind all these questions, but if a cap is two parallel plates, then why would there be a positive and negative marked on them.

Philip

Nigel Goodwin

Super Moderator
Philipc said:
I hope you don't mind all these questions, but if a cap is two parallel plates, then why would there be a positive and negative marked on them.

You are quite correct, a capacitor is simply two plates with an insulator (which can simply be air) between them. However, large values would be extremely large (think room sized!). To get large values down to a reasonable size the plates are made of metal foil and rolled up, the insulation (instead of being air, paper, plastic or anything permanent) is a chemical solution which the rolled up foil sits in. The entire device is designed to work with a DC voltage one way across it, this voltage actually creates the insulation between the plates from the chemical solution - they don't work too well without a DC polarising voltage (although non-polarised ones are availble).

k7elp60

Active Member
Power supply filter capacitors

Several posts ago I made the statment that the rule of thumb for a full wave rectifier is 3000uF per amp of load current and that 0.566 is the maximum load current you can draw. I could not remember why these values were in my mind so a did some experiments. Over the years I have built many linear power supplys and have a homemade load bank to test power supplies. This load bank is merely a power transistor on a large heatsink. I control the current drawn by the power transistor with an op amp. So it is a large variable resistor. This load bank is also used for testing rechargeable batteries. The advantage of the load bank is that as the battery discharges the current does not change as it would if I was using a normal power resistor.

The experiments I did consisted of a 12.6 volt CT secondary transformer,
rated at 12.6 VAC at 4 A,a 35 amp bridge rectifier, a 10,000uF capacitor and a 0.01 ohm 1% resistor. In both circuits the bridge input is connected to the 12.6 volt leads of the transformer. The + lead of the bridge is connected to the + of the capacitor. The load bank is connected directly across the capacitor leads. The - capacitor lead also has one lead of the 0.01 ohm resistor connected to it. I monitored the voltage drop across this resistor as it will be the charging current for the capacitor

The first experiment I connected the CT of the transformer to the other lead of the 0.01 ohm resistor. Here were the results with a load of 3.3 amp load.
Er= 0.112 volts or peak charging current 11.2 amps
Ripple= 2 volts p-to-p
Ripple valley= 6.3 volts
Er=0.132 volt or peak charging current 13.2 amps
Ripple = 2.2 volts p-to-p
Ripple valley = 6.0 volts

The second experiment was with the other end of the 0.01 ohm resistor connected to the - lead of the bridge.
Er =0.088 volts or peak charging current 8.8 amps
Ripple= 1.7 volts peak to peak
Ripple valley = 14 volts
With a load of 4 amps
Er = 0.096 volts or peak charging current of 9.6 amps
Ripple = 2.0 V p-to-p
Ripple valley 13 volts
In all the experiments took about 1/2 hour. There were short periods when I turned off the power to change connections, but the interesting thing was the the transformer was hot to the touch, which means drawing the rated current in dc from the transformer caused it to heat up. The figure of .566 of rated current was a figure I obtained several years ago from an engineer at Triad transformer corp. It was obtained after I had problems of transformers over heating. The other thing I noticed with the experiment was that the ripple valley was very close to the RMS rating of
the transformer, in both types of rectifier circuits and the capacitor = to 3000uF per amp.

I wrote this incase others would like to know my point of view
Ned

mozikluv

New Member
cap size

hi,

the formula is C = (Io/1.5)*(8.3^-3)

the letter "o" with the current symbol (I) is the rated current output

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