How many caps do I need to compensate for ripple current

#12

Joined Nov 30, 2010
18,217
Did you look at the math on that?!!!

I like my equasion better. Humans can actually use it.

t06afre

Joined May 11, 2009
5,934
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#12

#12

Joined Nov 30, 2010
18,217
http://en.wikipedia.org/wiki/Ripple_(electrical)

Here. I tightened up that Wiki link.
The Hammond link is good, and so is the information.

My problem with the Wikilink is that it presents 2C Er F =I as the first equasion and I debunked that in post #20. I can't trust the equasions that follow a false starting equation!

edit: Now my link doesn't work right and the original (by t06afre) works.

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t06afre

Joined May 11, 2009
5,934
http://en.wikipedia.org/wiki/Ripple_(electrical)

Here. I tightened up that Wiki link.
The Hammond link is good, and so is the information.

My problem with the Wikilink is that it presents 2C Er F =I as the first equasion and I debunked that in post #20. I can't trust the equasions that follow a false starting equation!
Then you use a full wave rectifier. The frequency of the output signal will double in respect to the input frequency.
By the way the wiki link is tricky to get correct due to ) at the end. In order to get correct you need to use BB codes
(URL)

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#12

Joined Nov 30, 2010
18,217
I already used a full wave rectifier. The Wiki page used a full wave rectifier. These things are obvious if you read the information available.

#12

Joined Nov 30, 2010
18,217
I just had a great idea! Instead of me going back to the 1870's and converting heat into B.T.U.s, how about somebody with a True RMS meter do some measurements? Or somebody that can simulate see if a sim program can come up with rms ripple voltage?

t_n_k

Joined Mar 6, 2009
5,455
I just had a great idea! Instead of me going back to the 1870's and converting heat into B.T.U.s, how about somebody with a True RMS meter do some measurements? Or somebody that can simulate see if a sim program can come up with rms ripple voltage?
You'll probably find that most circuit simulation applications have that feature already built in.

With respect to the mathematical analysis I posted - I wasn't suggesting you use it - many people share the information on this site and there are some of us interested in these things. No doubt someone with the time and patience could embed the equations in a spreadsheet.

wayneh

Joined Sep 9, 2010
17,027
Most references including the Wiki make simplifying assumptions, most notably that the ripple is "small" compared to the DC voltage. This makes the calculations - estimations, actually - simpler, but gives results that are flat wrong when that assumption falls apart, which is often. The detailed article is good at describing the problem but the author exaggerates the uniqueness and value of their solution, as it is merely exchanging one iterative solution for another.

Simply put, there is NO analytical solution. There are "close enough" approximations when the ripple is know to be small, and there is the spreadsheet solution when you need precision or the ripple is "big".

t_n_k

Joined Mar 6, 2009
5,455
Simply put, there is NO analytical solution. There are "close enough" approximations when the ripple is know to be small, and there is the spreadsheet solution when you need precision or the ripple is "big".
Agreed with your comments. And yes there is no avoiding an iterative solution.

I guess in the end it depends how accurate your calculations need to be.

The effect of transformer parameters is probably an important factor as well. I've observed some pretty awful waveforms on the secondary side at the rectifier input in real circuits. The high peak current in the charging phase can significantly distort the AC waveform during that part of the cycle.

I'd certainly like to see your spreadsheet. You can PM me and I'll give you an email address. Often a modern simulation program can do just as well - particularly as you can play around with the circuit parameters 'on the fly' to determine those of more significance and the degree to which they effect the overall performance.

#12

Joined Nov 30, 2010
18,217
In just the last week or two, I've seen the C Er F = I equasion with a coefficient of 1 and a coefficient of 2, both from rather sophistocated sources. I've been using the radical2 coefficient for decades, and building the circuit works, as I showed, within 1.4%

The "authorities" are publishing coefficients that are off by +/- 40% for the typical circuits on this site.

My conditions were 1 volt p-p ripple on a 10 volt supply because 10 volts DC was convenient and 1 volt of ripple is a reasonable goal for the kind of power supply that people on this site would use. Would the results change significantly if I built the circuit to use 100V DC? if I changed the current to 1680 ma instead of 168ua? if the ripple allowed was 3 v p-p?

In other words, where does the radical2 coefficient get significantly wrong, or "fall apart" for the denizens of this site? Can it be trusted for say, voltages from 3 to 50 and current from 100ua to 3 amps?

I can build those circuits, but I'm as lazy as the next guy. If you know where the limits are, I'd sure like to hear from you.

t06afre

Joined May 11, 2009
5,934
In other words, where does the radical2 coefficient get significantly wrong
What is the radical2 coefficient ?

wayneh

Joined Sep 9, 2010
17,027
I'd certainly like to see your spreadsheet. You can PM me and I'll give you an email address.
I'll dig it up and get it in shape for sharing. You know how it is, you develop it to your own satisfaction at the time and then usually stop short of making it simple and usable by others. That means when you look back some time later, it can be tough to figure out what the heck you were doing!

In the meanwhile here is more about what it does. The goal is to determine the effect of (a resistive) load on the time-averaged voltage out of a full-wave bridge rectifier, filtered with a simple capacitor.

First the analysis looks at the rectified sine wave from the bridge (blue line), and also at the voltage across the filter capacitor while it is discharging (green line). The voltage at the load is the higher of the two (red line). All of the ugly math is buried in determining where those two curves intersect, where conduction starts and stops.

Once you can calculate the voltage-versus-time profile, then you can calculate the time-average voltage. And you can set up a table in Excel to look at the effect of changing the RC time constant, by changing the load R, on the observed voltage. As you would guess, a large time constant (big C or R, or both) gives a tiny ripple and the measured voltage is the peak (1 in the chart, black line). A big load (small R) reduces the time constant, increases the ripple, and sends the average voltage down (red line). At the extreme where the cap isn't doing anything compared to the big load, you just measure the RMS of the sine wave, getting ~64% of peak (blue line).

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MrChips

Joined Oct 2, 2009
23,069
I'm a bit late joining in but here is my input for whatever it is worth.

Power Supply Design is standard fare in any Introductory Electronics Course. Just to make sure that I can remember how to do this I have done the following:

(1) Do the simple math.
(2) Check my results via simulation using CircuitMaker.
(3) Measure actual results on a test circuit.

The test transformer is a Hammond 165P25 CT 5A weighing 5.5 lbs.

Hammond is known to make reliable transformers that are conservatively rated. The voltage specified is the voltage out at full load. No-load voltage on this transformer is 27V.

Ok, let's do the math. We could do a rigorous mathematical analysis but there is no need. (See Post #21 for the mathematically inclined.) A simple first order approximation will give us good results.

Let us make the following assumptions:

Vac = transformer output at rated load = 25V
Vdiode = diode drop = 0.7V
C = filter capacitor = 0.01F
R = load resistor = 10
f = line frequency = 60Hz

Vac is the RMS value from the transformer. For fullwave centre-tap rectification, we divide this by 2 and multiply by square-root of 2 to give peak voltage. Then we subtract the diode-drop Vdiode.

Vdc = Vac*1.414/2 – Vdiode

Substituting the above values give us Vdc = 17V

Now we calculate the average load current:

Iavg = Vdc/R = 1.7A

We make a simple approximation that this charge must be supplied by the filter capacitor during one half-wave cycle. This time duration is one-halve the period of the mains frequency.

T = 1/2f = 0.0083 s

The charge supplied during this time is the average current multiplied by the time.

Q = T * Iavg = 0.014 Coulomb

Hence the voltage on the filter capacitor will drop by:

Vripple = Q/C = 1.4V

Now we need to calculate the peak diode current. To do this we have to estimate when the diode is turned on.

The voltage into the capacitor is

V = Vdc * sin(2*pi*f*t)

The capacitor will begin to recharge when the sine wave exceeds the load voltage, i.e. when

Vdc – Vripple = Vdc*sin(2*pi*f*t)

we solve to find t

t = arcsin(1 – Vripple/Vdc) / (2*pi*f)

This is the point approaching the peak of the sine wave when the diode starts to conduct.
This calculates to be about the 67 degree mark or 3.08ms.

We make the assumption that the diode conducts from about 67 degrees to 90 degrees,
that is, from 3.08ms to 4.16ms = 1.08 ms

tconduct = 1/(4*f ) – t

Assuming all the charge is replenished during the turn on time, we can calculate the peak diode current if only we knew the shape of the current pulse. In the best case, if the shape is rectangular, the area of the rectangle is given as

Q = Ipeak * tconduct

Ipeak = Q/tconduct

Ipeak = 0.014/0.00108 = 12.96A

If we approximate the shape of the current pulse to be triangular then Ipeak will be twice = 26A.

There you have it.

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#12

Joined Nov 30, 2010
18,217
and the coefficient K of the equasion K C Er F = I is?

wayneh

Joined Sep 9, 2010
17,027
There you have it.
That's your step 1, the simplified math using the standard assumptions. That approach falls apart when the ripple is "large". Any comments on steps 2 and 3?

t06afre

Joined May 11, 2009
5,934
and the coefficient K of the equasion K C Er F = I is?
As I said before it is no constant. In your equation the variable F is the mains frequency. Then you apply a full wave rectifier. The frequency out from the rectifier will double in respect to the input frequency. Hence the factor 2.

#12

Joined Nov 30, 2010
18,217
t06afre, if you could understand post#8 your answers would make more sense.

CDRIVE

Joined Jul 1, 2008
2,219
OH BOYS!!! In case you haven't noticed, I think you guys scared off the OP. I think his last post was #4.