Power supply tank cap

Been years since I've seen the big ARRL manual. In the following circuit, which cap (C1) value keeps voltage VDC above 12vdc given a 4 amp load:

______ ______ _________| |________| |___________________ +17VDC | T1 | | FWB | | | | | | === C1 _________| |________| |___________|_________

120vac |_____| 12vac |_____| _|_

/ / /

Anyone recall the ARRL formula?

It's understood from the book that tank C1 will raise the voltage to 1.414 (sqrt of 2) times the rectified voltage, here, approximately

17 volts, while smoothing the ripple to more easily dealt-with levels.
Reply to
Father Haskell
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"Father Haskell"

** There is no *simple* formula that give the whole "answer" - much depends on the performance characteristics of the transformer used. However, assuming the tranny is sized adequately for a 4 amp DC load, the ripple voltage on the filter cap can be estimated from: I = C dv/dt where

I = average DC load current,

dv = voltage drop during the non charging period

dt = the time duration of that period.

( note: for a 120 Hz supply, dt = 6 circa milliseconds )

So, if say C = 10,000uF

dv = I dt / C

= 4 x 0.006 / 1 exp -2

= 2.4 volts.

So, a 10,000 uF cap will peak at 17 volts and dip to 14.6 volts, 120 times per second with a 4 amp load.

....... Phil

Reply to
Phil Allison

Yeah it's a bit difficult to understand.

First we need a transformer rated much higher than the final load current, lets say 6A. We measure now its open circuit voltage, which we assume to be

13.2V rms, and calculate its interiour resistance (13.2-12)V/6A= 0.2 ohms. With the bridge rectifier, which consumes 1.4V and a capacitor we get an open circuit Vo= 13.2 * sqrt2 -1.4 = 17.5V

First lets think we have an infinitly big capacitor. There would be no ripple and the charge taken by the load would be balanced with the charge delivered by the supply. This would result in an O/P Voltage of approximately: Vo(1-sqrt(Ri/2Rload)) = 17.5V * 0.834 = 14.6V =Vinf Rload can be found by 14.6V/4A= 3.65R, instead of solving quadratic equations, you can do try&error to come close. We can now also dimension the required transformer power rating: Pt = 1.2 * Iout *(Vinf +2*Vd) The factor 1.2 is for a bridge rectifier, Vd is the diode voltage loss. Pt= 1.2*4*16=76.8VA which corresponds to a 12V/6.4A transformer. We can now find the peak-to-peak ripple voltage for an arbitrary capacitor C, I just give the formula: Vripp= Iout/(2C*fn) * (1-sqrt sqrt(Ri/2Rload)) it is the 4th-root With 10000uF we get:

4/1.2 * 0.483= 1.613Vpp of which approximately 2/3 are below Vinf 14.6 -1.1 = 13.5V minimum O/P voltage There are a few approximations involved, but we are very close what actually happens. So be aware: a 4A transformer will just give you 2.5A out and if you want 12V stabilized you have to use a LDO regulator. Additional power rating is needed for 10% overvoltage and a higher transformer voltage for 10% mains low.

bravo Ban, good day

Apricale, Italy

Reply to
Ban

"Ban"

** What is?

Nothing I wrote was difficult to understand.

** This lot is difficult to follow, or believe.

Lots of hidden assumptions re the transformer characteristics.

....... Phil

Reply to
Phil Allison

LOL

you hid the conduction time though. but your 2ms is pretty close.

I just measured a 24Vac 50VA transformer a couple of days back. I measured an 0.1V drop with 60mA load, giving

Z = 1.7 Ohms.

DCR = 1.4 Ohms, so

Xl = 0.964 Ohms and the secondary leakage inductance is

Lls = 3mH.

scarcely ignorable....

I solve these problems using energy relationships:

Emax = 0.5*C*Vpeak^2

Emin = 0.5*C*Vmin^2

Emax - Emin = Eload = Pload*Tcap

usually I know Vpeak and Vmin, so:

calculate theta_cond = acos(Vmin/Vpeak) = angle at which Vin = Vmin

calculate Tcond = (theta_cond/360)*1/Fac

calculate Tripple = 1/(2*Fac) = 8.33ms @ 60Hz

calculate Tcap = Tripple - Tconduction = time when cap supplies load.

calculate Eload = Pload*Tcap

usually I know Vpeak and Vmin, so:

calculate Emax = 0.5*C*Vpeak^2

calculate Emin = 0.5*C*Vmin^2

knowing Emax - Emin = Eload,

calculate Cmin = Eload*2/(Vpeak^2 - Vmin^2)

then I pick an actual cap, and run thru it backwards to work out Vmin, Tcond and of course Eload. repeat until it converges (very fast, one iteration is usually good enough)

For the OP, this is:

Vpeak = 17

Vmin = 12

Iload = 4A

Pload = 4A*(17V + 12V)/2 = 58W

theta = acos(12/17) = 45.1 degrees

Tcond = 2.1ms

Tcap = 6.2ms

Eload = 58W*6.2ms = 362mJ

now we can calculate the minimum cap required:

Cmin = 362mJ*2/(17^2 - 12^2)

Cmin = 5mF

sanity check:

Epeak = 0.5*5mF*17V^2 = 723mJ > Eload

Emin = Epeak - Eload = 361mJ (interesting aside: half the energy, so about 3/4 the voltage...)

Vmin = sqrt(2*Emin/C) = 12V. yay.

OK, say he picks 10mF:

Emax = 1445mJ

Emin = 1083mJ

Vmin = 14.7V

theta = 30 degrees

Tcond = 1.4ms

Tcap = 6.9ms

Eload = 6.9ms*4A*(17V + 14.7V)/2 = 440mJ (20% higher than before, its a constant current load)

so repeat the loop:

Emin = 1005mJ

Vmin = 14.2V

theta = 33.5 degrees

Tcond = 1.55ms

Tcap = 6.8ms

Eload = 6.8ms*4A*(17V + 14.2V)/2 = 423mJ

- now this only differs from the previous value of Eload by -4%. Just for fun, lets do one more loop:

Emin = 1022mJ

Vmin = 14.3V

theta = 32.8 degrees

Tcond = 1.52ms

Tcap = 6.82ms

Eload = 427mJ - OK, this has now changed by +0.87%, which is close enough to zero for me. Hell, -4% is good enough (eg consider electrolytic cap tolerance)

for this setup, with a 1.5ms conduction angle and 8.33ms ripple period, we can calculate RMS input current too:

- assume its a rectangle (it aint) thats Tcond wide.

- this rectangle must have the same charge as Iout*Tripple, so:

Ipeak = Iout*Tripple/Tcond = 4A*8.33ms/1.5ms = 22A peak

- rms of square wave = peak*sqrt(D)

Irms = 22A*sqrt(Tcond/Tripple) = 9.4Arms

of course this neglects the transformer impedance; the leakage inductance will spread the current pulse out, thereby dropping the peak.

For a bridge connected directly across the national grid however (eg off-line smps), this assumption is pretty accurate.

its also pretty clear why more capacitance is a BAD thing - it decreases the conduction angle, thereby requiring ever-higher peak currents in order to deliver the same amount of charge in a shorter time frame.

likewise, its also clear why rectifier-capcitor front-ends of off-line smps cause flat-topping of the AC line - its the very high peak currents.

Cheers Terry

Reply to
Terry Given

"Terry Given"

** = a truly obnoxious NZ criminal maniac

aka " The Pinball Wizard "

( deaf, dumb and blind ...... )

** I hid nothing - you lying pile of shit.

" note: for a 120 Hz supply, dt = 6 circa milliseconds "

The word "circa" allows a reasonable variation either side of this centre value.

** Funny how the rms value of the charging current pulse train stays nearly constant no matter how big the cap.

....... Phil

Reply to
Phil Allison

Terry, a very good analysis. All you have to do now is take into account transformer wire resistance, leakage reactance, diode resistance and capacitor resistance.

These factors matter very much in view of the very large peak current which flows during the short conducting period.

The last thing to do is calculate the power dissipated in the transformer, diode and the capacacitor. If within the components' power ratings all is then fine.

--
Reg.
Reply to
Reg Edwards

"Terry Given"

** = a truly obnoxious NZ criminal maniac

aka " The Pinball Wizard "

( deaf, dumb and blind ...... )

** This is just completely ***WRONG***.

And so is the whole stupid lot that follows from it.

....... Phil

Reply to
Phil Allison

"Reg Edwards"

** It is complete crap.

** The fool tried and failed.

** It is not THAT large.

Its rms value is about 1.6 times the resistive load equivalent.

....... Phil

Reply to
Phil Allison

I prefer to think of it as a crude approximation. Lls = 0.3mH according to my trusty HP3577. OK, its a very crude approximation. Ignoring Rprimary' wasnt a great call, if I add it in, the numbers stack up. not that it mattered in that case (I cared about Z not L)

none of the rest follows from it.

its funny, Phil, how you respond to technical posts. I'd think you could understand this stuff....

Cheers Terry

Reply to
Terry Given

Hi Reg,

thanks. its just back-of-the-envelope stuff really, but its surprising how many people dont know how to do it.

All you have to do now is take into

LOL :)

you forgot skin effect.

Barton has a lengthy section on rectifier-capacitor analysis, replete with a plethora of evil equations. Hell, a paper even appeared in IEEE Industry applications just last year.

most of my designs havent had transformers though - slap 'em straight across the grid :)

Ll and C tend to make it semi-sinusoidal too. if you account for the parasitics, spice can work wonders. if the models are OK (why do 10BQ100 models have Rs = 100 ?!).

years back, I got a tech to measure the RMS input current on a few drives. the 4A drive had a higher RMS input current (16Arms IIRC) than the 12A drive, which had 3% line chokes. At first I thought he'd measured it wrong (his response: "do I look like I'm f***ing stupid") until I did the aforementioned maths.

Michael Gaspari wrote a great paper (industry apps again), in which he analysed the effect of supply impedance on cap lifetime. no surprises, but it has a huge effect as it directly controls cap ripple current when there is no handy wee transformer to provide some L.

I just did some lifetime calcs for panasonic TS-ED, TS-HA and TS-HB caps. Interestingly enough, for the 35mm dia x 50mm caps, the product of ESR*Iripple^2 ~ 0.75 for 6 different caps.

Panasonic also mention that

L = Lrated*2^[(Trated - (Tambient + dT))/10]*(V/Vrated)^-6.5

for V >= 0.9*Vrated

and they give dT = 15C (OK, they bury it deep in a paper), so Rtheta =

20K/W for that cap.

the voltage lifetime multiplier is interesting too.

in my case, I used Trated, Tambient, Lrated and Ldesired (45,000 hours) to calculate the amount of ripple current I can bung thru the cap. I'll leave the maths as an exercise for the astute reader.

I also found that a 3,000 hour 3.53A cap gave me a slightly longer life (64% rated = 2.27A) than a 7,000 hour 1.93A cap (110% rated = 2.13A) but there wasnt much in it. except the 3,000 hour cap was 470uF c.f. 330uF for the 7,000 hour cap.

aint engineering fun!

Cheers Terry

Reply to
Terry Given

"Terry Given"

** Like you - f*****ad.

...... Phil

Reply to
Phil Allison

"Terry Given"

** = a truly obnoxious NZ criminal maniac

aka " The Pinball Wizard "

( = deaf, dumb & blind asshole )

** Psychopaths like Given area crude approximation to a human being.
** Like a criminal psychotic is a VERY crude approximation to a human being.
** It was completely FUCKING ASININE !!!!!

As was using an wild example only YOU know about.

As was failing to relate the example to reality.

** You are one vile, stinking LIAR - Given.

A slimy, scum of the earth kiwi POS who needs dealing with.

Watch out for me.

..... Phil

Reply to
Phil Allison

I was not referring to your post, but to my own explanation. And even an intelligent person like you didn't follow, so it is right. If I had commented about your post I would have written:

  1. what about the bridge rectifier?
  2. How can a 10mF cap charge to the peak voltage?

The charge will start when the momentary sin value is higher than the remaining voltage on the cap. The current will come to a peak short before the sin is max, because then the voltage difference is greatest and then will fall until the the decreasing sin equals the max. cap voltage. We have a constant load current so then the cap voltage lineary ramps down until the next half-wave. peak charge current is 14.5A = Vo/sqrt(Ri + Rload)

Do not believe, try to understand.

The transformer I looked up is an M85a with P=80W based on 1.2T peak and 40K temp rise, loss factor 9%. secondary 1.7mm dia. I also have simplified the diode voltage as constant, but the actual Ri is smaller(0.17R) and the 0.2R should make up for that.

--
ciao Ban
Apricale, Italy
Reply to
Ban

Phil, you're just jealous because you dont understand it. Perhaps this is why most of us here at SED get paid to design things, whereas you fix things others design. If you could dismount your psychopathic hobby horse, you might learn something.

Cheers Terry

PS you must be pretty stupid, to post comments like that in writing, in a forum that will last a very long time.....

Reply to
Terry Given

"Ban"

** Then why the ***HELL*** did you post the words **directly under** my post where I signed my name ???

The MOST BASIC RULE of usenet etiquette to post UNDER the words you are commenting on.

So - if you are NOT doing that, then f****ng do NOT !!

** Some pile of rabid drivel ?

No way.

** Yawn.......

.......... Phil

Reply to
Phil Allison

** ROTFLMAO

Given is completely insane.

........ Phil

Reply to
Phil Allison

Because it was the next available space, you started *your* supersmart response with the words:

and I did something similar. So please can you answer my questions?

  1. what about the bridge rectifier?
  2. How can a 10mF cap charge to the peak voltage?
--
ciao Ban
Apricale, Italy
Reply to
Ban

Yes Terry, but at least you *got* it and admitted it here, so that makes you a bit more smart than this Phil. It seems a bit funny to me that your tranny has more than 4 times the loss of the primary winding in the secondary, maybe there is another one in parallel?

--
ciao Ban
Apricale, Italy
Reply to
Ban

"Ban"

** That is not an anawer.

** Like hell you did.

** ROTFL

Go f*ck yourself - you asinine wog pig.

........ Phil

Reply to
Phil Allison

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