Of course rules of thumb are just that. Don Lancaster goes into some detail in his Radio Electronics Oct 1989 Hardware Hacker column to arrive at the rule of thumb, "In an 8300uF capacitor used in a full-wave line-operated supply, the volts of ripple will equal the amps of load current." [Emphasis his] Quite a difference with the 2200uF given earlier.
On a different subject, it wasn't made clear why a two-prong wall wart is OK, but a line cord to an enclosed transformer must be three-prong.
On a different subject, it wasn't made clear why a two-prong wall wart is OK, but a line cord to an enclosed transformer must be three-prong.
I don't see a difference at all, the C vs Amps equation is a linear one. I=C*dV/dt
A 1 volt RMS figure is a line in the sand E.g. starting point.
he might be using 1V p-p for his rule, plus some margin factor (waveform correction?) perhaps
Ive done the math from scratch, so it works for me.
Class I is 2 prong AC, double insulated build safety = most consumer gear
most DIY solutions use Class II, depending on mixing the two classes sometimes stray line related currents travel between boxes on the interconnects creating hum.
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He might be, but he's not.
He derives his figure from well known and accepted PSU theory before stating it.
In other words (and fewer too), he can use a two-prong line cord to a transformer in an enclosed box.
which is?
I=CdV/dt is THE basic definition of a capacitor
maybe you should do the math from scratch too.
1Vpp use 6,200 uF/A which isn't that far way from 8,300 uF, I just choose to remember 2200 a standard value. I reckon once ripple get smaller than a few volts p-p a further waveform correction is needed due to smaller conduction angle.
no no I wouldn't ever advise that, not in public forum. look were talking bout basic safety here with uninformed noobs that grab parts from Lord knows where.
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So...
C=0.0022
dV=1V ripple
dt= 8.33 milliseconds (120Hz)
I'll let you solve for I.
Yeah. Neither would I.
knock yourself out 🙂 Linear Power Supply Design see section called Capacitor Value , my rule of thumb
equation is from> 1.Voltage Regulator Handbook - National Semiconductor Corp. - 1981 Edition
equation is from> 1.Voltage Regulator Handbook - National Semiconductor Corp. - 1981 Edition
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You're the one the brought "THE basic definition of a capacitor" formula into the discussion. Instead of failing to see a difference, you should have given your source, which is ESP and not NatSemi, up front. They're rules of thumb, and I found the, yes, differences interesting.
what? no no I took their formula back down to the basic definition . look closer it's the same thing
I still fail to see any difference except their equation falls apart a low ripple voltages.Eg average, RMS, peak, all have another definition, so their accuracy matters too. The math always works. if you use the calculus meanings. the instantaneous current = slopes of the voltage waveform times capacitance!
you never disclosed the source of 8,300 or even defined it clearly!
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What?
I don't do calculus, but if the math always works, and it's the same thing, please explain the 3.78x difference.
If their equation falls apart at low ripple voltages, isn't it of much limited value, since (I would think) that is where it's needed most?
I don't do calculus, but if the math always works, and it's the same thing, please explain the 3.78x difference.
If their equation falls apart at low ripple voltages, isn't it of much limited value, since (I would think) that is where it's needed most?
its a rule of thumb jeez don't focus on the number understand how to use it.
I thought I already commented about differences , see post 44, concerning Vrms vs peak-peak = 2.8 for a sine wave but as the ripple voltage get smaller its not a sinusoid anymore the slope rises due to very small conduction time so you need another correction factor.
again, note> this rule of thumb is for using voltage regulators which allows for smaller caps and still gives low ripple on its output. Caps are expensive and bulky by themselves and can never compete with regulators for lowest ripple. Please read the ESP link for unregulated supplies and sizing caps, very practical real world advise. Huge caps mean very high peak charging currents to average out Iload for low ripple.
I thought I already commented about differences , see post 44, concerning Vrms vs peak-peak = 2.8 for a sine wave but as the ripple voltage get smaller its not a sinusoid anymore the slope rises due to very small conduction time so you need another correction factor.
again, note> this rule of thumb is for using voltage regulators which allows for smaller caps and still gives low ripple on its output. Caps are expensive and bulky by themselves and can never compete with regulators for lowest ripple. Please read the ESP link for unregulated supplies and sizing caps, very practical real world advise. Huge caps mean very high peak charging currents to average out Iload for low ripple.
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Ummm... I've noted the 'rule of thumb' thing at least twice already.
I'm focused on words. Yours and mine. And more words, like squirm, spin, & doubletalk. And, done.
wow really sofaspud
you cant struggle past the number differences? use your rule thumb then. IDC if you reject my sources, but you never explained your rule of 8300 uF, I suspect you cant.
"buy em the books, send them to school......"
you cant struggle past the number differences? use your rule thumb then. IDC if you reject my sources, but you never explained your rule of 8300 uF, I suspect you cant.
"buy em the books, send them to school......"
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No, I can't. And I'm big enough and honest enough to admit it. Rare virtues 'round here, to be sure. So I asked for some reconciling of the nearly 4x difference, based entirely on, "...they're not different... the math always works... it's the same thing..."
I never rejected your source. And if you suspect I can't explain the 8300uF rule, I didn't need to - you did it for me! C=dV/dt
Confucius say, " Not paying attention is bad budget decision."
no please stop taking bits and pieces out of context. that not what I said. I pointed out a 2.8 factor might be due from converting RMS to V peak to peak and the rest probably a waveform modifier due to lower ripple than a couple of volts. I don't know for sure cuz I haven't seen any definition nor any analysis of what you allude to. mine assumes higher ripple ~ 1VRMS his probably doesn't, so why would the numbers match exactly? sheesh
I also didn't disagree with the quote that ripple voltage increases linearly with I load, you can see that clearly in the equation I=C*dV/dt. nor that his C number is designs for a low ripple unregulated supply <1V p-p. they are most probably equal given the non matched conditions.
as one President used to say "there you go again"
wake up on the wrong side of the bed or what?
BTW it's I=C*dV/dt written on stone tablets somewhere...
^this tracks with the equation on the ESP site which is written in slightly different way. sorry perhaps you just cant translate it.
I'm done here unless you find anything new.
I=C*dV/dT
If frequency = 50Hz then dT = 10ms = 0.01s
Choosing a 2000uF cap = 0.002F and current = 1A
dVpp = I * dT / C = 1 * 0.01 / 0.002 = 5Vpp
i.e. 2mF/A gives ~ 5Vpp ripple.
What have I done wrong?
Repeating for 8m3F and 60Hz gives
dVpp = 1 * 0.0083 / 0.0083 = 1Vpp
Have I got this wrong as well?
The 2mF/A is a guideline for reasonable ripple on a supply for device that have adequate PSRR or don't care about quality of supply.
Many will use double the 2mF/A and some will double that again.
10mF/A is not that uncommon for users wanting very low ripple on the supply, eg for a phono pre that draws only 50mAdc, use 470uF .
If frequency = 50Hz then dT = 10ms = 0.01s
Choosing a 2000uF cap = 0.002F and current = 1A
dVpp = I * dT / C = 1 * 0.01 / 0.002 = 5Vpp
i.e. 2mF/A gives ~ 5Vpp ripple.
What have I done wrong?
Repeating for 8m3F and 60Hz gives
dVpp = 1 * 0.0083 / 0.0083 = 1Vpp
Have I got this wrong as well?
The 2mF/A is a guideline for reasonable ripple on a supply for device that have adequate PSRR or don't care about quality of supply.
Many will use double the 2mF/A and some will double that again.
10mF/A is not that uncommon for users wanting very low ripple on the supply, eg for a phono pre that draws only 50mAdc, use 470uF .
think about the waveform, your simple calc use a 50% saw tooth at half wave frequencies. use 110 or 120 Hz
also note the current through the capacitor will be equal to the load current for RMS values.
the actual charging current (in the equation) is a short duration so peak is calculated at the max slope of the wave form, the problem comes about when converting peaks into the average value load current.
also note the current through the capacitor will be equal to the load current for RMS values.
the actual charging current (in the equation) is a short duration so peak is calculated at the max slope of the wave form, the problem comes about when converting peaks into the average value load current.
10ms for 50Hz and 8.33 repeating for 60Hz, does use the 100Hz/120Hz recharge gaps. The gaps between the short recharge period is slightly less than than 10/8.33ms, usually around 85% to 95%
It's the draw down between the charging pulses that gives the ripple voltage. That draw down is usually a fairly constant current. Reduce the current draw to zero and the ripple follows suit. The only remaining current is leakage and that could be in the uA to nA range. Now the ripple will be so low that most of us (me included) will be incapable of measuring it.
Anyone else want to offer suggestions on where my example calculations have gone wrong?
It's the draw down between the charging pulses that gives the ripple voltage. That draw down is usually a fairly constant current. Reduce the current draw to zero and the ripple follows suit. The only remaining current is leakage and that could be in the uA to nA range. Now the ripple will be so low that most of us (me included) will be incapable of measuring it.
Anyone else want to offer suggestions on where my example calculations have gone wrong?
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exactly, the charging peak current must average out over one half cycle. that's why I used RMS ( area under the charging pulse )
No. The charge builds up during the charging pulse period.
After that is complete the capacitor supplies the current.
The charging is now off and the current draw is nearly constant DC.
No clever conversions to rms needed.
Just use the formula.
I asked if anyone else could offer suggestions on where the calculation had gone wrong.
After that is complete the capacitor supplies the current.
The charging is now off and the current draw is nearly constant DC.
No clever conversions to rms needed.
Just use the formula.
I asked if anyone else could offer suggestions on where the calculation had gone wrong.
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