Where did my math go wrong?

Thread Starter

erm

Joined Dec 29, 2010
3
Hi Guys,

Having a little trouble figuring out where I screwed up in a calculation. I have an under the counter light that comes with three 12 VAC halogen puck lights. The power supply is also a dimmer. The power supply converts 110 VAC to 12 VAC and has three levels of brightness. I was wanting to change the lights to LEDs, so I and purchased a bridge rectifier, but wanted to calculate the filtering cap for it. Everywhere I checked, says to use between a 1000 - 2000 uF cap. The calculations I found here give the formula:

Smoothing capacitor for 10% ripple, C = (5 × Io) / (Vs × f)

C = smoothing capacitance in farads (F)
Io = output current from the supply in amps (A)
Vs = supply voltage in volts (V), this is the peak value of the unsmoothed DC
f = frequency of the AC supply in hertz (Hz), 50Hz in the UK

Using this formula, I insert my numbers. I get .03472 F. That's 34720 uF! Did I mess something up? That's much higher than the 1000 - 2000 uF everywhere else recommends.

The power supply says it puts out 12 VAC and 60 watts MAX. 60 watts at 12 volts is 5 A max. 12 volts at 60 hertz and the formula shows 25/720 = .03472 F. Can someone tell me what went wrong? Thanks in advance.

Tony
 

kubeek

Joined Sep 20, 2005
5,793
Everything depends on the maximum permissible ripple voltage @ specified current, which is not accounted for in that equation.

Accutally it is mentioned in the link. Just a rule of thumb tells me that 2000uF for 5A current is very small and the reservoir cap will be empty much sooner than the end of the half-period. If I remember correctly we used to use 4700uF per amp of current, which is closer to the numbers that equation tells you. (for 50hz operation)
 
Last edited:

crutschow

Joined Mar 14, 2008
34,201
You don't really need a capacitor. Just feed the rectified AC to the LEDs (through a suitable resistor, of course). That will give 120Hz flicker on the LEDs but that's normally invisible to the average person.
 

kubeek

Joined Sep 20, 2005
5,793
You don't really need a capacitor. Just feed the rectified AC to the LEDs (through a suitable resistor, of course). That will give 120Hz flicker on the LEDs but that's normally invisible to the average person.
Normal light bulbs flicker at 100/120Hz too yet many people can see that, and LEDs are much worse because they don´t have the inherent smoothing caused by the slowly cooling filament of the bulb.
 

Adjuster

Joined Dec 26, 2010
2,148
You really have to be sure what kind of LEDs you will be using. Standard LEDs are DC only, and require series resistors* to be run from ordinary voltage supplies. It is normally considered inefficient to run single lighting LEDs from a 12V supply, series groups of two or three are more usual. *Or other more complex devices.

Some LED "bulbs" come with the necessary resistors built in, and even sometimes have internal rectifiers so that they can be used as direct replacements for tungsten lamps. These can be connected directly to whatever supply voltage they have been designed for - and models with rectifiers may be OK with AC. Don't assume you have this kind of device unless you are sure.
 

thatoneguy

Joined Feb 19, 2009
6,359
Normal light bulbs flicker at 100/120Hz too yet many people can see that, and LEDs are much worse because they don´t have the inherent smoothing caused by the slowly cooling filament of the bulb.
They will flicker about the same as CFL Flourescent bulbs do. It's not really noticeable until you wave a sheet of paper back and forth under one, or the rotor of R/C Helicopter, they have a strobe light effect.
 

kubeek

Joined Sep 20, 2005
5,793
They will flicker about the same as CFL Flourescent bulbs do. It's not really noticeable until you wave a sheet of paper back and forth under one, or the rotor of R/C Helicopter, they have a strobe light effect.
Some people have an eye disease with which their eyes move back and forth a lot more than in healthy people and they notice the flicker almost anywhere.
Apparently it is called nystagmus.
 

thatoneguy

Joined Feb 19, 2009
6,359
Some people have an eye disease with which their eyes move back and forth a lot more than in healthy people and they notice the flicker almost anywhere.
Apparently it is called nystagmus.
I did not know that it had a name! I Know some people notice it more and get headaches, and kids generally seem to notice it a lot more than adults as well.
 

Thread Starter

erm

Joined Dec 29, 2010
3
Thanks for the replies. Actually, these are 12 VDC LED light strips. There are 30 superflux LEDs on a PCB with resistors. They are rated to operate at 12 VDC. I just wanted to make sure I didn't have any problems connecting to the rectified source. So, now we're saying we don't need a cap?
 

thatoneguy

Joined Feb 19, 2009
6,359
Thanks for the replies. Actually, these are 12 VDC LED light strips. There are 30 superflux LEDs on a PCB with resistors. They are rated to operate at 12 VDC. I just wanted to make sure I didn't have any problems connecting to the rectified source. So, now we're saying we don't need a cap?
Depends on the power source. They'll flicker with full rectified mains, which may be noticeable. In a car (we don't support), they could do anything from flicker to burn out because of the spikes in the electrical system.
 
Hi Guys,
The calculations I found here give the formula:

Smoothing capacitor for 10% ripple, C = (5 × Io) / (Vs × f)

C = smoothing capacitance in farads (F)
Io = output current from the supply in amps (A)
Vs = supply voltage in volts (V), this is the peak value of the unsmoothed DC
f = frequency of the AC supply in hertz (Hz), 50Hz in the UK
Tony
I think the formula in not correct, If you apply this one
http://en.wikipedia.org/wiki/Ripple_(electrical)
you get about 420 uF, also for FWR you go for f= 120 with bridge rectifier
 

Adjuster

Joined Dec 26, 2010
2,148
I think the formula in not correct, If you apply this one
http://en.wikipedia.org/wiki/Ripple_(electrical)
you get about 420 uF, also for FWR you go for f= 120 with bridge rectifier
No, that is not correct. A 420μF reservoir capacitor is not near enough for a 12V 5A supply at 10% ripple.

That is a peak-to-peak ripple voltage Vpp =of 0.05*12V =1.2V.

The formula given in your reference for full-wave rectification is Vpp = I/(2fC) , from which C = I/(2fCVpp)

If I = 5A, f = 60Hz, Vpp = 1.2V, we get C = 5A/(2*60*1.2)

I make that a whopping 34722μF, just as the OP did in his original calculation.
 

Thread Starter

erm

Joined Dec 29, 2010
3
So the calculation was correct. I pulled the numbers from the power supply and those were the max ratings, not what the LEDs are pulling. This circuit is going in my kitchen, so no need to worry about the vehicle. If I try the circuit without the caps, I should be able to see if it needs a cap without burning anything up, right? Then try it with a cap afterwards?

I'm always more precautious when it comes to electricity. With plumbing, you can see where the leak is coming from, with electricity, you can't. Usually something has to start smoking or pop before you see a problem. Trying to avoid that. :)

Just out of curiosity, but aren't our bench top power supplies the same thing (120V to 12V with caps)? They don't have that big of a capacitor in them, do they? I'm just trying to power about 1.5 amps of LEDs.

Tony
 

Adjuster

Joined Dec 26, 2010
2,148
As others have mentioned, you may not need to smooth the supply at all. Before you go much further though, there is something you would do well to to check out. You may get too much voltage from the set-up you are considering, as follows:

  • The peak value of an AC voltage is √2 times (about 1.414 times) its normally stated (RMS) value. For 12V RMS the peak value is about 17V.
  • The 12V 5A transformer loaded to only 1.5A will give somewhat more than its rated 12V (but probably not much). You might then get something like 18V peak
  • The bridge rectifier will drop a bit of voltage, perhaps 1.5V, so perhaps the peak value of the pulsating DC would be about 16.5V. (All rough estimates - could be off a volt either way?)
So what's the problem? If you smooth this effectively with a huge capacitor, the voltage may be kept up close to the peak value, which may be too much for the LEDs. You would want to try it first, preferably loading it with something cheap like a 12V 21W car bulb. If the voltage measured too high, a power resistor could be added in series to knock a few volts off.

You might think it easier not to use the capacitor, but then it might be harder to say what effective voltage the LEDs were getting. It probably would be OK though.

If you do want to smooth it, since you have only 1.5A load, then you would need less capacitance, more like 10000μF, and remember that is for smoothing to 10% - some power supplies are less well smoothed. This simple formula is also a pessimistic. Having said that, because of the way LEDs work, if smoothing is required at all it needs to be fairly thorough.

Finally, the design of mains DC power supplies has evolved a lot over the years. Even the old-fashioned linear ones probably would not try to get the reservoir capacitor ripple extremely low, relying instead on regulation to reduce it further. Modern switching power supplies tend to start with a bigger rectified voltage and a lower current, so that less smoothing capacitance is needed. High frequency switching techniques allow voltage to be traded for current without any more really big fat caps being required
 
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