Ohms Law Calculation

[keeney123]

Re: Ohms Law Calculation

Example of how to calculate the filter cap value, capacitors have to supply the current during the off time of the diodes.
http://www.electro-tech-online.com/t...acitor.106374/

I think this guys formula may be in error. When considering RMS. The full wave is considered. That is above and below Gnd. One only has so much RMS power in this wave. If I rectify the wave the amount of power does not change except for the loss in rectification components. By this guy changing the frequency to 120 Hz the formula will not hold true as nothing was changed in RMS power. It is still the same consideration of power except now it is above the line. The ripple voltage needs to be consider as 120 Hz because this is the frequency of the ripple and hence the cap to smooth the ripple will be delta time for this sawtooth waveform. Because the waveform has changed I would say the RMS formula for the 60Hz of 120 Hz can no longer be used because we are not considering the full wave form but a component of ripple that has changed in nature from the original. I would think a calculus equation would have be derived to amount of energy giving at this time period because the area under the curve has changed?

[keeney123]

Re: Ohms Law Calculation

OK I see one mistake I made. The resulting Voltage from rectification is not a sawtooth it is a pulsating hump. So yes they can consider it the way this person has consider it. The sawtooth is a result of cap filtering. Sorry about my mistake. Still the power would be the same less the components I mention below.

[EasyGoing1]

Re: Ohms Law Calculation

Transformer outputs AC voltage.

The bridge rectifier converts AC voltage to DC voltage but without some capacitance, what you get is basically a series of pulses, all above 0v.

The peak DC voltage that comes out of the rectifier is equal to 1.414 x Vac - 2 x Vdiode where Vac is the AC voltage and Vdiode is the voltage drop on one diode in the bridge rectifier (typically between 0.7v and 1.1v, according to the datasheet of your bridge rectifier it's 1.1v )

Vdc peak = 1.414 x 40 - 2 x 1.1v = 54.2v ~

So you won't get 40v DC out of the bridge rectifier, but rather a DC voltage with a peak of about 55 volts. Also, keep in mind that transformers when idle typically can output up to 10% more AC voltage, only under load their output goes down to the rated AC voltage.

This is very important, because as you can see on that DC-DC converter board, at the input you have two capacitors rated only for 50v, which means if what you say it's true, eventually those capacitors will vent or blow up or go bad, because you're exceeding their maximum voltage rating by about 5v. At the very least, you would have to replace these with 63v rated capacitors.

Now of course, 120 times a second you'll have pulses of DC voltage going all the way up to around 55v and then slowly going down. Depending on how much capacitance you put after the bridge rectifier, you raise the minimum DC voltage so that it will always stay above some threshold.
With such DC-DC converters, I think your input voltage should stay at least about 1v above the output voltage, but they'll be more happy with voltage varying within a narrower range. So if you want 6v out, I guess you could power it with some DC voltage from 7v to 50v, but the regulator will probably be more happy with something like 40v..55v

How much capacitance you need, you can estimate using this formula :

Capacitance (in Farads) = Current (Amps) / [ 2 x AC Mains Frequency x ( Vdc peak - Vdc minimum desired ) ]

So for example, let's say you want 6v @ 5A from your DC-DC converter. The DC-DC converter probably converts the higher voltage to lower voltage with about 85% efficiency or higher, but let's make things easy and say 80%.
This means that in order to produce 6v x 5A = 30 watts of energy, it will have to be fed with at least about 36 watts (because 30 watts would be 4/5 of the total energy used due to the 80% efficiency, so since 4/5 is 30 watts that means 1/5th is 6 watts, so then 5/5 equals to 36 watts). Anyway, just to make things easier, let's just round it up and say we want to give the dc-dc converter more than it needs and simply work with 40 watts.

Now, right after the rectifier you have DC voltage with peaks up to 55v and some current (which depends on the VA rating of the transformer). You have to figure out how much capacitance you need to have there, so that the dc-dc converter will always be able to draw 40 watts from it.
One option would be to always keep the DC voltage equal or above 40v, if the transformer can output 1A of current. In this case, you have that formula :

C = 1A / [ 2 x 60 Hz x (55v - 40v ) ] = 1 / 120x15 = 1/1800 = 0.000555 Farads or 555uF

In your case, you have two 330uF in parallel at the input so it would be more than 555uF which would work out right.

Another option would be to use more capacitance to raise the minimum voltage higher, and lower the current. For example, let's say we want the minimum voltage of at least 50v, in which case we'd only need 40w / 50v = 0.8a of current to keep things running smooth

So C = 0.8A / 120 x (55-50) = 0.8 / 600 = 0.001333 Farads or 1333uF, in which case it's obvious your DC-DC converter has too little capacitance at the input.

Anyway, again, this is all assuming you want the dc-dc converter to have at least 40w of energy available at the input, so that you could have 6v at 5a (30w) and some reserves at the output. If you aim for lower maximum currents at the output, you can change the numbers and recalculate things using the two formulas above.

btw. Your transformer is rated in VA and has an AC voltage at the output. So if your transformer says it's 40v AC and 25VA, that means the maximum AC current at the output is about 25 / 40 = 0.62A
When this AC voltage is rectified using a classic bridge rectifier, the maximum DC current is not the same but rather smaller, and can be approximated using the formula I dc = ~ 0.62 x Iac , so for my numbers the peak AC current could be only 0.62 x 0.625 = 0.387A

So going back to my examples above, you may very think you could use 1A or 0.8A in your calculations, but if the transformer can only provide ~ 0.39A then you're screwed, you'd have to raise the minimum voltage by using more capacitance.

The 0.62 is a "magic constant" which works well for transformers in the 40-100va range, but is often a bit off for very small or much larger transformers.

so first of all .. if your transformer really outputs 40v ac, then really really really those 50v rated capacitors at the input of your dc-dc converter should be changed to some rated for at least 63v.

Alternatively, if your transformer has two secondary windings, you could connect those secondary windings in parallel for half the voltage but twice the current.

Mariushm,

Thank you very much for taking the time to write this... I'm going to re-read it a couple of times (for proper digestion ), but my first read through really has opened my eyes and helped me understand the fact that there is a lot more to consider where I was making very broad assumptions in my question. Of course I knew there were much more complicated physics involved in what I was asking, but I was hoping I could generalize and that would be sufficient for a generalized answer. But now I see why that's not even possible ...

Anyways, thank you, I do appreciate you taking the time to write this for me.

Mike