Power Supply Build for Router CNC

[Wobblycogs]

Thanks for that , I spent a whole evening trying to figure out if the 1.4 was supposed to be sqrt(2) and I couldn't find anywhere that definitively answered the question. In the end I assumed that people used 1.4 because after smoothing the average value of Vdc will still be slightly below Vmax (the highest voltage coming through the rectifier).

The next transformer "down" is this 24+24 model then below that is this the 22+22V. The mains in this house is normally around 235 to 237 volts so yes I would be a bit over-voltage. Does the output voltage change linearly with the input voltage? I'm thinking 235V into the 25+25 transformer would produce about 51V out (wired in series). Wish I'd put together a spreadsheet for the calculations now! Considering there's a 2.2V drop in the rectifier I think the 24+24 is probably the best choice in that case - I suppose I've just got to go away and crunch the numbers again.

I'll scrap the relay for the bleed resistor. Considering the price it would take 100 years to pay for itself in the electricity it saves.

I was thinking of trying to put a fan in the case as I have loads of 12V 90mm computer fans kicking around. I don't know how to power the fan from the circuit I have though. I only need a couple of watts so even if the way I get the power is really wasteful I don't suppose it'll matter.

[Jonathan]

Does the output voltage change linearly with the input voltage?

For the purposes of this discussion, yes near enough.

I'm thinking 235V into the 25+25 transformer would produce about 51V out (wired in series). Wish I'd put together a spreadsheet for the calculations now! Considering there's a 2.2V drop in the rectifier I think the 24+24 is probably the best choice in that case - I suppose I've just got to go away and crunch the numbers again.

If the transformer is wound for 230V to 25+25, then putting 235V in will indeed get 235/230*50=51V, however the tolerance is +10%, so 50*1.1=55V will be the limit. The rectifier drop is 1.1V, not 2.2V. The reason is (for the purposes of this discussion!) you only ever have 2 diodes conducting at a time and the forward voltage drop of two diodes will be about 1.1V. That means with 55VAC in to the rectifier, you can expect 55*sqrt(2)-1.1= 76.7V peak.

[m_c]

sqrt(2) essentially gives you the average value of a sine wave, and isn't electric specific.
It's officially the Root Mean Square equation for a sine wave - Root mean square - Wikipedia, the free encyclopedia

[Jonathan]

sqrt(2) essentially gives you the average value of a sine wave, and isn't electric specific.
It's officially the Root Mean Square equation for a sine wave - Root mean square - Wikipedia, the free encyclopedia

No, the average value of a sine wave, i.e. sin(x), is precisely zero. The RMS value is sqrt(2).

[m_c]

No, the average value of a sine wave, i.e. sin(x), is precisely zero. The RMS value is sqrt(2).

If you want to be such a pedant, a sine wave doesn't necessarily have an average of zero. It can be offset from zero. You've just assumed that all sine waves are drawn with zero as the base line.

As I'm sure you're aware, I meant that the average value for 90degree either side of a sine wave peak or trough i.e a 180degree section, is sqrt(2).

[Wobblycogs]

Thank you both, some things that we nagging me are a lot clearer now. I misunderstood the spec sheet for the rectifier and assumed it was 1.1V drop per diode in use.

I'm going to remove the relay before the bleed resistor but, from an academic point of view, how would you safely incorporate the relay? Whatever is supplying power to the coil would have to be completely independent of the capacitors. If I understand correctly current can't flow backwards through the rectifier so presumably power for the coil could be take from the between the secondary and rectifier.

There's just one thing I don't think I properly understand... why is the transformer 50*12=600W (or should that really be 600VA) not the 70*12 I calculated? I have a feeling this is an AC power vs DC power thing but that's right at the limit of my understanding of circuits.

[m_c]

Use a relay with a 240VAC coil and a set of NC contacts, and wire the bleed resistor through the NC contacts.
When 240 is applied, the contacts open, then when it's removed they shut and connect the bleed resistor.

[Jonathan]

If you want to be such a pedant, a sine wave doesn't necessarily have an average of zero. It can be offset from zero. You've just assumed that all sine waves are drawn with zero as the base line.

That's just changing your reference point - if it's got a DC offset then it's a sine wave with an offset, not a sine wave. I thought it was misleading to just say 'squrt(2) is essentially just the average value of a sine wave'.

As I'm sure you're aware, I meant that the average value for 90degree either side of a sine wave peak or trough i.e a 180degree section, is sqrt(2).

No that's a new one to me, as it's incorrect!
As I'm sure you're aware, to find the average value of a function you integrate it over the section in question, and divide by the 'length' of that section. So lets do it:
Let y=sin(x)
We want '90degree[sic] either side of a sine wave peak', so in radians the limits of out integral are 0 and pi and we integrate over pi:
Average=1/pi*Integral(sin(x)dx) from 0 to pi.
Average=1/pi*(-cos(pi)--cos(0))
Average=2/pi

2/pi is not equal to sqrt(2).

To get the correct RMS value of a function, you first square the function, then find the mean of the function (using the above method) then find the square root of it. Hence the name - root mean square.

[m_c]

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