Re: Stepper Power Supply. Question about Voltage

--- In CAD_CAM_EDM_DRO@y..., "kaynrc" <tom.kay@n...> wrote:

> Hello
>
> I am just getting around to selecting some transformers to build
> sparate little power supplies for Dan Mauch's 5 amp boards, and

each

> transformer is rated at 26 VAC, 5 amps. I plugged one in, and read

it

> at 29.2 VAC, which is fine by me.
>
> Then I connected the bridge rectifier and got a surprise, which is

a

> reading of 26 VDC. I had not yet hooked up the 10,000uf, 50 VDC
> capacitor. Because the VAC is supposed to be higher than the VAC

(by

> a factor of 1.41) I expected the DC voltage to be higher, around 35
> to 39 VDC. Again, it might be important that I did not attach the

big

> cap yet.
>
> So, what's going on? Will I notice a voltage increase when I add

the

> capacitor? Why is my DC voltage so low just taken right off the
> rectifier, since it's only suppose to consume around 1.7V? Is the
> transformer perhaps too small to supply proper voltage? And last,
> these 521 oz-in motors that I got from Camtronics (and the circuit
> boards) can run up to 40VDC, so am I simply going to operating

slower

> with somewhere around 25 to 30 VDC? Noticably slower?
>
> Thanks, Tom.

There's already been a post telling you what happens when you connect
the capacitor. The same should happen to you, and it's all normal.
Here's why.

I assume you measured the transformer with a DVM (or VOM) on the AC
scale, and the rectifier output on the DC scale. The AC is a pure
sine wave, and would give a true RMS (root mean square) reading on
most meters (some of them measure the peak and divide by 1.414, which
is accurate only for a pure sine wave. Others are called "true RMS"
meters). The DC range of the meter will average any voltage it sees
that isn't constant - if the variation is fast you won't see it
following the changes, just the "net change" over some period of time.
If you look at it with an oscilloscope, which has a much faster
response, you will see the changes as they happen. A typical DVM will
average over a fraction of a second, which is still many cycles of the
AC. The AC waveform, before rectification, is a sine wave with
amplitude (peak) equal to the RMS value multiplied by 1.414. The
average value of such a wave is zero, because the positive and
negative half-cycles cancel out. If you put your DC meter on the
transformer output (or the AC line, come to that), you will (should)
measure zero volts. However, the rectifier changes the polarity of
one of the half-cycles, so both half cycles "move" the DC meter in the
same direction, and you will now see a non-zero reading which is the
average of the rectified sine wave. Now we need some math. Without
going into the intricacies, we calculate the average in essentially
the same way we would do a statistical average, that is by adding up
the values of each sample and dividing by the number of samples. The
"samples" here are infinitesimal slivers of time, and the values are
the voltage we would measure during each sliver of time.
Mathematically, we integrate the function and divide by the period of
time over which we integrated. This is no different than finding the
average of any mathematical function. The sine function is incredibly
interesting, in that it is the it's integral (and derivative too) is
the same shape as the function, just displaced in time (it is called a
cosine function). But in any event, the integral of the sine function
in a quarter cycle (90 degrees or pi/2 radians)is numerically the same
as the amplitude (peak value) of the wave, and so the average is the
amplitude divided by the period of a quarter cycle, pi/2. Since the
amplitude is 1.414 times the RMS, the average is thus 1.414 /( pi / 2)
* Vrms, or 0.900 * Vrms (to 3 dp). In your case, you measured 29.2
RMS, which should have an average voltage of 29.2 * 0.9 = 26.28V.
This is what the DC meter should read.

You were close. The 0.28 volts is probably drop in the rectifier
diodes, or maybe the AC reading wasn't that accurate. The diodes will
drop some volts, but not a lot at the miniscule current drawn by the
average DVM. At "real" currents, the drop will be more like the 0.6,
0.7 V per diode everyone knows about, but it isn't truly constant.

BTW, the RMS value is supposed to be the voltage of a DC supply with
the same power as the AC wave. This is a little complex
mathematically, but it is what is useful. You can't use the
mathematical average, because as we saw that is zero over a complete
cycle, which is clearly not the power being delivered into any real
load. Interestingly, RMS calculation is like rectifying AC to make
DC, in that the square of a negative number (which is part of the
calculation) is positive.

Final point. Measuring AC is tricky, because most practical meters
are actually DC devices and barf on AC. The DVM does about the same
thing as your power supply also does to prevent the steppers barfing
on AC. It rectifies the AC, then uses a lighly loaded capacitor to
catch the peak value, which it then measures and divides by 1.414 in
order to display the RMS value. As I said earlier, the answer
obtained by this method will be a correct RMS value only if the sine
wave is pure. If it is half-wave rectified, or chopped by a triac or
MOSFET, or even pure DC, it will also not be correct. So you should
not measure your stepper current consumption with such an AC meter.
Probably, you will use a DC meter between the cap and the motors,
which will give you the true average current and voltage, but you
ought to be able to measure it as an AC current through the
transformer secondary. If a meter is a true RMS meter, it should read
the same as a DC meter on DC, and the same before and after the
rectifier. If yours reads 1.4 times the DC meter on pure DC, it is
peak-reading meter, so don't use it on non-pure sine waves. (Remember
that the RMS value is supposed to be the same as an equivalent-power
supplying DC voltage, so obviously the RMS value of a DC voltage is
just the average of the DC voltage.) Another test is to repeat the
rectifier (without the caps) output measurement, using the AC range.
It should be the same, but it will not be if it is peak-reading. It
will show you the 40-odd volts you were expecting, because it has its
own smooting cap inside.

Anyhow, this is probably boring you to death, but I think it will help
to know what is going on when you use an instrument. At a minimum, it
confuses, but it may be dangerous. Don't use a DC meter to measure
the wall socket then assume it is dead!

Cheers,
Jed nf1z