need optimum transformer specs for GGG ultra clean!

[sjaltenb]

hey, i know i have asked before, but i need the OPTIMUM input voltage (ac) for the GGG ultra clean power supply.

The power supplies will only be powering a single pedal (think spyder) @ 100ma

Only one power supply will need to put out 500ma, so i will probably use the recommended 24vac for that one. I will order the exact resistor values indicated, as I know that will be critical.

I need to know the voltage and current, (im guessing the current is 100ma), and I am hoping that 12vac is enough. Here is the schematic.



I need to know ASAP as I am having a transformer built to my exact specs

[brett]

Hi
you need to use a 24V AC transformer for any supply using this design, whether it's for 5mA or 500mA.
Note that for 500mA you'll need a heatsink on the LM317.
cheers
PS IMO this circuit is a bit of overkill.  Most of the "work" is being done by the rectifying diodes and that 470uF cap.  There's also technical advantages of using two or three filter caps in parallel, instead of one big one.  An LM7809 (with an Si diode on the ref leg if you want 9.7V), the same rectifier, 3 x 220uF filter caps, and the same diode protection will probably be as good (maybe better) be adequate for 99.99% of applications.  I'm sure that RG Keen will have something of the sort (ie elegant) over at geofex.com.

[slacker]

Why do you need a 24volt transformer?
If you full wave rectify it like shown in the schematic you'll get 36volts DC at the input of the regulator, that seems completely unnecessary if you only want 9 volts out.
A 9 or 12 volt transformer should work fine, or am I missing something?

[R.G.]

The "ultra clean power supply" is "ultra clean" in the same way that "Guarantee National Bank" is either guaranteed or national - it's a name, like "Fred's Power Supply".  This is not to say that it's a bad power supply, just that "ultra clean" is a bit enthusiastic. It's a workmanlike power supply done with a three terminal regulator, matching the chip maker's application notes. For info on how this is done, see http://www.national.com/ds/LM/LM117.pdf.

However, it looks to me like this was adapted from a variable power supply. 24Vac is way more than this design needs; that 317 is going to get much hotter and need a much better heatsink than would otherwise be needed.

Let's take a look at a more moderate design.
How much DC does the 317 need at its INPUT to produce a good output? Looking at the datasheet/app note, the 317 needs between 1.75V and 2.5V (depending on the output current) more DC at its input than at its output. Less than that and it cannot regulate. Let's play safe and say we always need 2.5V. So with a 9.35V output, we need a minimum of 11.85V at the input. That's the instantaneous minimum. With a capacitor filter, that's the minimum of the filter ripple voltage.

We know the max input voltage has to be less than 40Vdc because the 317 dies if it's more than that. That's at the peak on the DC filter cap at low current and high AC power line. The power line is often 10% high, so let's say that it's 40Vdc at the filter cap, plus 1.4V for diode drops in the bridge, total of 40.14V peak before the rectifiers. That' 40.14/1.414 = 28.38v rms. So the nominal transformer voltage is 10% less, or 25.8V. Hey - the designer just whomped on the maximum AC voltage for the transformer he could use and not kill the regulator. We can do a better (i.e. less power wasted) job by using a lower AC voltage transformer.

Back to the minimum DC input. We had 11.85Vdc at the bottom of the filter DC ripple. Since filter DC is a characteristic of how big your filter cap is, let's pick a ripple for a cap we like and then figure the transformer voltage. You have to know your DC output current to do this because ripple is linear with current. If your max current is 500ma, let's see what a 1000uF cap does. In a full wave bridge design, the capacitor gets charged every half-cycle, or every 8.67mS for 60Hz and 10mS for 50Hz. Going with 50Hz for the moment, and 500ma as you mentioned, the change in voltage on a 1000uF cap is according to the capacitor equation: I = C * dv/dt. Rearranging to find dv, we get
dv  = I*dt/C
or in this case,
dV = 0.5*10mS/1000uF = 0.2V
which is nicely small. Power supply designers often use 1000uF per ampere as a target. Notice that doubling the capacitor to 2000uF halves the ripple voltage, and halving the cap doubles the ripple voltage to 0.4V. But let's leave it at 0.2.

So on top of the 11.85Vminimum, we need to add 0.2V of ripple from the capacitor, for 12.05V peak at the cap. There are two diodes conducting to make this peak, so we need to add another 1.4V to this for 13.45V total.

In a perfect world we could just do the math for the rms of the transformer new, but we can't do that because the transformer voltage sags because of resistance in the windings and AC power line sag. AC power line sag is often 10%, sometimes worse. We'll use 10% because this is a non-critical application. The transformer sags too, because of the internal resistances. You may find this quoted as "regulation" on transformer data sheets, or you may not. In any case it's caused by the peak currents into the rectifiers dropping voltage through the resistance of the primary and secondary wire. Small transformers often have -15% regulation or worse. 8-10% is common for transformers as big as your fist, getting smaller (i.e. tighter voltage, less sag) as the transformers get bigger.

As an approximation, we'll take the perfect world minimum of 13.45V peak, and convert that to an RMS voltage of 13.45/1.414 = 9.51Vac. We then up that by the expected losses that we don't know, but are estimating: the 10% power line sag and 15% transformer sag to get a target voltage of 12.0327Vac.

It is likely that a small 12Vac will be the minimum you can use to get 9.35Vdc out of that regulator circuit with a 1000uF filter cap at 500ma.

One thing that is not all that widely known. the heating effect of a full wave bridge rectifier is hard on the transformers that drive them. To the transformer, the peak currents heat much larger than a sine wave current. This is a factor of 1.6 to 1.8 times the DC output. So you're going to need a transformer rated at 0.8 to 0.9A minimum.

In the end, a 12Vac 1A transformer, which is almost as common as dirt in the trade, will be just about perfect for your 9.35V 0.5A output in that circuit.

That is not to say that this is the only solution, nor is it the optimum one. Any solution that works and is cost effective is "optimum" unless you know other constraints.

I believe all of this is in "Power Supplies Basics" at GEO. I keep harping on people to go read all of GEO.

[George Giblet]

I'm with RG on the calculations.  One thing I can add is the current waveform from the transformer has large peaks which complicates the calculations.  However, this complication is undone to some extent by the 1.6 to 1.8 derating factor.  Also I'd allow 1V or more for the each diode since the diode voltage drop is high at the peak currents.

The optimum is not a simple thing to answer.  You can use a larger, lower voltage transformer to reduce heat dissipation in the regulator.  Or you can use a smaller higher voltage transformer and get away with smaller caps.  The large transformer might require a larger box which will cost more.  It all depends on what you want to optimize.  And what if you can get a big transformer on special, does that factor into optimum.

The 25Vac transformer is just crazy.

>  In the end, a 12Vac 1A transformer,

At the end of the day that says is all!

[sjaltenb]

Ok, i just learned a LOT!

Thank ya'll very very much!

Right now it looks like my transformer will have the following windings:

9x12vac @ 100ma (these will drive individual power supplies, which will in turn power SINGLE effects like fuzz, compressors, etc. nothing over a total rating of 20ma)
2x12vac@200ma (these will go to power supplies that will feed 3fx each (phaser, microvibe and tremolo, and the other feeds 3 EQs)
2x12vac @200ma (powering two tube drivers)
1x15vac @200ma (powering electric mistress)
1x12vac @1a(driving a power supply that will feed 4 external FX outputs)

Thank you very much for ya'lls help. I just dont want to order the custom transformer and it be the wrong type, etc. I am still working on a way to remove the power sections from the tube drivers and electric mistress so i can use better, outboard power supplies for those.