Linear Power Supply For Modular Synth Application.

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Transformer voltage rating.

The primary's voltage rating must be chosen according to the mains voltage of the country in which the power supply will operate.
The secondary's voltage rating is determined by:
Vsec=((Vout+Vdr+Vr+Vd)/CF)*(100/(100-imvd))
Where:
(Vsec) is the AC voltage at the secondary.
(Vout) is the voltage at regulator's output.
(Vdr) is the regulator's dropout voltage.
(Vr) is the ripple voltage in volts peak to peak.
(Vd) is the rectifier's forward voltage drop.
(CF) is the crest factor of the wave in the secondary.
(imvd) is the immunity to the mains voltage drop in percentage to the nominal mains voltage.

Let's see an example. Suppose that we want a transformer for our PS1 at +/-15V, a maximum current output of 0.5A and immunity to mains voltage drop up to 10%. What secondaries' voltage rating do we need?
Let's focus in one secondary:
(Vout) The voltage output is 15V, then:
Vout=15V
(Vdr) According to the “dropout vs junction temperature” plot in the LM317 data-sheet, the dropout voltage at 0.5A is below 2V. However, the LM317 data-sheet from Texas Instruments specifies as "recommended operation conditions" a minimum of 3V. We will follow this recommendation in this example, so:
Vdr=3V
(Vr) The ripple voltage is Vr=(0.007*Iout)/C. For this example we use a 4700μF capacitor with 20% of tolerance. Then, the worst case value is its minimum value, which is 4700-20*47=3760μF.
Vr=(0.007*Iout)/C
Vr=0.007s*0.5A/0.00376F
Vr=0.9Vp-p
(Vd) The forward voltage of the rectifier is 1V. You can see it here: http://www.fairchildsemi.com/ds/GB/GBU4D.pdf
Vd=1V
(CF) The Crest factor as we have said in a previous post is 1.3
CF=1.3
(imvd) And the immunity to mains voltage drop is 10%.
imvd=10%
We fit these data into our formula and Vsec is:
Vsec=((15+3+0.9+1)/1.3)*(100/(100-10))≈17V

So according to our calculations the minimum voltage rating we need is 17V.
We choose the next higher standard value which is 18V. This calculation was for one secondary. Thus, for two secondaries, the voltage rating we choose is 36VCT.

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Rectifier diode voltage rating. Peak Reverse Voltage.

To calculate the voltage rating of the rectifier diode, or Peak Reverse Voltage, we must consider the maximum peak voltage across the diode that can happen under the worst case conditions. The two worst case conditions that we consider are:
1.When the mains voltage rises and,
2.When the power supply has no load because the voltage rises due to the voltage regulation of the transformer.

Let's put an example, suppose that we have a 50VA 36VCT toroidal transformer which has a typical voltage regulation of 13% and suppose that the mains voltage rises up to 10% over its nominal value.
Let's focus now on one secondary. The voltage at the secondary is 18VAC. This specification refers to the nominal voltage at the primary. So, if the voltage at the primary rises a 10%, this will also happen at the secondary. Therefore:
18+18*0.1=19.8VAC.
As for the voltage regulation of the transformer, the voltage will rise a 13% when there isn't any module connected to the power supply. Thus:
19.8+19.8*0.13≈22.4VAC
Now we can calculate the peak voltage. When the power supply has no load, the wave, theoretically, is a sine whose crest factor is ≈1.414, thus:
Vpeak=22.4*1.414
Vpeak≈31.6V.
We ignore the forward-voltage of the rectifier diode, so:
For a split rectifier configuration like the PS1, the maximum peak reverse voltage across the rectifier diode would be 31.6*2=63.2V
So, the peak reverse voltage rating must be equal or higher than 63.2V
For a full wave or dual full wave rectifier configuration, the maximum peak reverse voltage across the rectifier diode would be 31.6V
So, the peak reverse voltage rating must be equal or higher than 31.6V

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Now we could add a derating factor. But once we have calculated the voltage rating considering the worst case conditions, it could seem redundant to apply a derating factor over it. This is up to you. For this example we apply it when the voltage at the secondaries is at its nominal value. The derating factor for a rectifier diode voltage rating is 2*.
Following the previous example, for a split rectifier configuration it would be:
Vpr=2*dF*CF*Vsec
Vpr=2*2*1.414*18V
Vpr≈102V
Where:
(Vpr) is the Peak Reverse Voltage rating.
(dF) is the derating factor.
(CF) is the crest factor of the sine wave.
So, the peak reverse voltage rating must be equal or higher than 102V.
For a full-wave configuration it would be:
Vpr=dF*CF*Vsec
Vpr=2*1.414*18V
Vpr≈51V
So, the peak reverse voltage rating must be equal or higher than 51V.

Capacitor voltage rating

To calculate the voltage rating of the filter capacitors we must consider the same worst case situations as for the rectifier.
Following the previous example, if the voltage at the primary rises a 10%, the voltage at the secondary would be:
18+18*0.1=19.8VAC.
If there is no load, due the transformer's voltage regulation, the voltage at the secondary would rise 13%:
19.8+19.8*0.13≈22.4VAC
Thus, the peak voltage would be:
Vpeak=22.4*1.414
Vpeak≈31.6V.
We choose the next higher standard value which is 35V.

We could also add a derating factor. The derating factor for a capacitor voltage rating is 1.3*.
So, the voltage rating of the capacitor would be:
Vc=dF*CF*Vsec
Vc=1.3*1.414*18≈31V
We choose the next higher standard value which is also 35V.

Regulator voltage rating

Consider the same worst case conditions for the regulator voltage rating.

*About the derating factors

The derating factors I'm suggesting are from sars.org.uk:
http://www.sars.org.uk/
Document

OK, and with this we covered the voltage at the power supply. We have seen how it behaves in different parts of the power supply and how to calculate the voltage ratings of the components.
I will write (I wish soon) the third part, the Power Dissipation, that is the most important and that I consider the most fascinating part.

[capicoso]

Hey, you answered me on other threads, but i'll ask here better.
So if i want a 36VCT that's capable of give 1A real, the VA should be higher than 64 right? I really don't know how much i'll need, right now i just have 3 modules... And this cabinet will be 16U, but'll make others of the same size( so i can transport them easier than a huge cabinet...) What would you suggest, using a big one like 80VA or more for two cabinets, or smaller 54VA one for each cabinet... ? I i don't think this first cabinet will need more than 600mA so with a 54VA 36VCT 1A or 1.5A? i'm fine i think...

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[capicoso]

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