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 Forum index » DIY Hardware and Software
Linear Power Supply For Modular Synth Application.
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Osal



Joined: Aug 16, 2011
Posts: 145
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PostPosted: Thu Jan 17, 2013 2:02 pm    Post subject: Reply with quote  Mark this post and the followings unread

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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Osal



Joined: Aug 16, 2011
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PostPosted: Mon Jan 21, 2013 11:20 am    Post subject: Reply with quote  Mark this post and the followings unread

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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Osal



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PostPosted: Fri Jan 25, 2013 10:42 am    Post subject: Reply with quote  Mark this post and the followings unread

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.

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capicoso



Joined: Nov 19, 2012
Posts: 127
Location: Argentina

PostPosted: Thu Mar 14, 2013 12:24 pm    Post subject: Reply with quote  Mark this post and the followings unread

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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Osal



Joined: Aug 16, 2011
Posts: 145
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PostPosted: Thu Mar 14, 2013 1:14 pm    Post subject: Reply with quote  Mark this post and the followings unread

Quote:
So if i want a 36VCT that's capable of give 1A real, the VA should be higher than 64 right?
Yes, the minimum advised ratio Isec/Iout is 1.8

Quote:
I really don't know how much I'll need, right now i just have 3 modules...

You could do an average of the modules power consumption from this document in modularsynthesis.com :
http://www.modularsynthesis.com/modules/parts/DJB-Power.pdf

Quote:
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...

I think that you are referring to a power supply using LM317/337 regulators. Using the standard regulators LM317/LM337 the bottleneck of the design is the power dissipation, if it is passive. Doesn't matter the current rating of the regulator or how big is the transformer.
I would suggest to use small power supplies up to 0.8A instead of big ones. This is also a way to start and after you will know better what you need. When you finish the first cabinet you will be able to do current measurements, etc...

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capicoso



Joined: Nov 19, 2012
Posts: 127
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PostPosted: Thu Mar 14, 2013 1:25 pm    Post subject: Reply with quote  Mark this post and the followings unread

Osal wrote:
Quote:
So if i want a 36VCT that's capable of give 1A real, the VA should be higher than 64 right?
Yes, the minimum advised ratio Isec/Iout is 1.8

Quote:
I really don't know how much I'll need, right now i just have 3 modules...

You could do an average of the modules power consumption from this document in modularsynthesis.com :
http://www.modularsynthesis.com/modules/parts/DJB-Power.pdf

Quote:
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...

I think that you are referring to a power supply using LM317/337 regulators. Using the standard regulators LM317/LM337 the bottleneck of the design is the power dissipation, if it is passive. Doesn't matter the current rating of the regulator or how big is the transformer.
I would suggest to use small power supplies up to 0.8A instead of big ones. This is also a way to start and after you will know better what you need. When you finish the first cabinet you will be able to do current measurements, etc...

thanks
yes i'll use lm317/337 regulators, the mfos adj power supply to be more specific... I'll use a 54VA one then. 54VA 36VCT so i can use 0,8A... so i just ask for a 54VA 36VCT transformer right? What i'm not sure if what amperage should i ask for, 1A or 1.5A? that's the thing i didn't understand well, if i ask for a 54VA 36vct, the secondaries current is given by that relation 54/36... 1.5A right... so specifing the sec current is a bit redundant right?
thanks again
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Osal



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PostPosted: Thu Mar 14, 2013 2:01 pm    Post subject: Reply with quote  Mark this post and the followings unread

capicoso wrote:
if i ask for a 54VA 36vct, the secondaries current is given by that relation 54/36... 1.5A right...
Yes this is correct.
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feggster



Joined: Sep 12, 2011
Posts: 51
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PostPosted: Sun Jul 28, 2013 8:03 am    Post subject: Reply with quote  Mark this post and the followings unread

copied from other thread..

I'm in the process of building a power supply for a small amount of modules (nicholas woolaston's simple vco's etc)

my transformer : http://uk.farnell.com/pro-power/ctfc20-18/transformer-20va-2-x-18v/dp/1780851

20VA, 2 X 18V
Primary Voltages: 2 x 115V
Secondary Voltages: 2 x 18V
Current Rating: 556mA
Power Rating: 20VA
Input Voltage: 115V
Mounting Type: Chassis
Output Voltage: 18V
Secondary Current Nom: 556mA
Transformer Type: Isolation

I used your calculations for the fuse current and ended up with a figure of 129ma.... so I'm guessing I use a 125ma slo blow fuse.
I noticed in the calc for the fuse there is a 'difference' in actual values..
Quote:
the current rating of the fuse (Ifuse) is, as we said:
Ifuse=Ip*1.5
Ifuse=0.218*1.5= 0.327A
In this case a 315mA blow slow fuse will do the job.

whats the safe margin for this?
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Osal



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PostPosted: Sun Jul 28, 2013 3:21 pm    Post subject: Reply with quote  Mark this post and the followings unread

Quote:
whats the safe margin for this?


Hey feggster thanks for post it here.

The reference of this is "The Art of Electronics" by Horowitz and Hill, where says that the current rating must be higher than 1.5 times the current on the primary.

The correct way to avoid any confusion would be to choose the very next higher standard value to the result of the calculus.

However, for that example, I considered that 315mA was close enough to the result of the calculus. There is not a "safe margin" neither it is a very critical value.

A downsized fuse will blow up one day when you turn on the power supply. A very oversized fuse could be very dangerous, if it not blows up in a short circuit condition. A little bit oversized fuse (depending on the limiting current of the regulator), could not blow in case of a short circuit after the regulators and this could stress the transformer (depending on its size).

For your application 125mA or 160mA is correct.

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feggster



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PostPosted: Mon Jul 29, 2013 4:45 am    Post subject: Reply with quote  Mark this post and the followings unread

thanks, that makes perfect sense I will stick with 125ma fuses for my application, thanks again for the work you have put into all this.
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Osal



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PostPosted: Thu Jan 30, 2014 8:25 am    Post subject: Reply with quote  Mark this post and the followings unread

The ripple at regulator's input (II): The size of the capacitor.


We have already discussed about the ripple voltage in the post "The ripple at regulator's input" in the previous page and know that there is a linear relation between the ripple voltage and the filter capacitor value.
Vr=(0.007*Iout)/C
But... what is the optimal ripple voltage at the regulator's input? and, what is the capacitance required to achieve it?
A small ripple voltage at the regulator's input (versus one large) implies the following:
-A smaller ripple at the regulator's output. (Good)
-Higher immunity to mains voltage drops. (Good)
-Higher voltage at the regulator's input, thus more power to dissipate. (Not good)
-A larger capacitor.
----A larger capacitor use to have higher current ratings (Good)
----We can expect a larger size and higher cost. Although I don't consider these in our context, where we may build a few power supplies and that we assemble them in big cases.

Let's see, through an example, how much the ripple affects to those parameters. Let's pick up several capacitor values and estimate several parameters and write it down in a table
The example is again our PS1 assuming the following:
General:
Voltage Output =15V
Current Output=0.8A
Forward Rectifier Voltage Drop=1V
Crest Factor of the wave at the secondary=1.3
Voltage across one secondary=18VAC.
To estimate the power dissipation:
A high mains fluctuation of 10% over its nominal value.
A capacitance 20% over the nominal value of the capacitor due its tolerance.
To estimate the maximum thermal resistance required for the heat-sink:
Ambient Temperature=40C
Junction to Case Thermal Resistance=5C/W
Case to Heat-Sink Thermal Resistance=1C/W
To estimate the immunity to a mains voltage drop:
A capacitance 20% below the nominal value of the capacitor due its tolerance.

Looking at the attached table:
From the 2200μF to the 10000μF rows:
Look at the immunity columns. There is a great improvement (From -5% to -16%) or (-9% to -20%)
However, if we look at the column of the heat-sink's thermal resistance required, it decreases 1C/W.
From the 10000μF and the 22000μF rows:
There is not remarkable variations in the immunity nor in the thermal resistance.

Consider, also, that each time that the capacitance is doubled, the ripple voltage is halved.
Then, notice that once the estimated maximum ripple input is smaller than 1Vp-p, to double the capacitance does not produce considerable variations.

So what?

One rule of thumb could be to calculate the needed capacitance for a maximum ripple voltage of 1 Volt p-p. (considering capacitor tolerances)
C=(0.007*Iout)

But there is a more easy rule of thumb:
Give at least 10000μF (or the very next higher standard value) for each 1 Ampere of power supply output.
Examples:
For a 0.2A power supply give at least 2200μF for the filter capacitors.
For a 0.5A power supply give at least 5600μF for the filter capacitors.
For a 0.8A power supply give at least 8600μF for the filter capacitors.
etc..


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