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Linear Power Supply For Modular Synth Application.
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Osal



Joined: Aug 16, 2011
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PostPosted: Wed Jan 18, 2012 11:32 am    Post subject:  Linear Power Supply For Modular Synth Application.
Subject description: And other audio applications.
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Hello,

I think the Power Supply together with the Power Distribution and the Housing is the important part of a modular synthesizer.

Last spring I started studying it, and I have run several tests, including measurements of temperature of transformers and TO-220 packages, measurements of currents and voltages in different parts of the power supply, under different situations and configurations, different transformer sizes, different capacitor sizes, different rectifiers, etc...

So, this is intended to be a guide in order to help understand and build a power supply for a modular synthesizer.

I would like to start with a few words concerning safety: Dealing with mains electricity is dangerous and could result in electric-shock including death. You can only build a power supply if you are aware of this and if you are actually qualified in order to do it. Basic things, like wearing shoes with insulating outsoles or putting the free hand in the pocket when working with the other hand close to mains electricity to avoid currents passing through the chest, must be known.
Normally you always work in the power supply having it disconnected from the mains electricity. If you need to work with the power supply energized in order to test it, all cables or parts with mains electricity must be correctly insulated. The housing case must be connected to earth if it is conductive.

Use IEC connectors like (farnell) JR-101-1-FRSG-03 and crimp connectors like 3625515(farnell code) and properly insulating covers like 3625515(farnell code) for a neat and safe result. This kind of connectors allow you to disconnect the cable physically when you are wiring the mains electricity. In addition, the cable must be removed in order to place or replace the fuse. The insulated crimp connectors make sure that all parts with mains electricity are correctly insulated.

Another danger regarding power supplies is the risk of fire. Shorting accidentally the secondaries of a transformer without fuse or with an overrated fuse can result in fire. Especially if the power supply is placed in a wood case. By the way, flammable cases must be avoided.

­So take in account all of the above and be careful; anything that happens is under your responsibility.

The tests and experiments were done mainly in a bipolar linear power supply using regulators LM317/LM337, more specifically in a PS1.
http://www.electronic-sea.net/ps1.html
However, conclusions are also applicable to full-wave bridge rectifier followed by any linear regulator power supplies.

The comments that follow are focused on the aspects that I found relevant and on the common misunderstandings regarding power supplies that you can read over the net. Any question is welcome.

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Last edited by Osal on Wed Feb 08, 2012 9:34 am; edited 4 times in total
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Osal



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PostPosted: Wed Jan 18, 2012 2:39 pm    Post subject: Reply with quote  Mark this post and the followings unread

Power Supply Current Rating

The current rating of a power supply is determined by current ratings of all its components. These are easy to calculate except from the heat-sink. Passive heat-sink for regulators must be approached with care. We will discuss this later.

Lets start by looking at each component at a time:

Fuse

The fuse must be slow-blow type, also called anti-surge type, so as to avoid that it blows up when turning on the power supply due to the inrush current.
To calculate the current rating of the fuse you can use as a rule of thumb, multiplying the rated current in the transformer's primary by 1.5.
How do we know which is the current rating of a transformer's primary?
Let's take as an example a 50VA 36VCT transformer and 230VAC of mains voltage.
From these data we know that:
The voltage in the transformer's primary (Vp) is 230VAC.
The voltage in the transformer’s secondary (Vs) is 36VAC.
We can calculate the current rating of the secondary (Is) as follows:
50=V*A
50=36*A
A=50/36=~1.39
Is =1.39A
Now, we know that the ratio of the current in the primary to the current in the secondary is inversely proportional to the ratio of the voltage in the primary to the voltage in the secondary.
So the current rating of the primary (Ip) is:
Ip/Is=Vs/Vp
Ip=(Vs/Vp)*Is
Ip=(36/230)*1.39
Ip=0.218A
Finaly 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.

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Last edited by Osal on Wed Feb 08, 2012 8:31 am; edited 1 time in total
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Osal



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PostPosted: Wed Jan 18, 2012 3:52 pm    Post subject: Reply with quote  Mark this post and the followings unread

Transformer.

Sometimes, there is the misunderstanding that the current in the transformer's secondaries is the same as the current in the regulators' output. This is not correct.
The bridge rectifier splits the alternating current in the secondaries in two components, one alternating and one direct. The capacitor allows the alternating current to pass (come and go) through itself, from the rail to the ground, and from the ground to the rail, but it doesn’t allow the passing of the direct current, which goes to the ground through the regulators and the load.

Ratio Isec/Iout

Calculating the current rating of the transformer needed for a determined output current is complex. So I have done measurements for different rating toroidal transformers: 15VA, 25VA, 30VA, 40VA, 50VA, 60VA, 80VA and 100VA. The results showed that the ratio of the current in the secondaries to the current in the regulators output (when the transformer outputs its current rating and when it is at its rise temperature) is approximately 1.6. So Isec/Iout=1.6
However, several transformer manufacturers advise us to use the ratio Isec/Iout=1.8. I stick to this manufacturers' advise. (This is true for full wave rectifiers and dual complementary supplies; in other types of power supplies the ratio could be different)

For example, for a 50VA 36VCT transformer, what is the maximum current we can obtain in the power supply's output (Iout) ?
First we calculate the current rating in the secondaries (Isec):
50=V*A
50/36=A=1.39
Isec=1.39A
Then, we know that Isec/Iout=1.8
1.39/Iout=1.8
1.39/1.8=Iout
Iout(max)=0.77A
So, for this transformer the maximum output current we can obtain is 0.77A.

Another example: For a power supply that must output 1A (Iout=1A) and a 36VCT transformer, what is the minimum VA rating for the transformer?
Isec/Iout=1.8
Isec=1.8*1
Isec=1.8A
Then,
VA=V*A
VA=36*1.8=64.8
So, to output 1A you need a transformer equal to or bigger than 64.8VA

Increasing ratio I sec/I out : Derating the transformer.

Derating the transformer implies the following:

1) The transformer works at a lower temperature. This is good for itself and for the other electronic components that are close to it. Lower temperatures increase life and reliability of electronic components and in some cases ensure better performance. This is a benefit.
2) The transformer increases its voltage in the secondaries due to the transformer regulation. This way you can achieve a power supply with more tolerance to mains voltage drops but also more heat to dissipate. For example, using a 36VCT 100VA transformer to output 1A, I measured and estimated 18.8VAC for each secondary instead of 18VAC.
4) The inrush current and the peak current increase due the lower resistance of the transformer windings. However this is not that important in these small power supplies, as we will see later.
5) A bigger transformer implies more weight, size and cost.

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Osal



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PostPosted: Thu Jan 19, 2012 3:47 am    Post subject: Reply with quote  Mark this post and the followings unread

Bridge Rectifier:

Maximum Average Forward Current Rating:
Full wave rectifiers conduct two of their four diodes at a time, so each element is one cycle on, one cycle off. This means that the maximum average forward current rating is Isec/2 for each element.

For example, if you have a power supply with a current of 1.8A flowing in its transformer's secondaries , this is the current that will also flow through the bridge rectifier. So, in each element will flow half of the current, that in this example would be 0.9A.

However, the cost of the rectifiers doesn't prevent us to use higher current rated ones. A good choice could be a GBU4A which can stand up to 4A for each element. In addition it has a very high peak forward surge current rating and the GBU package simplifies the PCB layout.

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Dougster



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PostPosted: Thu Jan 19, 2012 8:11 am    Post subject: Reply with quote  Mark this post and the followings unread

Not trying to rain on the parade, but those GBU parts are huge! At least compared to four 1N4001 diodes, plus they're considerably more expensive.

But if you really need 4A per diode instead of 1A...

Regards,
Doug

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Osal



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PostPosted: Thu Jan 19, 2012 9:11 am    Post subject: Reply with quote  Mark this post and the followings unread

Where I wrote GBU4A I meant GBU4B

Hello Doug,

You have GBU 4A packages from 0.34 British Pound at Farnell. This is not considerable to me.
In addition it is important its high inrush current rating. Yes, you can estimate it and choose a smaller rectifier accordingly, but why you want to do it?
Finally, simplifies the PCB layout.
These are the criteria I considered to choose this bridge rectifier. However you can consider other criteria and decide accordingly. I'm not saying use it. I'm trying to explain how the whole thing work. Decisions are up to you.

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Dougster



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PostPosted: Thu Jan 19, 2012 9:37 am    Post subject: Reply with quote  Mark this post and the followings unread

Osal wrote:
However you can have other criteria and decide accordingly.

Yes, there are many trade-offs. I think that is the problem! There are too many ways to make a power supply... Laughing

I bought a reel of 1N4004 diodes and I think they were less than two cents (US) each. That's part of the reason I use them everywhere. The trade-off I made was to simplify my parts stock... Wink

Regards,
Doug

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PostPosted: Thu Jan 19, 2012 9:49 am    Post subject: Reply with quote  Mark this post and the followings unread

Yes and this is to design, decide trade-offs Very Happy
Other important aspect is to derate electronic components. It adds a wide extra protection to the component, also component does work at less temperature, improving its reliability and life expectancy.
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PostPosted: Thu Jan 19, 2012 9:55 am    Post subject: Reply with quote  Mark this post and the followings unread

Hey Osal,

thanks for sharing. i will print out this thread as a condensed and thorough building guide! great value to me.

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Osal



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PostPosted: Fri Jan 20, 2012 9:30 am    Post subject: Reply with quote  Mark this post and the followings unread

You are welcome Fonik.

Filter Capacitors:Ripple Current Rating.
Ripple Current (Ir) is the alternating current that passes through the capacitors (don't confuse it with the ripple voltage).
You can calculate the ripple current knowing the current in the secondaries (Isec) and the current in the output (Iout).
We know that the sum of the ripple current (AC) and the current in the output (DC) equals to the current in the secondaries (AC).
In order to sum RMS AC current and DC current we use the following formula:
RMS AC+DC=sqrt(AC^2+DC^2)
For example for a power supply with Isec=1.8A and Iout=1A we would have:
Isec=sqrt(Ir^2+Iout^2)
1.8=sqrt(Ir^2+1^2)
1.8^2=(lr^2+1)
(1.8^2)-1=lr^2
sqrt((1.8^2)-1)=lr
lr=~1.5A
So, through the capacitors is flowing 1.5A. However we need to add a margin to this value. Ripple current generates heat inside the capacitor. Lower temperature ensures longer life and reliability of the capacitors.
Therefore, as a rule of thumb we can simplify the previous calculus with:
RMS Ripple current rating => Iout*2.5
The Capacitor's Ripple Current Rating is specified in manufacturer's data sheet.

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PostPosted: Sun Jan 29, 2012 7:56 am    Post subject: Reply with quote  Mark this post and the followings unread

Regulator's current rating

The regulator current rating is specified by the manufacturer in its data-sheet.
However, the regulator's current rating will be limited by the power to dissipate since, Pd=(Vin-Vout)*Iout+(Vin*Ig). It will be discussed later.


A word about Inrush current.

There are two kinds of inrush currents when turning on a power supply; one is the current needed to magnetize the transformer and the other is the current needed to fill the filter capacitors.
The inrush current will affect the components which precede the component that has caused it.
However the inrush current will be limited by the transformer's windings resistance. And as you will see this is not that important with the small transformers that we usually use.

Let's see which are the affected components:
The fuse is a slow-blow type, also called anti-surge, which is designed to stand the surge current.
The mains switch will see a maximum inrush current limited by the primary's resistance.
The bridge rectifier will see a maximum inrush current limited by the secondaries' resistance.

For example, for a 100VA 36VCT toroidal transformer, which has a primary with resistance of 18.7 Ω and whose secondaries in series sum up to a resistance of 0.7 Ω, we have:
Max Inrush primary current = 230/18.7=12.3A
Max inrush secondaries current = 36/0.7=51.4A
Consult the data-sheet of the switch and the bridge rectifier you are to use, to be sure that they stand this possible maximum inrush current.
For example, a standard IEC switch can stand up to 10A of continuous current, so it will easily stand a surge of 12.3A. Some switch's data-sheets include the inrush current rating.
A bridge rectifier, such as a GBU4B, can stand surges up to 150A.


OK, until here, I covered the current in a typical power supply used for modular synthesizer application.
I hope this will help you. Smile

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Paradigm X



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PostPosted: Mon Jan 30, 2012 7:24 am    Post subject: Reply with quote  Mark this post and the followings unread

Thank you Osal, for these very helpful notes. Doug also.

Ive read once but will read again.

Very Happy
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Osal



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PostPosted: Mon Jan 30, 2012 8:52 am    Post subject: Reply with quote  Mark this post and the followings unread

Paradigm X wrote:
Thank you Osal, for these very helpful notes. Doug also.

Ive read once but will read again.

Very Happy


Thanks Paradigm X , Someone is helping me to rewrite it in a better English, So I guess that the new version will be more understandable. If you have any question let me know.
More writings regarding power supplies to come.

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Dougster



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PostPosted: Tue Jan 31, 2012 6:20 pm    Post subject: Reply with quote  Mark this post and the followings unread

Osal wrote:
Someone is helping me to rewrite it in a better English, So I guess that the new version will be more understandable.

That's great! I was going to volunteer to help edit your writing, but I haven't had time to finish my own writings yet... Laughing

I asked about having a dedicated section here on E-M for tutorials and writeups, so maybe we can make this the best place on the Internet to learn about DIY audio electronics!

Regards,
Doug

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PostPosted: Wed Feb 01, 2012 5:47 am    Post subject: Reply with quote  Mark this post and the followings unread

Dougster wrote:
I asked about having a dedicated section here on E-M for tutorials and writeups, so maybe we can make this the best place on the Internet to learn about DIY audio electronics!
thumb up
Good idea Doug!

Thanks a lot Osal, helps a lot! Smile

Back to study

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Osal



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PostPosted: Wed Feb 08, 2012 9:21 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks Dougster, It would have been great. It is appreciated. However I have already corrected it and edited my previous post. I guess it is more understandable now.

You're welcome wmonk

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Osal



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PostPosted: Fri May 18, 2012 4:58 am    Post subject: Reply with quote  Mark this post and the followings unread

Until now, we talked about the current in the power supply. Specifically, about the current ratings of the components. Now we'll talk regarding the voltage, and later regarding the power dissipation.

But first, let's review some points:

Fuse Current rating, a more easy way to calculate it:
From the second post we know that:
Ifuse=Ip*1.5 and,
Ip=(Vs/Vp)*Is
This is the same as,
Ip=(Vs*Is)/Vp
so,
I fuse=(Vs*Is)/Vp*1.5
Notice that (Vs*Is) is actually the (VA) specification of the transformer. So,
I fuse=(VA)/Vp*1.5
It is very easy now! Let's say that you want to know the fuse current rating for a 80VA transformer to be used in Europe where the nominal mains voltage is 230VAC.
80/230*1.5=0.522A
So, choose a slow blow fuse with a standard value close to it.

Derating components:
Derating is to overrate a component to operate it at lower temperature thus increase its life expectancy and reliability. I mentioned it previously , but I think it deserved a post for itself because it is very relevant. I wanted to remark it.
Notice that I didn't quantify it for the rectifier, so consider I rectifier≥Iout*1.3

Summary:
Fuse current rating =(VA)/Vp*1.5
Transformer current rating ≥ Iout*1.8 or,
Transformer VA ≥ Iout*1.8*Vs
Bridge rectifier current rating ≥ Iout*1.3
Capacitor Ripple current rating ≥ Iout*2.5
Regulator current rating is specified by the manufacturer but depends on the power dissipation.

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PostPosted: Mon Sep 10, 2012 3:47 pm    Post subject: Great stuff! Reply with quote  Mark this post and the followings unread

Thanks very much Osal for these posts! It's very hard to find these design rules/guidelines elsewhere (or at least I haven't found any).

Please post more when you get the chance!
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PostPosted: Thu Sep 27, 2012 1:27 pm    Post subject: Reply with quote  Mark this post and the followings unread

Thanks synchromesh.
Ok, let's follow with the voltage.

The Voltage

Let's start talking about the nature of the mains electricity and the purpose of the power supply.
Mains AC voltage is not stable. It has fluctuations. The mains voltage can rise over, or drop below, its nominal value.
The purpose of the power supply is to obtain a suitable DC voltage from this mains AC voltage.
Thus, the power supply must be designed considering both, when the voltage rises and when the voltage drops.

Mains Voltage Rise

The voltage ratings of the power supply's components and the power dissipation of the voltage regulator must be calculated for the worst case scenario, that is when the mains voltage is high. Normally, they are calculated considering a +10% of voltage rise over the nominal mains voltage.
To protect the power supply to higher voltage rise, it should be added an over-voltage protection circuit.

Mains Voltage Drop

The voltage regulator needs a minimum voltage at its input in order to warranty correct voltage regulation. If a mains voltage drop brings the voltage at the regulator input below this minimum, the voltage regulation is lost.
Usually the power supply is calculated to be immune to a voltage drop -10% of its nominal value. However, sometimes mains voltage can drop beyond this -10%, depending on the quality of the mains electricity.
This depends on countries, on areas (rural, industrial, urban, etc...), on time (day, night, mass events, etc..), on source (mains electricity, local new green energies, etc...).
It is important to know about the mains electricity where the power supply will operate.
The immunity to mains voltage drop is defined by:
imvd%=((Vout+Vdr+Vr+Vd)*100/(Vsec*CF))-100
Where:
(imvd) is the immunity to the mains voltage drop in percentage to the nominal mains voltage.
(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.
(Vsec) is the AC voltage in the secondary.
(CF) Is the crest factor of the wave in the secondary.

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PostPosted: Sat Oct 06, 2012 4:03 pm    Post subject: Reply with quote  Mark this post and the followings unread

Transformer's voltage regulation

Without going into details, the voltage at the secondaries drops as the load increases.
This is the transformer's voltage regulation and is defined by:
Voltage regulation % =(Vnl-Vnom)*100/Vnom
Where:
(Vnom) is the secondaries' nominal voltage.
(Vnl) is the secondaries' voltage at no load.

The nominal voltage at the secondaries is specified at full load (Always consult the manufacturer specifications). Therefore we can expect a higher voltage at the secondaries when there is no load.
It must be considered when calculating the voltage ratings of the components that follow the transformer.

Let's put as example a toroidal transformer 50VA 36VCT. In its data-sheet we can read that it has a 13% of voltage regulation.
A 50VA 36VCT transformer can provide a maximum of 1.39A. Therefore, the voltage at each secondary is 18VAC when the transformer is providing 1.39A.
As the current decreases the voltage increases.
When there is no load the voltage is 13% higher than its nominal value. So, it is 18+0.18*13=20.34VAC at each secondary.

When using a derated transformer in a power supply, the nominal voltage at the secondaries is also higher than its nominal value. We can take profit of this to achieve more immunity to the mains voltage drops, but also we must consider it when calculating the power dissipation at the regulator.
To estimate the actual nominal secondary voltage for whatever load, we measure the mains electricity voltage at the same time than the voltage at the secondaries and then the actual nominal voltage at the secondaries is:
Vsn=(Vs*Vpn)/Vp
Where:
(Vsn) is the actual nominal voltage at the secondaries.
(Vs) is the measured voltage at the secondaries.
(Vpn) is the nominal primary voltage.
(Vp) is the measured mains voltage.

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PostPosted: Sun Oct 07, 2012 3:12 am    Post subject: Reply with quote  Mark this post and the followings unread

Forward voltage drop of the rectifier diode

After the bridge rectifier, the wave's peak loses one or two rectifier forward voltage drop depending on the rectifier's configuration.
In the case of a dual center tapped rectifier (like the PS1) there is one diode voltage drop from ground to Vin Peak:
Vsp- Vd = Vinp
In the case of a full wave rectifier (like the PS2) there are two diode voltage drops from ground to Vin peak:
Vsp- 2Vd = Vinp
Where:
(Vsp) is the voltage peak of the AC voltage at the secondaries.
(Vd) is the rectifier's voltage drop.
(Vinp) is the peak of the rectified voltage at the regulator's input.

The forward voltage drop of a diode rectifier is about 1V. For example, the very well known 1N4002 has a typical forward voltage drop of 1.1V. Consult always the manufacturer specifications! Smile
If we want lower forward voltage drop we can use Schottky rectifiers which its forward voltage drop is about 0.5V.

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PostPosted: Mon Oct 15, 2012 9:00 am    Post subject: Reply with quote  Mark this post and the followings unread

The ripple at regulator's input

To a given load, the capacitor determines the ripple voltage at the regulator's input.
The ripple voltage in volts peak to peak is defined by:
Vr=(Δτ*Iout)/C

Where (Δτ) means deviation in time and refers to the time in which the capacitor is discharging.
To estimate the ripple, usually is assumed as worst case scenario that the capacitor discharges during all the cycle. However, in this context, according to my measurements, the longest discharging time is the 70%. of the cycle.
If the worst case scenario is a mains electricity frequency of 50hz, this 70% of the cycle, equals 0.007s. We can use it as a factor and the formula simplifies with:

Vr=(0.007*Iout)/C

Where:
(Vr) is the ripple voltage in volts peak to peak
(Iout) is current output in amperes.
(C) is capacitance in Farads. (Keep in mind that the electrolytic capacitors have usually a 20% of tolerance.)

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PostPosted: Mon Oct 15, 2012 9:49 am    Post subject: Reply with quote  Mark this post and the followings unread

Regulator dropout voltage

The regulator needs an input voltage higher than the output voltage. The dropout voltage is the minimum difference within the output and the input voltage to ensure correct regulation.

There are three types of regulators regarding this dropout voltage: Standard, Quasi-LDO and LDO. The standard regulators have the larger dropout voltage (2V ~ 3V) and the LDO (Low Dropout Regulator) the smaller ( less than 1V)
http://sva.ti.com/assets/en/appnotes/f4.pdf

One considerable characteristic is that the Voltage dropout increases as increase the current through the regulator.
Look, for example, at the LM317 regulator data-sheet, in page 6, the plot of the dropout voltage http://www.ti.com/lit/ds/symlink/lm317-n.pdf

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Osal



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PostPosted: Mon Jan 14, 2013 11:46 am    Post subject: Reply with quote  Mark this post and the followings unread

This had to be posted after the "Transformer's voltage regulation" in order to follow an order from the mains electricity to the power supply output:

The Crest Factor of the wave at the transformer secondaries.

The Crest Factor is the ratio of the peak of a wave to its RMS Voltage.
CF=Vp/Vrms. This is needed to estimate the peak voltage after the rectifier.

Under no load, the wave at the secondaries is like the one at the mains electricity. Theoretically it should be a sine, which crest factor is sqrt(2) ≈ 1.4
In a rectifier-capacitor circuit, as more current flows through the rectifier, more the sine-wave is distorted reducing its crest factor.

It can be estimated measuring the true RMS Voltage and the peak voltage.
For example, look the following measurements in a PS1 with a 50VA toroidal transformer:
When I out=0.77A...CF=1.27
When I out=0.5A.....CF=1.35
When I out=0A........CF=1.4

So, as an approximation, we can use CF=1.3

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PostPosted: Wed Jan 16, 2013 10:18 am    Post subject: Reply with quote  Mark this post and the followings unread

Power supply output.

Finally, at the power supply's output we have the desired DC voltage.
The voltage regulator takes care to keep the voltage at its output constant at the same value.
This is true as long as the voltage at the regulator's input doesn't drop below the sum of the dropout voltage and the output voltage, the junction temperature is below its maximum rating, the maximum current output is not exceeded, etc. Take care of the voltage regulator.

OK then, we started from the mains electricity and through the power supply we arrived at its output. Now that we have seen how the voltage behaves in each part of the power supply, lets go and talk about the voltage ratings.

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