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Talkin' PSUs

Printed From: Graham Slee Hifi System Components
Category: DIY AUDIO
Forum Name: Free Audio Designs and Info
Forum Description: Graham shows his competitors how to design and gives you some great projects to build
URL: https://www.hifisystemcomponents.com/forum/forum_posts.asp?TID=5062
Printed Date: 28 Mar 2024 at 8:47am
Software Version: Web Wiz Forums 12.01 - http://www.webwizforums.com


Topic: Talkin' PSUs
Posted By: Graham Slee
Subject: Talkin' PSUs
Date Posted: 25 Oct 2020 at 8:59am
Talkin' PSUs

A PSU1 in kit form has long been hinted at, but the responsibilities I'd have to take for your safety are too much. Therefore, I thought it best to suggest a circuit, and then it's your responsibility to ensure your own safety, should you wish to build it.

The circuit shown here is not the PSU1, although it has similarities. I have assumed the reader has sufficient know-how to understand it. This topic is meant to be educational rather than a DIY project and can cover as many aspects of power supply design as you want.

24 volt DC linear power supply




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Replies:
Posted By: Sylvain
Date Posted: 25 Oct 2020 at 6:00pm
A little elaboration of the design diagram, the transformer, current regulation and output in the light of the 5v DC design experience now under DAK shop, would be gratefully appreciated, please.  


Posted By: Ashleip
Date Posted: 19 Jan 2021 at 8:34pm
Sorry this is probably a basic question....

From memory, output transformer winding configuration (+ to +, - to -) gives an output voltage of 20v. Although does the voltage regulator not need Vin > Vout to supply the 24v? 

If my memory is wrong and the wiring of the output windings gives 40V, why is Vin chosen to be the maximum rated voltage for the 7824 when the data sheet suggests stable 24v Vout from Vin between 27 and 37V?

Thanks! 


Posted By: Graham Slee
Date Posted: 19 Jan 2021 at 10:44pm
Originally posted by Ashleip Ashleip wrote:

Sorry this is probably a basic question....

From memory, output transformer winding configuration (+ to +, - to -) gives an output voltage of 20v. Although does the voltage regulator not need Vin > Vout to supply the 24v? 

If my memory is wrong and the wiring of the output windings gives 40V, why is Vin chosen to be the maximum rated voltage for the 7824 when the data sheet suggests stable 24v Vout from Vin between 27 and 37V?

Thanks! 


Ratio of RMS to V peak?


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Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 4:00am
Now I have an audience of two; I'd best start to flesh out this design.

Painting by numbers never turns out well - you need to become your artist. It is better to have somebody teach - even a crappy teacher like me - so you gain the power to make your power supply.

The above circuit works, but only just, providing the nominal values apply. It will work better on a British mains supply (which are not just exclusive to Britain) and probably really badly on a European mains supply.

Ashley points out the voltage regulator drop-out voltage, it needs =/>26 volts DC to function, but he didn't change AC to DC.

The current from the 20V AC EMF charges C1 via a diode on either end. The datasheet says they drop 0.65 volts each. Think of it as an entry fee. The capacitor receives the voltage after a 2 x 0.65-volt discount.

That's even worse; we're now down 1.3V to 18.7V - this will never work - he wants >26V, but he's only got 18.7!

That's right, I only have 18.7 volts, but that's AC; it's rated in RMS (should be lower case, by the way).

The RMS is the root mean square, which involves the inverse of the square root of 2. A given peak voltage is multiplied by the inverse of the square root of 2 to show its RMS value, provided it's of a sinusoidal shape.

The inverse of the square root of 2 is approx. 0.707, but we don't know the peak. We only have the RMS, so we need the inverse of 0.707 to multiply the RMS. 1/0.707 = 1.414, and so 18.7V RMS becomes 26.44 volts peak.

26.44V being >26V, we're good to go.

The output voltage is a nice steady, regulated 24V, but at what current? C1 stores charge, but how much can it give up before its voltage drops below 26V?

It can give up 0.44V. So what is the current we can take out before that 0.44V is used up?

To help us here, we need a formula, and the charge (Q) formula is Q = CV (capacitance times voltage), but we need to relate it to current (I) and the charging rate of the rectified AC (T).

I will not bore you with proof of how I and T are derived, except to state what might be evident that Q = IT (current by time, current x time), and if Q also = CV, then CV = IT.

We want to know how much current (I) we can use before the 7824 regulator ducks out between rectifier peaks.

If CV = IT, then CV/T = I (by dividing both sides by T to leave I).

We know the capacitance of C1, which in Farads is 0.0047. We know the 50Hz half cycle timing, which is 1/100, or 0.01, and we have 0.44V to throw away.

So I = 0.0047 x 0.44 / 0.01 = 0.2068 A, which is near enough 200mA.

I think that's enough for a first lesson.


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Posted By: Sylvain
Date Posted: 20 Jan 2021 at 6:56am
Truly appreciate your guidance but between the mathematics and formulas and technical expressions ...
i value the information but this spectator audience is a mere mortal.......and I chose to paint by numbers at least I have something at the end ....


Posted By: Ashleip
Date Posted: 20 Jan 2021 at 9:26am
Doh! What a schoolboy mistake forgetting to convert from rms to peak.

Thanks Graham - great way to learn by going through like this, rather than painting by numbers where you can take what you learn and apply it to other uses.

So... let's say for example you wanted to increase the potential current capacity of the power supply to 1A, based on I=CV/T you have two options, increase the transformer voltage or increase the capacitance. As both C and V are proportional to I, if we want 1A, approx 5x 200mA, we need either 5x 4700uF or 5x 0.44V. I'm assuming the latter is preferable (as 23000uF seems quite large!) so you would want to swap the 20V transformer for a 22V unit, which will give us a 3.3V over the required 26V, and therefore I=1.5A?


Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 9:58am
Originally posted by Sylvain Sylvain wrote:

Truly appreciate your guidance but between the mathematics and formulas and technical expressions ...
i value the information but this spectator audience is a mere mortal.......and I chose to paint by numbers at least I have something at the end ....


Sylvain, may I point you at the Reckoning of Time, by the venerable Bede;

"Should someone rather less skilled in calculation nonetheless be curious about the course of the Moon, we have also for his sake devised a formula adapted to the capacity of his intelligence"

(Translated from the Latin, and written in England very nearly 1,300 years ago!)

Even I, would have a hard time working through the subsequent formula, which, incidentally, brings to mind a Dean Friedman song.

I will try my best to do a modern-times "Bede simplification," but please do not judge me harshly as the capacity of my intelligence to explain it so, doesn't look to be all that great. Wink



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Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 10:10am
Ashley, I wish I had gold stars to hand out, but you'll just have to accept this virtual one. Star

I will simply answer yes to your third paragraph for the time being. You have grasped the nettle!

For my part I must try and take this steady because I currently have the man-flu (the original camel/rhino coronavirus).

For now I will say that in picking on a 24 volt regulated supply, I think we could learn more than if we'd picked another voltage.

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Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 12:52pm
Painting by numbers: primer 1

You need a calculator with a √■ key, and a  χ-­¹  key

With the √■ key you can obtain the value of the square root of 2 by typing the sequence

√■ 2 =

With the χ-­¹  key you can obtain the inverse of the square root of 2 by typing the sequence

√■ 2 = χ-­¹ =


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Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 1:17pm
So, let's try to get 1amp output from a DIY 7825 based power supply using only off-the-peg components.

First, let's take a look at the 7824. Absolute maximum input voltage is 40V (other 78 series are 35V I believe).

The minimum input voltage is 26 volts, but that's for a perfect 7824 - good luck! I'd suggest 26.5 volts, but that's in the ripple voltage gutter.

New term warning! "ripple voltage gutter."

Best explain that next post, before we can really continue.


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Posted By: Graham Slee
Date Posted: 20 Jan 2021 at 4:56pm
Ripple voltage is peak voltage minus the gutter voltage of 26.5 V DC.

Vripple = Vpk - Vgtr

It's easier to test different ripple voltages, so let's restate that as gutter voltage plus ripple voltage equals peak voltage.

Vpk = Vgtr + Vripple

We find ripple voltage using CV = IT by rearranging it by dividing both sides by C, such that

V = IT/C

Now I = 1A and one times anything is whatever that anything was, so for I = 1A, we can forget I, so

V = T/C

For a full-wave rectifier (who'd use anything less?) T = χ-1 x 100 pulses per second, which is 0.01, so

V = 0.01/C

The original value of C (C1) is 4700uF, which is 0.0047F, so

V = 0.01/0.0047.

Do we need a calculator? Can we still do schoolkid maths? Canceling 0's we get

V = 10/4.7

10/5 is 2, so we can guess at 2.1 volts, and it's actually 2.13 volts.

Remembering what this is all about, obtaining Vpk, then

Vpk = Vgtr + Vripple = 26.5 + 2.13 = 28.63V

Therefore any input between 28.63 and 40 volts will work to produce 1 amp.

But now we need to know the transformer voltage.

√■ 2 = χ-¹ = 0.707

So 28.63 x 0.707 = 20.24V

But we now add the entry fee, the rectifier voltage drops, which are now hit by inflation!

The more current we draw, the larger the silicon is stretched (not exactly right, but I'm teaching Sylvain too), so it might now be 0.7V per diode or even 0.75V.

We have two diodes at any one time, so we must add 1.5V. Thus the AC voltage has to be 20.24 + 1.5 = 21.74V.

As Ashley noted, we could use twice the capacitance and reduce the ripple voltage by half. If we use 10,000uF, then instead of

V = 10/4.7 = 2.13volts, we have 10/10 = 1volt

Adding Vgtr, 26.5, we need 27.5V peak, and 27.5 x 0.707 = 19.44V

We add a 1.5V entry fee, and the transformer AC is 20.94V.

It looks like this is going to be awkward?


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Posted By: Sylvain
Date Posted: 20 Jan 2021 at 6:29pm
Thank You thank YOU


Posted By: Ashleip
Date Posted: 20 Jan 2021 at 8:18pm
Great stuff! Really enjoying this unravel!


Posted By: Graham Slee
Date Posted: 21 Jan 2021 at 8:54am
Meanwhile, at "new Babel," this they began to do...

They began to harmonize the World's voltages. The elephant in the room was Britain. Continental Europe had 220V, and Britain must be brought into line, but faced with replacing every substation transformer in the land, Britain refused.

Having caused a log-jam in the law-making machinery, they did what they always do - it was fudged. It was decided to "adjust" by statutory instrument all voltages in the EU to 230V.

The voltages in continental Europe remained at 220V, and the voltages in Britain remained at 240V, but the "230V adjustment" included an "accuracy annex," which allowed 230V the greater latitude of +/-10%.

In olden times the latitude was set to the achievable tolerance, which was +/-6%, and you could buy transformers with taps for 220V or 240V. Now it's 230V, period.

Therefore the off-the-peg transformer primary is rated for 230V, but as there is no such rated supply voltage, we must design for a supply voltage that may be 207V, or it may be 253V or anything in between.


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Posted By: Graham Slee
Date Posted: 21 Jan 2021 at 10:00am
Deciding on the transformer, part 1

When boiled down, my last post sets upper and lower extremes of what we might expect.

Using a 230V transformer in GB, the voltage might be 10% up, but if you were reading this sur le continent, it might dip 10% down.

Taking 253 and dividing it by 207, we get 1.22, which means +/-10% is +22% if we design for the worst possible case.

We have to tighten this design and establish the lowest AC voltage that keeps the 7824 in regulation.

Using a 4700uF smoother, we can accept a minimum of 22V AC. Using a 10,000uF smoother, we can accept a minimum of 21V AC.

Those are the secondary voltages we must have at 207V mains input. If not, a continental user on a cold dark winter night might find ripple on the 7824 output.

Working upwards from 207V to 230V, we find we must multiply by 1.11111 (not 1.1, as the 10% latitude would suggest).

22V thus becomes 24.4V, and 21V thus becomes 23.3V.

We can obtain a 24V transformer off the peg, so it looks like we have to use the 10,000uF smoother. We have established the conditions for 207V use.

There is a further spanner to throw into the workings, transformer regulation. That is the percentage over-voltage of the transformer outputs, with no-load. Should the 7824 input exceed 40VDC, it will probably blow.

Low wattage (low VA) transformers have poor regulation, but as we want to achieve 1-amp, the transformer we choose isn't going to be amongst the lowest wattages - this is what we now must look at.


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Posted By: Graham Slee
Date Posted: 21 Jan 2021 at 11:40am
Deciding on the transformer, part 2

In voltage, you can relate RMS to Peak (or DC) by √■ 2 and do the inverse using √■ 2 = χ-¹

Now you learned it on the calculator; we can start thinking of it as √2 and 1/√2 (it's easier for me to explain).

With current, there is no equivalent authoritative ratio (maybe they've done one since?). There is, however, a rule of thumb ratio, which happens to coincide with the Fibonacci ratio. The important thing about it is it works!

There are no calculator keys for the current ratio, so you'll have to memorise it. IAC to IDC = 0.62; IDC to IAC = 1.62.

If you want 1-amp DC, you will require 1.62-amps AC.

In the last post, the transformer secondary voltage was decided at 24V. If we multiply that by the AC amps, we get 24 x 1.62 ≈ 39.

39 is the required power, but modern convention says we must call it VA and not watts, which is how we used to think of it.

Looking in the RS online catalogue, the nearest VA rating is 50VA for an E-I laminated clamp mount transformer. It costs £23.66 including VAT. There is a toroidal transformer available for £62.65.

If the power supply is to live in its own cabinet, I can see no reason for a toroidal transformer, so I would choose the laminated one and save almost £40.

My findings from my 1970s design indulgence blog regarding circulating currents in paralleled secondaries sway my choice to a dual 12V output, which I can wire in series for the required 24V.

Looking at the RS page: https://uk.rs-online.com/web/p/chassis-mounting-transformers/0504672/ - https://uk.rs-online.com/web/p/chassis-mounting-transformers/0504672/ , we can click on the datasheet link and find the information for part number 504672.

It's a lot bigger than the 12VA transformer of the "teaser circuit" on page 1, and it won't fit in a PSU1 case, but we decided we wanted 1-amp.

The datasheet shows a no-load voltage of 13.3V and 13.3/12 ≈ 1.11 - it has 11% regulation. We can now work out the maximum input voltage the 7824 will contend with.

The maximum mains voltage allowable in the GB is 253V, 230V plus 10%. It used to be 240V plus 6%, and it still is! Therefore the maximum mains voltage is 255V.

255/230 ≈ 1.11. We also have 11% regulation (for when the power supply is left on with no load), and so 1.11 x 1.11 ≈ 1.23. Therefore the 24V AC can go as high as 29.5V AC.

Let's see what it could be after the rectifier. We first remove the "entry-fee" of two diode-drops, but this is on no-load, and the diodes only need to pass quiescent current for the 7824 and the discharge resistor R1. Assuming we use regular silicon rectifier diodes, the diode-drops will be 0.6V. We take way 1.2V from the 29.5V and are left with 28.3V AC.

We have to take the peak voltage, which we do by multiplying 28.3V by √2.

The answer is 40.05V DC.

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Posted By: Sylvain
Date Posted: 21 Jan 2021 at 2:06pm
Deciding on the transformer, part 1
Deciding on the transformer, part 2

Thank You very much for the structure ........paragraphing and sequencing the work in a logic rationale  BUT       it will take 24 hours and 3 readings to digest each paragraph......BUT GOOD STUFF 


Posted By: Ashleip
Date Posted: 21 Jan 2021 at 5:22pm
Another idiot question incoming (and might also be answered in your 70s page?)

The reason you'd want to avoid an EI transformer for a PSU mounted within the amplifier would be to avoid electromagnetic fields inducing mains hum on the input signal. Presumably though, the same magnitude of mains hum will be induced on the DC output of the PSU as would be on an incoming audio signal (assuming same location of audio / DC cables). 

Is it right that this magnitude of mains hum induced in DC signal is negligible because it is insignificant in relation to the 24V output, whereas this same magnitude on a phono signal (e.g.7mV) is significant? 


Posted By: Graham Slee
Date Posted: 21 Jan 2021 at 10:21pm
Well, it's true that wire has inductance, and laying a wire in an AC magnetic field, you'll get some of that AC on it. In fact, if you were to run the signal input parallel to a mains wire, it should pick that up too.

In fact, if you put any transformer in a loop of wire, the loop will pick that up. Changing the ground wiring (yes, the ground) on the '70s project recently, so it didn't encircle the transformer (just one turn), the S/N improved by 6dB (the noise halved). Might you have noticed the diagram shows a round transformer?

The most effective means of coupling "hum" is to mimic a transformer. At least put it into a loop. Windings in a loop are very effective at picking up hum. It's a bit similar to an induction loop hearing system but in reverse. Instead of a hearing aid coil, you might use a magnetic cartridge. You could sit that inside an inductive loop, such as a house mains wiring.

You might even sacrifice on the altar - I mean, put the turntable on the top shelf of the hi-fi rack right above a power amp. Some people even arrange the DECT phone right next to it. Others might have a lighting dimmer switch on the wall behind it.

Heck no! You need a toroidal transformer, dude! Listen to the marketing; it's much safer than me.Wink

(edited to include emoticon)



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Posted By: Sylvain
Date Posted: 22 Jan 2021 at 8:36am
Changing the ground wiring (yes, the ground) on the '70s project recently, so it didn't encircle the transformer (just one turn), the S/N improved by 6dB (the noise halved). Might you have noticed the diagram shows a round transformer?

hum!!!!!!!

So is the solution in 'placement' and ''lay out''of wires and beware of the magnetic field of Torioidal !!!!!and looping power leads ...


Posted By: Ashleip
Date Posted: 22 Jan 2021 at 9:02am
Tongue need a good map to navigate the marketing campaigns! 
Good to hear that for something like this EI is perfectly good and no need to spend 3x as much for a heavy doughnut.


Posted By: Graham Slee
Date Posted: 22 Jan 2021 at 9:16am
So much for the interest in the design process. The significant bit about 40.05 volts wholly ignored because the topic must descend into a debate about the incredible beauty of toroidal transformers. And there was me thinking you'd like to learn something. You know it all already!

But I will continue for the sakes of those who might eventually come along with unprejudiced minds.

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Posted By: Ashleip
Date Posted: 22 Jan 2021 at 9:24am
I had noticed this, but got sidetracked... What is it about transformers eh?!
As this is only slightly above 40, can we change diodes to increase the drop slightly or is that a stupid idea? Do we need to go looking for a transformer with <1.1 voltage regulation, or is that a pretty standard value?
(Edited this post because I hadn't properly read the previous post Tongue)


Posted By: Graham Slee
Date Posted: 22 Jan 2021 at 9:48am
The point was 40.05 volts.

The sustaining voltage - the absolute maximum on the datasheet for the 7824 voltage regulator, is stated as being 40V DC.

We're supposed to learn the difficulty in designing a mains power supply, having a regulated output, for operation on all 230V voltages.

I started this topic because of the number of members who have asked me about a PSU1 in kit form. Then there is another group who wonder if a PSU1 can be used to power other items, and if not, what can I do to assist them?

Rather than pushing all the bits into a bag and selling it as a kit - a lethal instrument due to its mains connection - perhaps some education into the hows and whys might result in something the DIY'er can make better use of.

So, back to the 40.05 volts calculated earlier. That is the maximum DC input on the voltage regulator we can squeeze it down to for a mains supply at the top of the specified grid voltage range while ensuring it doesn't drop out of regulation at the bottom of the specified grid voltage range.

The crucial questions here should be about how we can ensure reliability. The reader could be in some European country where the mains power demands take the voltage right down or somewhere in industrial England where a large substation now supplies a smaller domestic demand. The latter can be subjected to 255V AC.

The DIY enthusiast cannot easily obtain a transformer of the exact voltage required (which is around 23V) and must work with what is available.

We should be asking, will 40.05V damage a 7824 voltage regulator?


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Posted By: Graham Slee
Date Posted: 22 Jan 2021 at 10:19am
A diode is NOT a stupid idea. (Thumbs Up)


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Posted By: Graham Slee
Date Posted: 22 Jan 2021 at 7:50pm
If we put a forward-biased diode just before the 7824 input, the off-load voltage drop is roughly 0.6V, and that adds 1.5% into the "component tolerance pool." It might still reach 40V, but we've made extra allowances for inaccuracies, which inevitably creep in everywhere.

We also have to put some faith in the electricity grid.

The datasheet doesn't tell us the conditions of the absolute maximum voltage, whether it's on-load or off-load. If it said on-load, then we know that off-load, the input voltage will be higher. Unfortunately, it doesn't say.

The datasheet also notes, "Absolute maximum ratings are those values beyond which damage to the device may occur. Functional operation under these conditions is not implied." So, it says not implied, so better not conjecture.

At the low mains voltage end, we had about 0.7V going spare, plus the 26.5V minimum 7824 input. Theoretically, the 7824 input can go as low as 26V, but take a look at the output tolerances on any 7824 datasheet. Every little helps!

The diode has made it so the rectifier circuit centres-up with the 7824 input range for all expected grid conditions.


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Posted By: Graham Slee
Date Posted: 23 Jan 2021 at 8:23pm
The non-technical layman understands electricity due to it being a greater part of modern living.

He understands Voltage that, in most cases, it appears between two points. The live and neutral of a mains plug and the pointed end and the blunt end of a battery are two examples.

In a mains derived DC power supply, both types of electricity exist. The input is from the live and neutral of a mains plug. The output mimics that of a battery.

A battery contains charge, and in use, the charge flows in the circuit it powers, and that flow is called current.

As the charge is consumed, the current flow diminishes, and eventually, all the charge has gone.

The smoother or reservoir capacitor of a mains derived DC power supply acts similarly to a battery. It doesn't have the long term charge of a battery. Instead, a rectifier charges it, and the load discharges it. The rectifier is constantly applying charge, and the load is continuously discharging it.

ripple voltage
(image copied from Morgan Jones)


The transformed and rectified waveform is like a series of upward going loops. The top of each loop is the peak value of the AC voltage.

VDC = √2 x Vac

The charge is used up in current flow in the circuit powered by the power supply.

The smoother capacitor voltage decays, only to be met by the next loop. These loops are 0.01 seconds apart.

Rather than the DC voltage being a straight line over time, it ventures up and down by an amount called ripple voltage.

The ripple voltage is directly related to the current being drawn from the power supply. If we don't draw any current, then this DC voltage is a straight line over time, but as soon as we draw current, the Voltage starts to ripple.

The letter used for charge is Q, and it stands for coulombs. Charge is related to Capacitance (the storage of charge) and Voltage.

Q = CV

Charge is used up by current draw over time, so

Q = IT  (I stands for current in amps, T stands for time in seconds)

Therefore Q is related to both, so

Q = CV = IT

If we want to find Capacitance and know what ripple voltage is acceptable and the current the load draws, we also understand that the rectified waveform repeats every 0.01 seconds. We can use

CV = IT

Rearranging it to obtain C on its own involves dividing both sides of the equation by V.

The V's of CV/V cancel to become C. On the other side, we have IT/V.

C = IT/V

If we want to know the ripple voltage, then we divide both sides by C

V = IT/C

So now, even the layman should have a better idea of designing power supplies.

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That none should be able to buy or sell without a smartphone and the knowledge in how to use apps


Posted By: Sylvain
Date Posted: 24 Jan 2021 at 9:53am
Yes  Thank you for consideration to the 'non technical laymen'' and the step by step  structure .....enthusiast like me will benefit in their development and understanding.....but ' bite size'' at a time.


Posted By: Graham Slee
Date Posted: 24 Jan 2021 at 12:04pm
My last post was necessary to understand the manipulation of the smoother capacitance. It is also essential to know where the charge is "held."

If current is the flow of charges around a circuit, then that current must return to where the charge is kept (Kirchhoff conservation laws).

The main charge is kept in the smoother. We can use one capacitor or use any number in parallel to make up the value determined by the formula.

To ensure the voltage to the 7824 remains within its input range for all grid supply voltages, we determined a maximum ripple voltage of 1-volt.

C = IT/V

= 1(amp) x 0.01(sec) / 1(volt) = 0.01 Farad (which is 10,000 uF)

But what if we don't have a 10,000uF capacitor? Say we have two 4700uF capacitors we can put in parallel, making 9400uF. Will that suffice?

Using the rearranged formula

V = IT/C

= 1(amp) x 0.01(sec) / 0.0094(Farads) = 1.064 volts

Will 0.064 volts make that much a difference?

Let me answer that question by stating that electrolytic can capacitors (which are the only type having the capacitance we require) have tolerance of +/-20%.

It is possible (although not very likely) that a 10,000uF capacitor could be 8,000uF.

Let's run that through our formula.

V= IT/C

= 1(amp) x 0.01(sec) / 0.008(Farads) = 1.25 volts

Likewise, 9400uF would become 7520uF, and the ripple voltage 1.33V.

Based on discussions so far, we might see sufficient latitude and trust in hope by using two 4700uF capacitors that both can't be -20%, or we could wish to reach perfection. Being pragmatic, will we need precisely 1 amp output? And if not, we can risk a third of a volt on a "brown-out" voltage of 207V on a 220V supply grid. If the grid voltage is 240V, there is no need to worry.


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Posted By: Ashleip
Date Posted: 26 Jan 2021 at 9:15am
Aaah - so am I right in saying that by spec'ing two capacitors in parallel you reduce the likelihood of the combined capacitance being close to the limit of the stated tolerance in comparison to if you only had a single capacitor then?

Just catching up on the last few posts and trying to understand how this would change if the design parameters changed. You say 40.05 is fine and that we should have faith in the grid - it's unlikely that this situation would actually occur. At what point would we question our faith? Say we are designing for 250mA, that means we need a VA >= 9.72 (1.62 x 24 x 0.25). So we could use a 12VA, 1 x 230 / 2 x 12V transformer from the same range as that you posted. On the spec sheet this transformer has a slightly higher no load voltage of 13.6V. This means that (24 x 1.11 x 13.6/12-1.2) x sqrt(2) = 41V. Would we still put faith in the grid in this instance, even with this slightly higher exceedance? (I think the mains voltage would need to be >249V to exceed 40V at the 7824 so I'd be inclined to say yes we should probably still have faith, but I have no experience to base that on!)

As you say there is a lot more to the PSU design than meets the eye! Thanks for all the info and help so far!


Posted By: Graham Slee
Date Posted: 26 Jan 2021 at 1:06pm
I've tried to communicate a number of precautions, including the excellent suggestion of a series diode (I'll forgive you for beating me to it).

When we run out of precautions, we then hope in faith. I can quote, "cursed is he who trusts in man," but we work in an imperfect system, so all we can do is our best.

We could fire a crowbar circuit to blow a fuse to protect the voltage regulator, but they have a notoriety for unwanted triggering and might fire in a survivable situation. We live in a digital world where we think everything can be protected, but that world runs on an analogue foundation highly dependent on nature.

The 7824 could be classed as a terrible design considering there's only 15.5V practical input latitude, but I've not seen any improvement elsewhere. Politicians can use their "power" to take over from engineers as technical decision-makers, and they reveal their stupidity, but there's little we can do about it. Politicians can rule man, but nature has its own ideas.

The 7824 was quite viable when first designed. Still, when a dangerous bunch of dreamers decide on a fictitious grid voltage and tell designers they must create things to make their crazy idea work, what can the designer do (apart from curse the idiots).

Obviously, we must design properly. Politicians also make the safety laws, and some designers work to them; some ignore them, and others are oblivious to such laws.

Working with a European certificated test lab in the redesign of the PSU1 in 2005, much of what is in (or will be in) this topic was discussed at length. Every imaginable scenario had to be answered, and where rejected, had to be re-thought and presented again. The only thing allowed to fail was the product itself, provided it failed safe.

Things stop working, and that's a fact of life. We try to design-in longevity, but nature throws its limitations at us. We cannot always design a universal solution. Still, we can demonstrate the bones and provide as much knowledge as possible to enable others to feel more confident in having a go.

We all got where we are today because of handed-down knowledge. The human race has been very successful at doing that. Today, what was intended to carry that on - the Internet - has become a place of darkness. Instead of useful information for our continuation, opinions that serve minimal purpose abound.


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That none should be able to buy or sell without a smartphone and the knowledge in how to use apps


Posted By: jfaust
Date Posted: 26 Jan 2021 at 10:49pm
Guys I just wanted to stop in here and give you my take on a 24vDC power supply for the US.

First let me say I've really enjoyed Graham's lessons and the whole flow of the process.  I also watched the evolution of the design so I'll be taking you through my process and choosing some findings that came out of the process. So please don't think i'm poking fun, it's just as things evolved.

1. The US standard for power is 120 +/- 5% at 60Hz, so 114 to 126 vAC and the 60 HZ translate into a time standard of 0.00833.
2. Output of the psu should be .5 amps

So let's get started.  About half way through you decided we needed a transformer with 24 volt output so that's what we're working with.

Now for the diodes.  Our first encounter was a .65 vDC drop each so 1.3 vDC total.  That was followed by .75 vDC each for 1.5 DC, and the last iteration was 6.0 vDC for a 1.2 vDC drop.  I'm choosing worst case, so .75v vDC each.

Looking at 24vAC -1.5 vAC we get 22.5 vAC RMS.  Please excuse my incorrect capitalization for this entry, I'll work hard at getting it right in the future. Anyway, we want to know what that the peak to peak will be, so 22.5 X 1.414 = 31.82 vAC, right in the sweet spot for a 7824 (which does have a .5amp option). When looking at the minimum 26.6 voltage for the 7824 we see we have 31.82 vAC - 26.6 vAC for a 5.52 vAC margin.  So far so good.

Then going on to Ripple Voltage.  Here I return to our lessons and chose our final C of .0094f.
  Using V=IT/C we get 0.5 X 0.00833/.0094 = .4431 Volts
  To find Vpk we use Vpk = Vgtr + VRipple so we get 26.5 Volts + 0.4431 volts for 26.93 vAC
  Converting to RMS 26.93 vAC X .707 = 19.04 vAC RMS
  Adding back in our .75 vAC diodes we get 19.04 vAC + 1.5 vAC for 20.45 vAC RMS so we're good to go.

Looking at the Transformer.
  First looking at Input Voltage spread. Low voltage 120 vAC / 114 vAC = 1.0527
  Low Voltage conversion gives us 20.45 vAC X 1.0526 = 21.53 vAC RMS at the low voltage
 
  Now we determine what our output current needs to be so IDC to IAC =1.62 hence .5 IDC X 1.62= .81   IAC.

Opps, can't find the rat of my notes so Ill return later.


Posted By: jfaust
Date Posted: 02 Feb 2021 at 8:46pm
Hey Graham, Sylvain, and Ashleip

As we proceed on, I think it best to forget everything in my last post after the first sentence.

A quick recap. My initial project was modify the final design here for North American power and a 0.5 Amp output. Time was the only constant that changed relative to the formulae Graham presented.

For example, Ripple Voltage.

UK & EU:  Using V=IT/C we get V=0.01/0.0094 which equals 2.13 Volts.
North American: T=0.0083, I=0.5, and C=0.0094 the result is .44 Volts.

Now a question for Graham.

I reviewed the data sheets for the 1N4002 from three manufacturers and found the worst case Instantaneous Forward Voltage to be about 0.88 Volts with a junction temperature of 25 C dropping down to about 0.72 Volts at 150 C for .5 Amp.  First, this was taken from the Forward Voltage chart in each data sheet, is the correct source for our entry fee for the rectifier?

Secondly, not knowing what the expected junction temperature, I used the worst case of 0.88 Volts.  Is this correct?

Moving on to the transformer.

Sizing: 0.5 Amp X 1.62 (DC to AC) =0.81 Amps AC
           24 Volts x 0.81 Amps =19.44 Amps AC so a 20VA transformer should handle the load.

There were several 120 Volt mains with two 12 Volt secondaries out there, so not a problem.

The changes to 0.5 Amps, the North American power standard, the forward voltage drop, and a 0.5 Amp 7824 all worked out.  The design was a go.

Then a bad thought, could the design work both here and in the UK using a 230/115 mains transformer.  The answer was Yes even with the desired secondary voltage occurring near the bottom of North American power spread of 114 Volts to 126 Volts.  The upper end output voltage still was under the 40 Volt top end of the 7824.

A smaller version of  Grahams transformer, the RS 504-280 meets the requirements with only having to change the transformers mains connection configuration. 

Now more for Graham.

With the changes being made are the 10k 1W R1 and 1R 1W R2 still correct?
The same question follows for C2,C3, and C4.

This has been enlightening.




Posted By: Graham Slee
Date Posted: 03 Feb 2021 at 10:51am
Maybe a slip of the calculator, but I get 1.06 volts instead of 2.13 volts. At 0.5A, it will be 0.53V ripple, close to your 0.44 volts, but I get your meaning.

And yes, half the current, half the transformer VA rating will suffice, and 20VA is closest.

The 10k 1W resistor is there to discharge the capacitors when switched off, basically to prevent them from discharging through you. It is an arbitrary value that steals little current but brings down the voltage across the capacitors within a reasonable time. T = CR = 100 seconds, to around 0.1 of the voltage. The rating of 1W is to keep it relatively cool. I find 5 to 10 times calculated wattage works.

C2 and C3 help stabilise the regulator from instability (v. high-frequency oscillation). Still, we'll have to redraw the circuit with a diode drop between C1 and C2 (in series with the regulator input), which is there because we have around a volt too much.

Just a gut feeling, but after that diode, the beneficial value of C2 might need to be 1uF, and a 50V ceramic might fit the bill. The diode has an impedance, and the 1uF value might polish-off some HF rubbish. The connections should be kept short, at least 1 inch, to keep the inductance low.

C3 on the regulator output is always suggested on the datasheet to be 0.1uF, but the reason is never fully explained. An experienced EMC examiner told me that the 78 series often cause EMC test failures due to oscillations in multiples of MHz, but my grey cells have forgotten the exact frequency by now. The capacitor acts as a capacitive load on an op-amp's output, and the 78's are in-fact op-amps with a pass transistor. 0.1uF might well quell oscillations, but once the capacitor lead inductance nullifies its existence, at that frequency, it can oscillate again.

Philips circuits often placed a small resistance before the capacitor, so we might find that taking C3 to the other end of the resistor helps more.

The resistor value will depend on the voltage drop you're willing to put up with. At 0.5A, one ohm will drop 0.5V. If the resistor is taken too small, it might not prevent capacitive load oscillation. Maybe we could risk 0R33 and only lose 0.16V?

C4 is another arbitrary value. The regulator high frequency rejection only goes so far. The 78's go-to about 1MHz, so to match, often, a 10uF tantalum is recommended (you'd need a very old datasheet to see that). I would not risk tantalum at this voltage, so I chose 10x its value in an electrolytic. The voltage rating of 100V places its dissipation factor in the ideal region (lowest DF).


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That none should be able to buy or sell without a smartphone and the knowledge in how to use apps


Posted By: jfaust
Date Posted: 04 Feb 2021 at 6:10pm
Just wanted to thank everyone for this learning experience.  As Graham puts it, I had to revive some old grey cells, but it was worth it.

Graham, after all this can you tell us how our design would compare to the 24Volt green wall wart?



Posted By: Graham Slee
Date Posted: 05 Feb 2021 at 10:21am
Originally posted by jfaust jfaust wrote:


Graham, after all this can you tell us how our design would compare to the 24Volt green wall wart?



I can tell you now.

The quality of switched mode depends on the sum of its parts, and there ain't many inside a wall wart!


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That none should be able to buy or sell without a smartphone and the knowledge in how to use apps


Posted By: Sylvain
Date Posted: 06 Feb 2021 at 5:16pm
The quality of switched mode depends on the sum of its parts, and there ain't many inside a wall wart!

 One Hifi cable and accessories add 150% percent mark UP to public Counter shelf Switched mode DC power supply. All for a new cable and one capacitor upgrade for Medical standard grade unit......the demand for these units is enormous and not just for the new generation Computer industry but when a dedicated and Regulated bespoke Hifi Switched Mode Power supply comes ....
 



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