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

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?

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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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.

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.

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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.

Edited by Graham Slee - 26 Jan 2021 at 2:25pm]]>

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!

Edited by Ashleip - 26 Jan 2021 at 9:16am]]>

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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Yes Thank you for consideration to the