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Power Supply Quiz (Lousy participation you guys!) The Answers, and a free P.S. Simulator I wrote

Got a short quiz for anyone willing to give it a go.

> > These questions are for a standard power supply voltage doubler > circuit. Although this quiz applies to any normal component values > (except very large capacitors or transformers with large series > resistance), let's say the transformer voltage is 10VAC RMS, .2 ohm > secondary resistance, 2 capacitors of 1000uF each, a 50 ohm load, and > 1N4004 diodes. Assume the power supply is switched on at the > zero-crossing of the power line, and the line voltage and the load do > not change over time. > > 1. Does the maximum surge current from the transformer occur on the > first half-cycle of the line?

No. It occurs on the second half-cycle.

> 2. How much more (or less) is the surge current on the 2nd half-cycle > compared to the first? You can say the same, a little, or a lot. Have > to include if it's more or less though.

It's more, by a little. More than any other half-cycle.

> 3. What about the surge current in the 3rd half-cycle? How does it > relate to the other cycles?

The 3rd half-cycle has the lowest surge current of them all.

> 4. How many half-cycles does it take for the power supply generate its > maximum DC voltage?

The peak output voltage of the full wave doubler occurs on the 2nd half cycle. It's a higher peak voltage than any other half-cycle.

I got this from a power supply simulator program I wrote last weekend. I didn't want to believe what I was seeing, but I couldn't find anything wrong in the software so I built one and put a scope all over it. Sure enough!

A screen shot from the simulator for this little quiz is here: files.me.com/wingsy/8um1mw Don't let those funny letters scare ya. It's a PNG file from my MobileMe account at Apple's me.com web server.

If you want a copy of the simulator, get it here:

Mac: files.me.com/wingsy/59fmbv Windows: files.me.com/wingsy/64trdl

Both of those are zipped folders containing the app and the docs.

Reply to
Wingsy
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"Wingsy"

** Many would have sat back cos your Q was ambiguous on the issue of WHERE the surge current is being measured.
** True only of the secondary current AND in the isolated case of switching at a zero crossing.

Why the HELL did you omit to specify DIODE current ???

Why did you say " maximum surge current from the transformer " when you meant something ELSE ????

** Because the load current doubles when the second cap gets its first charge.
** According to your program.
** According to your program.
** Shame that is not convincing proof to anyone but you.
** Tell us - how did you switch the transformer on only at voltage zero crossings ???

Did you switch the primary or the secondary ???

Makes a big difference to the result.

..... Phil

Reply to
Phil Allison

"from the transformer" means "from the transformer" secondary, i.e., where the current is coming from, as opposed to the primary, where it is being applied.

Like I stated. (But the primary current surge should occur at the same time as the secondary, no?)

So, rather than attempt to answer the question, you claim the test is ambiguous? If you could tell me how the current from the transformer is not the same as diode current then perhaps I could understand your confusion. I said current "from the transformer" because I meant current in either diode, not just "diode current" ... cause then you'd be asking "Which diode?"

Wrong. It's because the uncharged cap, during the period between the

1st half cycle and the 2nd, is slightly charged in reverse polarity as the charged capacitor pumps current through it and into the load. Then the 2nd half cycle has to charge that cap starting with a slight reverse charge.

Yes, and according to scope measurements of a real working doubler.

Yes, and according to scope measurements of a real working doubler.

Why is it that when someone shows you something that you didn't already know it causes you to become all snarky and irritated? Do you have a confidence problem with your knowledge (or lack thereof)) of electronics?

If you must know... I switched the variac on & off several (sometimes many) times for each measurement I made.

Tell us why.

Reply to
Wingsy

Your sine waves in your psu program pdf are distorted (flat as you say) because of the way current is drawn; sharp small conduction angle pulse E = IR.

You can look up Schade for rectifier analysis. Here's Onsemis rectifier HB which has examples of using curves that Schade derived pg.63.

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Reply to
Hammy

Yes, I realize that. The simulation software even shows the effects of this in its graph of the secondary voltage waveform. But my primary AC power, straight out of an unloaded wall socket, and with no load of any consequence being on in the rest of the building, STARTS OFF with flat tops. I believe that is what accounts for the diode currents in a real power supply (powered from this building's AC) being less "peaky" than the simulation.

Interesting read, thanks. Haven't seen that before. But it has nothing in that chapter about factoring in the voltage drop of the diodes across their current ranges. He does subtract 1.8v (2 x 0.8v) from peak secondary voltage in the full wave bridge design example, where he uses

25v pk from his secondary when using a 19vac transformer. That supplies 26.8v pk, but he used 25v in the calculations. I assume he dropped 1.8v for the diodes.

The hardest part of my PS Simulator was determining the actual voltage drop across the diodes as the capacitor is being charged, because the diode voltage drop depends on the current, and the current depends on the voltage drop. I used an iterative method, refining the result until the difference from one iteration to the next was really really small. To be accurate I think you've got to account for the variable voltage drop when designing low voltage power supplies. I think that's what has caused me to slightly miss my mark every time I design one. I either overkill it a little, or underkill it.

Reply to
Wingsy

FWB: High voltage: Vo = 1.4 * Vrms Low voltage: Vo = 1.4 * Vrms - 2 Irms = 1.5 to 2 * Idc

C = 2000uF / A at 12V, 60Hz, so C = Idc * 2000uF / (Vdc * F/60) = 0.12F * Idc / (Vdc * F)

Or in the case of this thread, figure a bit more for a doubler (Vo =

2.7-2.8*Vrms, Irms = 4*Idc).

Sure it's fun to say it generates 18.397V, but any more than two digits is automatically defeated by parts tolerance, line variation and plain old good engineering practice.

Tim

--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
Reply to
Tim Williams

Here is the actual scope image from a full wave doubler. It's the same parameters as the simulation image in my original post except this transformer is 5v, not 10v. I cranked the gain way up on the output voltage display (purple channel) in order to get better resolution on the voltage peaks. That's why the baseline isn't visible, it's cranked way down below the screen so the peaks would be visible. The 120vac primary voltage is the blue channel.

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Reply to
Wingsy

I'm wondering why you wrote a program instead of using a Spice variant.

Reply to
a7yvm109gf5d1

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