Series Connect Lab SM Power Supplies

All I did was look at and copy what he wrote, moving the decimal. I did no calculations.

Not even a guesstimate. I would have seen it easily, but itwasn't worth my time, sort of like you. So like I said... All I did was rinse and repeat (move the decimal), thinking that was his mistake.

Sorry, but you dolts are "wrong again". Let alone the fact that it isn't 0.15 V either. Funny that none of you 'noticed' that.

And you think they are going to drop nearly 3/4 of a volt each?

PSU's can be connected in series only if the grounds are isolated. That is, the Negative of each supply has to be floating. Else what happens, if a negative is grounded and you connect the positive of one supply to the negative(ground) of the other you are directly shorting that positive terminal to ground. It will behave exactly as if you simply shorted the + to - on a single PSU. It will either burn up or trigger some over current protection.

For non-load required supplies: One simple way to check this is to connect them in series as usual but insert a larger resistance(e.g., 1k 10W) inbetween the + of one supply and the - of the next. Do not load the final supply. If they are floating there will be no voltage drop across the resistor. Reducing the resistance should, theoretically not matter and setting it to 0 should get you where you want.

GND--[-PSU+]--R--[-PSU+]

But if the PSU's are internally grounded you'll actually end up with this

GND--[-PSU+]--R--GND--[-PSU+]

in the first case no current will flow through R and the second case there will be.

For loaded supplies: In this case you will have to use a low enough resistance(and high enough W) to load each supplies

GND--[-PSU+]--nR--[-PSU+]--R--GND

But if the PSU's are internally grounded you'll actually end up with this

GND--[-PSU+]--nR--GND--[-PSU+]--R--GND

If the PSU voltage is V, for example, then you have V---nR---V---R---GND for the first

which results in 2V/(n + 1)R current flowing through the resistors which gives the voltage n/(n+1)*2V across the nR resistor.

In the second you have

V---nR---GND

which gives a voltage of V across the nR resistor.

By choosing n < 1(preferably

Why does that not surprise us?

John

I didn't say, but a silicon PN diode could drop that, or more, at 30 amps.

John

Since Skybuck appears to be even less intelligent but less anti-social, and meds can do that, I suspect he is Skybuck off the meds.

Now he's going to say I'm clearly wrong because of the time of the posts, but I can't even see most of them.

More like a volt for a 6A rectifier at that current.

Best regards, Spehro Pefhany

At 30A? ;-)

I didn't say it would do it for very _long_.. ;-)

Best regards, Spehro Pefhany

Strange claim.

Dont forget that some power diodes drop upto 1.5v each at full current, 0.6v is only the knee voltage.

NT

Because you are a presumptuous twit.

No, you are clearly nothing more than a complete and utter Usenet retard. It has nothing to do with the time of the posts, you stupid ditz.

A "6A rectifier" at 30 Amps would be a small explosion, and that's about it.

Nah, for a second or more it would be well within specifications. Then it would overheat if the current continued.

Best regards, Spehro Pefhany

ave your

=A0I

Not much point in series-connecting supplies anyhow--it takes 5 to barely double the output, per WrongLander.

-- Cheers, James Arthur

I like these for clamps:

http://61.222.192.61/mccsemi/up_pdf/1N4148X%28SOD-523%29.PDF

Ifsm =3D 2A, 1A @ 1s.

That's a pretty mean thing to do to an SOD-523 though.

-- Cheers, James Arthur

How could he forget? He never knew to begin with.