9V 1A schematic needed

Larry Brasfield wrote: [..snip...]

You're the dodger pretending to have not understood that you were called due to complete the thresholding of the circuit. No one was saying that your trivial 4KTRB trash was in error- you were asked to do something with it, and you could not. But go ahead and revise the history- that would be typical of a total farce and sham like you.

No kidding, Sherlock, do ya' think? What a damned sorry-assed p.o.s.

Why don't you post a copy here- so we can have a good laugh.

[snip]

I'm with Larry on this one. If we draw a box around the capacitor and regulator, then the charge going into the box, integrated over one period has to equal the charge coming out of the box, integrated over one period. This is just conservation of charge.

The integral of the charge going out is fixed, because we have a linear regulator. So as we reduce the conduction angle, we must have a larger current for a shorter time, and because of the nature of the RMS calculation, it seems to me that it is impossible for Irms to go down when we do this. What am I missing?

--Mac

Mac doesn't need any help from the likes of you.

I don't think "impress" is the word for it- elementary trig integration is not all that arcane. Post it here and let us all have a looksee. What I can see so far from your generalizations of the results is that it is not very sophisticated and your understanding is pretty basic- like along the lines of "this gets longer makes this smaller" descriptive crap to be expected of a nobody.

I read in sci.electronics.design that John Fields wrote (in ) about '9V 1A schematic needed', on Sun, 20 Mar 2005:

Again, in UK, asking to 'close the account' gives the bank the problem of making sure it goes to exactly 0.00, even if at 23:59:59 they add bank charges.

We already shot down your so-called rationale as a typical troll dodge.

Not really- we understand you all too well.

[...snip evasion and dodging...]

You are a fake- that much is clear. If you could come close to finishing the detector easily, it would have require far less than the 10,000 words of bs excuse you have since posted. You can't because you are a fake.

PLONK Again!

average. Is that right? Im^2 is a component of Irms^2? That doesn't look "independent" to me...and when you further particularize it to the charging situation here where Im=Load Current, you are even more wildly off the mark. Hey! Link us to that learned treatise you posted in rec.audio- we want to see more of your "wisdom". Or do you prefer to keep dodging on that one too?

Is this what you're talking about?

I wouldn't bet on that. Evidence is that Fred knows a lot is not any sort of dummy when he stays on his meds. There also seems to be some evidence that he doesn't like you.

In article , Active8 wrote: [...]

I quote "Errors in our favour are not errors"

Have you ever had a bank give you extra money?

....should be in that Irms^2 expression.

"Active8" wrote

"SABER" is not free! BUT LTspice is! same result. ALSO: Three other Engineers verified my finding.

I do not use three terminal regulators, they do not work well with 100 volt plus inputs or from -55C to 125C!

Besides, No problem with the power supply. Did replace the POR though, it oscillated at high temp! (Not my design)

Ooooh my- am I reeling from the insult of being P-LONKED by such a personage....

I had a bank in San Jose credit my account for over $400 that wasn't mine, in 1968.

Told them about it, they said they'd correct their error.

This went on for months with no correction.

When I went back to Motorola I closed the account and walked with the money ;-)

...Jim Thompson

are

currents

John, leakage inductance in rectifier transformers is a subject of interest to me. I am moving house right now and I need time to collect my thoughts anyway, so I would like to post or email you in a week or two when I can make more sense.

Roger Lascelles

Big damn difference between your weak Savg

There are several examples in there, some of them boiling down to an open-loop emitter follower tethered to a zener reference. One of the designs there, the 53V version, uses feedback and I like that one as a basis for a 9V supply.

I'm not a designer, just a hobbyist, but here is how I'd proceed from that 53V design:

(1) I like the idea of using the Darlington topology, with a cheap and readily available 2N3055 for the power pass transistor. I might also like this because I still have a few laying around.

(2) The other transistor can be a 2N3904 or 2N2222. At Ic=1A for the

2N3055, I'd expect no better than beta=30, so the Ic on this other transistor would need to be about 33mA. Since Vce might be anywhere from 2.5V to perhaps 6V or so, this suggests that the transistor should be able to dissipate at least 1/4 watt, itself. I think both of those mentioned above can do this ... but not much more. So watch this.

(3) 1A is a lot of current. Radio Shack sells only two kinds of transformers (12.6V and 25.2V) and its their 12.6V ones that come in

1.2A and 3.0A versions. With the 12.6V transformers, you'll need to use a bridge rectifier and with an average current of 1A, you'd be better to select the 3A version since the peak currents will probably be in that area.

(4) At 1A average load, a 12.6V transformer is going to need a pretty seriously sized filter capacitor to get up to and hold a supply rail for the circuit. The peak voltage would be about sqrt(2) *12.6V or

17.8V, less a pair of diode drops at high current, or about 2V less to say 16V or so. There are books with charts to help you figure out exactly what size the capacitor should be for some desired ripple (you can work it out with trig) but my rough guess takes about 1/3rd of the 60Hz time (say, 5ms or so) and then takes some desired ripple voltage (let's say 1V here) and calculates C=dt*I/dV, which would be 5ms*1A/1V or 5000uF. Radio Shack sells 4,700uF at 35V. That should be okay.

(5) The feedback method shown in the 53V supply on the web page looks reasonable to me. It's basically a zener being compared to a portion of the output voltage by way of a transistor that then pulls the Darlington base lower (more current) if the voltage climbs much above the reference point. However, I'd just use a zener from 6.2V to 9.1V or so, not the 15V (of course.) Call it a 6.2V zener for now. Here again, the transistor can be a 2N3904 or 2N2222, I believe.

(6) The zener needs a current through it. I'd go with 1mA. The data sheets often spec things at 20mA (Izt) but 1mA has worked well in the past for me. You can redesign for larger Izt, if you want. This current establishes the Ic for this transistor comparator and that, together with the minimum estimated difference between Vin and the base of the Darlington pair provides the size of the collector resistor. With a ripple of about 1V, going let's say from 15V to 16V, the minimum Vin will then be 15V. The Darlington pair will require some 1.25V or so, so the base should be at 9V+1.25V or about 10.25V. The difference is then 4.75V. At 1mA, this suggests a 4.7k resistor in the collector. I'd go with even less... a 3.3k here, just to be absolutely certain that at least 1mA was available for the zener.

(7) The base of the comparator transistor needs to divide the output voltage down. And it would be nice to have some slight adjustment range, so I'd recommend a potentiometer in here, as well. With a beta of about 200 or so for the transistor, I'd estimate that a current through this divider that is about equal to the zener current should be okay... another 1mA or so. With 9V on the output, this suggests about 9k ohms total. A little less would be okay. In any case, this seems kind of convenient for a 1k potentiometer (common value) to give us some variable range for adjustment. The base voltage should be

0.7V above the zener voltage, so about 6.2+.7 or about 6.9V. Actually, because of the light base current, it will probably be a little less than this. This suggests that we want the grounded side resistor to be 6900 ohms (assuming 9k total) and the high side to be the remainder, or 2100 ohms. However, with a 1k pot in the middle, and set to the midpoint, we'd take about 500 ohms off both sides, so we'd like 6400 and 1600, perhaps. Using 5% resistor values, I'd select 6200 ohms and 1500 ohms and put the 1k pot in the middle of the two.

(8) As shown in the 52V supply on that web page, a small cap would also be nice across the zener.

(9) On the output, we should have another filter capacitor. I'd use a large value, perhaps 100uF or so.

All this arrives at a design looking like:

,------, ,-----| |-------+----------+-------------+------------, | | |(+) | | | | | | | --- 4700uF \ | |

12.6V | AC | --- 35V / 3300 | | AC | | | \ |/c NPN | | |bridge|(-) | +-----------| 2N3904 | '-----| rect |--, --- | |\e | '------' | /// |/c NPN | |/c NPN | ,-| 2N3904 '----------| 2N3055 --- | |\e |\e /// Vout | --+-----, | | | | | | | | /---/ | +---- Vout \ | / \ --- 10uF | 1500 / | --- --- --- 100uF \ | |6.2V | --- | | | | | | | --- --- --- \ | /// /// /// 1000 /