power supply 2

By the way, if the output is 25v, and the resistance of the coil in the secondary is 2 Ohm, does that mean I am pulling 12 amps on the secondary ?? if so no wander its getting hot, but the transformer is rated 2 Amp.

ken

On Mon, 17 Jul 2006 19:40:52 -0400, in message , "Ken O" scribed:

The coil resistance is irrelevant, or at least it should be, because the transformer should be seeing only AC current. The current must be deduced by calculation of the parameters of the circuit after the rectifier. The schematic you've included does not show what is going on; I can't make any sense of it. Could you try to be a little more specific about your circuit description?

--
Love is like a dying ember
And only memories remain
And through the ages I\'ll remember
Blue eyes cryin\' in the rain

Normally, the coil resistance drops only a few percent of the total voltage when the transformer is operated within its ratings. Do you have an amp meter that reads over 2 amps that you could put in series with the battery being charges, to see what current is charging it? You also probably don't need that big capacitor between the rectifier and the battery, and having it makes the transformer hotter and raises the charging current a bit. I suspect you do need some resistance in series with the battery (or in series with the primary of the transformer) to set the charge current to below what the transformer is rated to supply.

I'm not sure why its getting hot but remember that if you are dropping the voltage on then you are increasing the current.

you you have

120/25 ~= 5

Your secondary is drawing 5 times the current then the current on the primary side. So make sure the transformer is rated to handle this. You can use a fuse that is ~1/5(not sure exactly how to compute fuse rating cause many people say different things...) of the maximum current that you will draw form your secondary side circuit.(fuse, ofcourse, goes on primary side). You'll want a slow blow fuse because there will be an inrush of current to charge the cap.

Now, if each cycle you are discharging a large portion of the current then on recharge you will be pulling a large portion of current through the transformer. So you shouldn't use any larger of a capacitor than needed.

You really should make some measurements though and see just how much power is being dissipated in the transistor and transformers.

If your trying to discharge the full value of a large cap very quickly with a very small time constant then theres a lot of peak inrush of current and over time the avg current could be larger than what you anticipated.

Maybe your design calls for a larger transformer and transistor?

--
So far, who knows?

On Mon, 17 Jul 2006 17:39:33 -0700, in message , Alan B scribed:

On re-reading, this sounds confusing. I'll break it down.

1) The resistance of the coil should be negligible compared to the coil's inductive reactance; that is to say, when pulling AC through the transformer, the secondary will have an impedance that should be significantly greater than its simple DC resistance. If you are letting DC run through the transformer, then you have a circuit design flaw.

2) You must "do the math:" calculate the expected average current in the post-rectifier circuit, and make sure your transformer, rectifier and filter are all rated to handle that calculation.

3) Is that a battery in your circuit? Why do you have both a DC power supply and a battery? Where are your power inputs to the circuit you have drawn? Is your transistor switch a direct short from Vcc to Vee when on? What is the duty cycle? Is that capacitor you have drawn part of the switch circuit, or part of the power supply filter? Is what you have labeled "ground" actually attached to anything, or was that just a handy schematic graphics tool?

--
Love is like a dying ember
And only memories remain
And through the ages I\'ll remember
Blue eyes cryin\' in the rain

"Alan Bullshit"

** Oh dear - wot a f****it.

......... Phil

"Alan Bullshit"

** Nonsense.

The secondary source impedance of a mains frequency transformer is

*dominated* by its resistance.

Only the primary winding acts like an inductor when the tranny is UNLOADED.

....... Phil

On Tue, 18 Jul 2006 13:14:43 +1000, in message , "Phil Allison" scribed:

Er, how is the example in question an unloaded transformer?

--
Love is like a dying ember
And only memories remain
And through the ages I\'ll remember
Blue eyes cryin\' in the rain

"Alan Bullshit Fuckwit"

** How pathologically asinine.

This PITA is a completely anencephalic public menace.

FUCK OFF !!

...... Phil

On Tue, 18 Jul 2006 13:29:52 +1000, in message , "Phil Allison" scribed:

A simple "I don't know" would have sufficed.

hi I replaced the capacitor with a 680uf 200v. I did not need 200v but that was the only one over 50v I could find. The transistor is acting normal now, not too hoe, The transformer is not getting as hot as before but I still find it too hot. I cant measure the current. I am using the impulses from the 555 timer to charge it. Unless there is another way beside using an amp meter, please let me know

ken

Okay.

Have you tried with no capacitor? No particular need to have smoothed DC to charge a battery. After you get the transformer temperature down, we can get into what you have to do to power the 555 from the same rectifier.

If you can find a very low ohm resistor, like .1 ohm, you could put that in series with the battery, and read the average DC voltage across it. The average current is 1/R times the voltage you read.

But I think you need something like an automotive head lamp in series with the battery, to act as a current limit (with a visual indication of high current).

I removed the capacitor:

-The battery is charging a bit slower then before

- The transistor is cool

- The transformer is getting warm just like the small capacitor used before.

the transformer is a 115v to 25.2 v rated 2.0 A So this some out to putting a 12.6 Ohm resistor @50 watt. But this is AC not DC.

beside adding a fan next to the transformer , I do not know what to do. and why it is getting that warm.

Ken

A 25.2 VAC winding puts out 1.414 times that much voltage at its peaks. So the rectifier puts out about 27.6 volts after the .12 volt rectifier drop.

What is the nominal voltage of the battery being charged?

Transformers got warm / hot on account of I^2R heating mostly.

That indeed suggests the load current is greater than the transformer was designed to deliver.

Graham

I have to turn off the charger , the transforme was getting very hot. The transistor is still cool. But the last design charges a lot slower without th capacitor. The battery being charged is a deep cycle marine battery with 650 cranking amps. Not sure what you mean by nominal voltage

ken

"Eeyore" wrote in message news: snipped-for-privacy@REMOVETHIS.hotmail.com...

How do I turn down the voltage on the secondary? Should I just put a resistor between the capacitor and the battery being charged?

ken

(snip)

If you measure the battery voltage with a meter, when it is charged, what do you measure?