I am trying to build a supply using lm 338 regulator.
For the bridge,i used a BR68 bridge rectifier.
before soldering all components on pcb, i checked the bridge o/p and it was a pure dc 18-19 v(as i am using 18 v transformer).
but, after completeing the circuit, the o/p is constant and it is 26 V DC and 54 V AC....I am just unable to think how this happened??
also...as i turned the 4.7k pot,it started to burn!!..
its just irritating when something like this happen...

 

Hello,

You made a dangerous mistake.
There is a connection to the mains that needs to be removed.

Greetings,
Bertus

 

@bertus

OK..
but i didnt connect it that way...i have made it the same way u said...the transformer o/p is isolated...i didnt connect it to the mains...
i used a simple 18-0 V transformer.

 

Bertus is right; that connection was dangerous.

However, it won't simulate properly, as SPICE will see the primary side of the transformer as floating. You can get around that by adding a 10MEG resistor from the primary side to GROUND - just as long as you understand it's only there to allow the simulation to run.

Note also that the MOV (V1) you're using is rated for 275v, however the P-P voltage of the incoming AC will be higher than that. If you multiply 240V x 1.4143, you get > 339.4v P-P. You should use a MOV rated for 350V, or it will cause the fuse to blow during normal operation.

As far as the output of the transformer:
You say it is rated for 18VAC.
It should measure about 18VAC. Then after the bridge rectifier (without the capacitor), it would measure around 17V DC, but it would be rippled DC. When you add the cap in, you would suddenly be reading about 24V DC. This is because the cap charges up to the peak voltage of the rippled DC.

You are using 120 Ohms for R1. This means there will be at least 10mA flowing through R1, and perhaps as much as 10.833mA, because Vref (the difference between Vout and ADJ) may vary from 1.2v to 1.3v between regulators.

If you have the pot set for a higher resistance to get more voltage out of the regulator, you might have up to 22v or so across the pot. Since P=EI, E=22V, I=10.8mA, you might be dissipating 237.6mW power in the pot. You would need a pot rated for 1/2 Watts.

You can reduce the pot power requirement by increasing the value of R1, however this means that you will not get guaranteed regulation out of the LM338 until the minimum load of 10mA has been met. The output voltage may be somewhat higher than it is supposed to be until the minimum load requirement is met.

Let's say you increased R1 to 330 Ohms, and let's say that Vref is the nominal 1.25v.

I=E/R, so I(R1) = 1.25/330 = 3.7878...mA

There is also a small amount of current that comes from the ADJ terminal. Nominally, it's about 50uA, but it might be as high as 100mA. Anyway, let's just add 50uA to the result we got before:
3.7878mA + 50uA = 3.8378mA
Now, since E=IR, and R=E/I, we can figure out that if E=1V, and I=3.8378mA, for every 260.5 Ohms of resistance between ADJ and GND, the output will increase by 1v. The minimum output voltage will be the same as Vref, or 1.25v (nominally).

If you wanted a maximum of 22v out, and Vref is 1.25v, then you would need a pot that is (22v-1.25v)*260.5 = 20.75*260.5 = 5.4k (rounded off)

A plain 5k pot would give you a maximum output of 19.2v+Vref, or 20.4v. (roughly)
Maximum power dissipation in the pot would be 19.2v x 3.8378mA =73.7mW; double that for reliability's sake ; around 150mW.

 

That was a superb explanation from you sir.
Thank you very much!!

means...u r saying that if i need o/p voltage only till 20V, i can use a 5K pot but i wud need to increase my wattage rating for that....

i used the following pot-->



I dont know it is of what watt.
What would you suggest as the rating for the pot?


I just burned another pot of 4.7k.
But i think that pot have a rating of 2 W..
So why do they burn?

 

If you change R1 to 330 Ohms, you will need a 5k pot rated for 150mW or more.

If you keep R1 at 120 Ohms, you will need a 2k pot rated for 1/2W or more.

I do not know what your pot was rated for. You would have to find the manufacturer's datasheet for it. But, since you were only able to use about 2/5 of the range of your 5k pot, you would have to de-rate the manufacturer's specification, as they give the specification for when the pot is set to maximum resistance.

For example, if your pot is rated for 250mW (1/4W), but you are only able to use 2/5 of the resistance of the pot, the effective rating would be 100mW.

 

Thanx a lot sir.

I will change my pot tomorrow.
Hope i will get it from the store.

Thanx again.

 

You burned up a 4.7k 2W pot?

You must be connecting something incorrectly, or you are trying to change the pot setting too quickly.

It would help if you posted a photo of how you have things wired.

 

No sir, I didnt burned a 2 W pot.

I was just saying that i burned another pot.

also i said that i think a pot would be of 2W.

anywz, I didnt get how I get 50V AC at the o/p?

 

Ack! 50VAC at the output??

I somehow missed that part.

Unplug the transformer from the wall.

See if there is continuity between the primary and secondary side of the transformer.

Make certain that you have the BR68 bridge rectifier hooked up properly; AC connections to the transformer, and proper polarity of the DC outputs to your filter cap(s)

 

More on the bridge rectifier - don't trust the markings on it. I have purchased bridge rectifiers before where the markings were applied incorrectly. You can verify each pin to every other pin using the diode test function of your multimeter.

 

As i had used a 10,000uF(35V) cap on the pcb, after unpluging the mains, the cap stored charge and it showed 25V dc and also 50 V ac.....i didnt get how a cap stores ac voltage??while measuring the voltage across the cap terminals, i accidently shot the two rtacks with the measurment probe and it gave a bigh spark with a loud noise.

Strange!

is my br68 damaged??

 

OK, it sounds like your meter is an inexpensive one, and it is out of calibration or broken.

If you are measuring 25v DC, with an inexpensive meter you should measure about 17.7VAC. Inexpensive meters do not read true RMS. Expensive meters that measure true RMS cost around $100 USD or more, and would measure 0VAC across an un-powered capacitor, whether it was charged or not.

A capacitor does not store AC. Once power is removed, it will remain at a constant DC level, but due to internal leakage, it will slowly discharge over time. Since your cap is quite large, the discharge may take a while.

You can safely discharge your capacitor by using a 5k Ohm resistor to short out it's leads. It will take a while for the cap to discharge, because even at 25V, the current through the resistor will only be 5mA. You can speed up the discharge time quite a bit by using two 12v automotive light bulbs of the same type wired in series, and connect them across the capacitor. Side marker lamps are small and would work quite well.

 

sir,
i came to a conclusion that my bridge is damaged....
i used br68, its rating is 800V reverse voltage and 6A.
i dont understand why its damaged?
can u plz suggest me a bridge rectifier for this application or which diodes shud i use to make a bridge?
will 1n4001 work or will i have to try 1n4007?

 

I'd missed the 35V rating on the capacitor before.

This is not enough. You need a 65V rating or better for the capacitor; otherwise it's leakage current will be too high, and it will fail quickly.

Your bridge may have been damaged by the high initial surge current when trying to charge the 10,000uF capacitor.

Or, it could have been defective prior to installation.

1N400x series diodes are only rated for 1A current.

 

can i use a 3300uF(65V) cap of for this??
the shop i buy from does not have 10,000 cap of 65V
he says nobody uses it now-a-days....
he sugested me 3300uF cap...
can i use it??


and about the diodes.(for bridge)..which diodes can i use for 5A capacity?

Thanx.

 

Certainly, you can use the 3300uF cap. You can use two or three of them in parallel if you wish.

Ask the shopkeeper for a rectifier bridge that is rated for at least 5A continuous current; 10A would be better, and at least 100PIV.

 

Thanx a lot sir...

i will rebuild it and then post....

Thanx again....

 

Sir,
I brought a 6A diode, Made bridge, used 3300uF cap,Added a 1W POT.... this time, the output is 0V AC...but still, the pot burns...Am really worried...
Plz suggest something.

 

How much DC voltage do you have over E1 (It normal to mane caps with C)?