why is my full wave bridge overheating?

This rectifier should be able to handle all the current that the regulator can put out. I think something in the relay driver circuit (that is not shown in this schematic) is causing the trouble. Draw up the whole circuit and you might even see it, yourself. If not, I'll take a look at it.

1) how much current are you taking, how much volts have you got at the input of the 7805? Amps * V drop across the 7805, ie watts should be less than the max dissipation for the package 2)if you use a single rectifier you will get less volts at the 7805, so less heat, but more ripple 3) use a 9v transformer will be better

4) the jpg shows that the ground of the cap is not connected to the minus of the rectifier, it should be.

5)put a small capacitor after the 7805

apart from that seems ok

martin

It seems to me that the 24V AC secondary voltage is pretty high to feed a 5V regulator. The regulator will be dropping a rather large voltage and drawing a large current for even modest current loads.

Right. It'll be the *regulator*, not the bridge that's cooking. Which, funnily enough is what it says if you 'mouse over' it.

Also the reservoir cap voltage is too low btw !

Something like a 12 or 9 Volt wall wart would fix it.

Graham

Additionally from what other people told you, I would strongly suggest to use a 35V cap (at least), not a 25V.

24V AC * 1.414 = 33.9 Volts! You are massively overstrssing that poor capacitor... I'm wondering that it has not already taken off :-)

HTH Wolfgang

--
From-address is Spam trap
Use: wolfgang (dot) mahringer (at) sbg (dot) at

Red hot? Incandescent? Not at 180 F... Is "red hot" an exaggeration or is it literally true, that it's glowing red?

Well, for one thing, your 24 V AC, rectified, is probably about 34 volts peak, too much for a 25-volt capacitor, so that capacitor is going to fry soon.

In the meantime... You're asking the 7805 to produce about a 29-volt drop (from 34 to 5 V DC) at a current of maybe 200 mA (just guessing based on what you say about 9-volt batteries). That means you are asking it to dissipate 29 x 0.2 = 5.8 watts. That's too much.

Get a different transformer so the input voltage to the 7805 is much lower. Also, put the 7805 on a big heat sink.

Further to what I just said, I assume it's the 7805 that is running hot. If the rectifier is running hot, I don't know why, unless the capacitor is starting to break down and is about to explode!

If your schematic is true of your circuit... add a .1 capacitor at both the input and output of your voltage regulator... I built a circuit like that one before and, using a scope, found that the regulator is oscilating..... causing your heat problem.... Larry ve3fxq

slow_mac snipped-for-privacy@yahoo.com wrote:

In message , dated Sun, 13 Aug 2006, slow_mac snipped-for-privacy@yahoo.com writes

You have about 25 V dropped across the 7805. You don't say what current it's supplying, but the fact that it's getting too hot shows that the power loss in the 7805 is excessive.

If you can't replace the 24 V transformer by one with fewer volts out, you need a **power** resistor in series with the **input** to the 7805. Without knowing the current, I can't tell you a value, but you need to drop about 17 V across it. Sounds like a 10 W to 20 W resistor may be required.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Try using this program:

It will suggest a heatsink for your design.

--- You need to cut the voltage going into the 7805 down to around 8VDC.

Right now you've got 24VRMS from the transformer being rectified and smoothed, with the result that you've got about 32VDC on the input to the regulator _and_ that poor cap!!!

If you want 5V at the load then you need about 8V at the input of the regulator for its headroom, plus another 1.5V for the diodes, so that's 9.5V. Assume you need 10% above that for ripple and you've got 10.4V, and another 10% for low line gets you a total of about

11.5VDC. That's 8.1VRMS, which is what your wall-wart should be putting out with low mains. Add that 10% back in, and that becomes 8.9V, which should be its output with full load. There are a couple of ways to fix the problem, but what I'd do would be to get a 9VAC wall-wart rated for the maximum current your load will draw.

-- John Fields Professional Circuit Designer

1. Do you have or can you get a measurement of DC current through one of the DC output leads of the bridge?

1a. How many operating hours does a 9V battery last?

If you go through a battery in 4 hours, you could be pulling 160 mA. With a voltage drop (2 diodes) maybe 1.6 volts, that's possibly .24 watt. The temperature sounds only a little high for the smallest size bridge rectifiers that I have seen if the bridge is on a board connected by thin traces or thin wires and has things surrounding it that will stagnate the air around it.

1. How is that 7805 regulator doing? Is it getting screaming hot?

24 volts rectified/filtered may be 32-33 volts, give-or-take, and could possibly exceed - even if only at the peaks of the ripple - 35 volts. I am under the impression that the maximum input voltage of a 7805 is 35 volts.

Possibly the 7805 is breaking down.

And, even if the 7805 is not breaking down, if it has to drop 28 volts at .15 amp that is 4.2 watts - which will require a heatsink, and preferably not the smallest one you can get for a 7805.

If you measure the temperature of the heatsink or the heatsinking tab of the 7805 with a non-contact thermometer, beware that those thermometers do not do a good job of reading most bare metal objects. They do read painted metal well, a piece of masking tape on metal well (if large enough to fill the device's "field of view"), and black anodized aluminum at least fairly well.

I would suggest a lower voltage transformer if you can get one. If your

24VAC one has a center tap, I would use one end lead and the center tap of the secondary instead of both end leads.

- Don Klipstein ( snipped-for-privacy@misty.com)

...

--
Oops... If your load\'s going to be drawing its maximum current all
the time, I\'d get a wall-wart rated for twice that current.

1. The 25 volt capacitor has probably boiled a good bit of the electrolyte off and is probably drawing a hell of a leakage current. Don't question whether it is toast or not. Toss it and put in a 50 volt capacitor. YEs, you can get by with a 35 volt capacitor, but you are running it right to its limits -- and it wouldn't take but a burp on the supply line to run you over the top.

1. It is possible that the 7805 is oscillating. That draws a hell of a lot of current when it happens. In general, a 100 nf (0.1 uf) capacitor RIGHT AT the input terminal of the regulator RIGHT TO the ground pin of the regulator will tame the oscillations down.

2. You are probably dissipating somewhere between 3 and 5 watts, which is a fair amount of power. That device needs something on the order of a 20C/w heat sink to stay under the limits on the data sheet. Somewhere I thought you said you put a heat sink on it, but didn't say what you were using for a heat sink.

1. Push come to shove, if your load is constant you can split the heat between a series input resistor and the regulator. Cement power resistors are nickel and dime parts.

Jim

In message , dated Sun, 13 Aug 2006, Wolfgang Mahringer writes

Nah! He's wired it in backwards; that's why his diode bridge is getting hot. (;-)

A few days ago, I re-formed a 10uF 25 V to 55 V in an emergency. I took it slowly; an hour from start to finish, monitoring leakage all the time. It leaked a little less when I finished than when I started at 25 V, but of course it might not have lasted very long. It's replaced now with a real 63 V part.

--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

Exactly why the title of this thread can't be correct. It either has to be the regulator that is over heating, or some part of the circuit, not shown, is shorting out part of the bridge rectifier.

OK, I did not mouse over these things. I was not expecting temperature readings in the mouseover messages - I see something new so often!

If the regulator is dissipating a fraction of its max power dissipation,

180 F (about 82 C) is actually OK for temperature of the heatsink tab of the regulator - if it is accurate. I see a lot of thermometers read low on things like that for any of 3 reasons:

1. The thermometer does not have thermal contact to the part being measured to overwhelm heatsinking of the temperature-sensing part of the thermometer by other parts of the thermometer.
2. The thermometer heatsinks the part being measured enough to screw up the temperature reading.
3. The thermometer is a thermal infrared non contact one that is reading low from the item of interest being too small (width half an inch plus an inch per foot of distance from the "muzzle" of the "ray gun" for my model), or from reading bare metal.

Good looking - I did not catch that! I only looked at the schematic for a few seconds... That could make the bridge rectifier hotter, the transformer overheat, or simply blow! If it fails short, the transformer could burn up pretty bad!

Come to think of it, 2200 uF is a big filter capacitor for input if the current is low, especially if a regulator follows. Big filter capacitor means spikier current waveform through the rectifier and transformer windings (regardless of whether or not regulation follows the capacitor). If the current waveform through the transformer windings is very spiky, then the RMS value of the current can be a few times the average current. If the DC current draw is 100 mA, then the RMS current flowing through the transformer secondary could be as much as .4-.5 amp or so, give or take a fair amount - and a meter that does not brag about being "true RMS" will not show that. A 300 mA 24V transformer in this application could overheat if the DC current draw is 100 mA! (Its winding resistance will probably spread out the current spikes to get RMS current down to 350 mA I would guess.) A

500 mA one could get close to its limit if the capacitor is on the high side of its tolerance range or if the load current is more than 100 mA. And beware that with mild to moderate overload, a transformer may do its overheating awfully slowly! I would use a 1000 uF capacitor with a 12V transformer, or a 470 uF one if I had to use a 24V transformer. Heck, with a 5V regulator and .2 amp current draw and a 24V transformer, 100 uF is adequate!

And watch for those Radio Shack transformers (I mean ones with wire leads in and out rather than wallwarts) that get hot when you baby them and get cooking hot when you are barely within their limits!

- Don Klipstein ( snipped-for-privacy@misty.com)

This will make things worse for the transformer, another item that I would be concerned with, for 2 reasons:

1. The current waveform through the transformer secondary will be bigger less frequent spikes with a lower duty cycle, and with higher RMS current for same average current. RMS current through the transformer secondary may already be a few times the DC load current, and this RMS current is what determines I-squared-R heating of the transformer widnings.
2. This will put a DC component in the current through the secondary winding, equal to the load DC current. If this is significant (even a fraction of the transformer's output current rating) it can add a significant DC component to the magnetic field in the transformer's core, in addition to the AC one that with usual design is already close to the limits. The core could saturate and primary current could get out of hand. This excessive primary current could have a somewhat spiky waveform where the actual RMS current (what makes the widnings heat up) is 1.5 or 2 or more times that indicated with a meter that does not brag about "true RMS" AC measurements.

I think that is a typo on his part - I hope!

- Don Klipstein ( snipped-for-privacy@misty.com)

Oh yes, I see... ;-)

Ugh...you are a real cap-killer...hehe

regards, Wolfgang

--
From-address is Spam trap
Use: wolfgang (dot) mahringer (at) sbg (dot) at