Maker Pro
Maker Pro

why is my full wave bridge overheating?

24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

Pokey.
 
J

John Popelish

Jan 1, 1970
0
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

Pokey.
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.
 
M

martin griffith

Jan 1, 1970
0
On 13 Aug 2006 11:20:11 -0700, in sci.electronics.design
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

Pokey.
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
 
G

Greg Neill

Jan 1, 1970
0
John Popelish said:
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.

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.
 
E

Eeyore

Jan 1, 1970
0
Greg said:
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
 
W

Wolfgang Mahringer

Jan 1, 1970
0
Hi Pokey,

24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F.

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
 
M

mc

Jan 1, 1970
0
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

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.
 
M

mc

Jan 1, 1970
0
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!
 
L

larya

Jan 1, 1970
0
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
 
J

John Woodgate

Jan 1, 1970
0
In message said:
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.
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.
 
H

HKJ

Jan 1, 1970
0
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

Try using this program:
http://www.miscel.dk/MiscEl/miscelPowerSupply.html

It will suggest a heatsink for your design.
 
J

John Fields

Jan 1, 1970
0
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

---
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.
 
D

Don Klipstein

Jan 1, 1970
0
24V AC to 5V DC. I uploaded comments plus schematic here:

http://www.flickr.com/photos/33356032@N00/214189669/

Works OK but gets red hot and I'm afraid to run it long. I measured
the temp at 180 F. It connects to a small circuit that has a relay and
a microchip in it which used to be powered with a 9 volt battery. I
was going thru a battery every day or two and I'm tired of buying them.

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.

2. 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 ([email protected])
 
J

John Fields

Jan 1, 1970
0
On Sun, 13 Aug 2006 14:18:28 -0500, John Fields

....
 
R

RST Engineering \(jw\)

Jan 1, 1970
0
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.

2. 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.

3. 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.

4. 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
 
J

John Woodgate

Jan 1, 1970
0
dated Sun said:
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 :)

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.
 
J

John Popelish

Jan 1, 1970
0
Greg said:
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.

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.
 
D

Don Klipstein

Jan 1, 1970
0
Right. It'll be the *regulator*, not the bridge that's cooking. Which,
funnily enough is what it says if you 'mouse over' it.

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.
Also the reservoir cap voltage is too low btw !

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

- Don Klipstein ([email protected])
 
D

Don Klipstein

Jan 1, 1970
0
martin griffith said:
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

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.
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.

I think that is a typo on his part - I hope!
5)put a small capacitor after the 7805

apart from that seems ok

martin

- Don Klipstein ([email protected])
 
W

Wolfgang Mahringer

Jan 1, 1970
0
Hi John,

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

Oh yes, I see... ;-)
A few days ago, I re-formed a 10uF 25 V to 55 V in an emergency.

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

regards,
Wolfgang
 
Top