why is my full wave bridge overheating?

On my last visit to the local dump I found several 5V switchmode wallwart units that had been thrown out with the equipment they supplied. Some of these units were rated for outputs of several amperes and were smaller and lighter than lesser rated iron transformered wallwarts. Small light more accurately regulated than most wallwarts and very efficient - what more could you want?!

If you have a municipal refuse container site nearby its well worth a look as these units turn up much more often than they used to!

You have two problems:

1) As noted in other posts the working voltage of C1 is way too low.

Calculate the PEAK voltage output of your 24VAC transformer.

With an RMS voltage of 24VAC the peak voltage is (24 * 1.414) or about 34 volts peak.

2) The voltage drop across the 7805 is 29 volts, this is too high.

At a load current of 300mA this will result in about 10 watts of power the regulator must dissipate.

With a large enough heat sink this may just be acceptable but your best solution is to use an input voltage closer to 8 or 9 volts peak.

Have you connected the diodes the right way round?

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and

them.

Also bear in mind that the transformer will produce 24V at its 'rated load'.

With small transformers lightly loaded that voltage could easily be 20% higher. That would be ~ 40V !

Graham

OK here's an update. First, thanks for the generous feedback.

It is the regulator that's getting hot, not the bridge.

I chucked the 25V cap and found a 470uf 63V cap in my box and put that in there (it does have the negative pin connected to ground - I will update schematic). I also found a .001k ceramic disc cap and placed it between pins 1 and 2 of the regulator.

I'm not sure what the value of a ".001k" cap is. I also have these:

104Z 50v, 102K 1kv, 103M KCK, 104k 1KV, and one that just says 331. I understand the 1kv = 1000 volts but I'm not sure what the values are. If elected president of the narcoleptic neophyte society, my first legislation would require manufacturers to print all values in microfarads so a .001uf cap would say ".001uf" on it.

I don't really have anything to measure oscillations. If I hit the lotto I will run straight to the Fluke dealer to pick up a $3500 Scopemeter. I'd love to get one of those and sign up to be a cadet at the Scopemeter academy. Then I could look at oscillations and zoom in and decode words in an RS232 data stream and be delighted beyond imagination.

Back on earth, I found a cement block power resistor in my box, a 10W

2.7 ohm. I put it in series between the bridge's positive DC output and the regulator. Surprisingly the cement resistor does not even get warm (but it obviously is not blown because the circuit powers up). I think I may have this wrong as I have seen a load resistor on some schematics that connects to ground. Should it go one end to the bridge and one end to ground or in series as I have it now?

The transformer is actually a 26v AC 300ma output and it feels OK to the touch with no appreciable increase in temperature.

My 7805 is in the TO-220 package and the heatsink is one of those small clip-on types. I cannot seem to locate a picture of this type. Something similar would be this:

I do have some white thermal contact paste on it.

There is a reason I'm using a 24v AC transformer. Mainly it's because I don't want to run two power supplies and the item ultimately being powered via relay requires 24v AC. I have to be cautious as to how much I reveal of the entire circuit because -sooner or later it seems like someone always winds up saying "just go out and buy one off the shelf, Pokey, you idiot" and I'd like to make this work.

It would be nice if I had a way to measure the current being consumed but my cheap DMM does not have this function. I found these specs from the relay's data sheet:

Minimum Contact Load: 10mA @ 5VDC. Initial Contact Resistance: 100 milliohms max. @ 100mA, 6VDC.

Nominal Coil Power: Contact rating 3 = 200mW. Contact rating 10 = 450mW.

Throw in a friendly 2n2222, a midrange pic processor, and a max233 and you pretty much have it. I will try to work on a complete schematic with the crappy but free software that I have.

Speaking of DMM, why do I read 10V AC at the 5V DC regulator output? I thought the bridge and filter cap knocked off any AC.

The term "red hot" is misleading as the part does not actually glow. I will now reference it as "bitch hot" which should clear things up.

Also, this isn't "let's go spend $50 at digikey" to fix the problem. In my dreams, I order my $3500 Scopemeter from digikey and become a cadet. This is more like "let's see if Pokey can dig enough parts out of his junk box to make this relatively simple thing work reasonably".

BTW, the new reading after doing all of above is 170F at the heatsink tab so there has been an improvement, but still seems high. This is measured with a non contact thermal thermometer so accuracy is what it is.

I think I may have a 12V or 13V zener diode around. What about replacing two of the 1N4004s used in the bridge with two zeners? I think this would then reduce the voltage to the regulator.

Pokey.

It's 0.001uF or 1nF ( 10% tolerance - that's the K ) and it's distinctly on the small side.

The value in pF is the first 2 digits plus N number of zeroes for the third digit. So 104 is 100000 pF =100nF = 0.1uF

The resistance is too low for it to warm appreciably.

In series. It's a 'volt dropper' not a load.

That'll take 40mA @ 5V for the coil which is peanuts really. Although, with your input volts @ 26V ( did you measure that value btw ? ) it'll result in 1.25W dissipation in the 7805 which is still fine ( esp with a small heatsink ) but your micro will take some more of course.

That may be your oscillation !

Stating temps in degrees C is preferred in science and technology btw. I think only the USA still uses Fahrenheit. The regulator won't actually die at that temp by a long long way. They're good for an internal temp of 125C.

Then the zener(s) will get hot instead ! You'd be better off putting one between the electrolytic cap and the regulator. At 40mA it'll dissipate 480-520mW which is the absolute max for a small zener.

Graham

Do you have any non polarized capacitors? A non polarized capacitor can be put in series between your transformer and the bridge, this will drop some voltage without creating the heat and losses of a resistor. A non polarized capacitor can be found in a speaker crossover circuit. Someone on the group has the knowledge to calculate the proper value, I would put in the 22uf 63v cap in my junk box and then decide if it needs to be smaller or larger. Mike

JUST GET A 9 V WALLWART.. grrrrrrrr, whats the average salary of a dude here, and the discusiion is a bloody 7805!

Lets get back to the WAR, beer, t*ts, BBQ formulae, 42, poisoning pigeons in the park, but not 7805s

sorry Pokey:-)

martin

Good move, will reduce transformer heating (not that I am sure there is or will be an actual problem there) as well as being good by having a voltage rating exceeding the voltage it will experience.

Less important, but I have always been able to get a scope for less by an order of magnitude to a bit more! (Think used scopes that are not really modern!)

In series. Also, I think 2.7 ohms is so low as to make very little difference. Maybe something in the 10 to 100 ohm range will do that job.

Try both before and after the filter capacitor - before is better with a lower value (as well as will reduce RMS current drawn from the transformer by lengthening output current pulses).

Good - same story after a couple hours running?

That's an awfully small heatsink that I estimate to be good for half a watt or a watt - minor improvement upon no heatsink at all.

Actually, I consider that a good reason! Now just get a bigger heatsink for the regulator!

Two guesses:

1) Is this the same if you switch leads of the DMM? Do you get the same reading if you put a small capacitor (under 1 uF and nonpolarized) in series with one of the leads of the DMM? If so, this means the 7805 is oscillating. It is widely suggested to have capacitors (preferably largish ceramic like .1 uF) from 7805 input to ground and 7805 output to ground, as well as to keep ground paths nice and short and thick within a circuit having an IC or for that matter an amplifier stage. Some multimeters have an "output" jack which has a capacitor to the normal "+" jack. That jack is used to get AC-only voltage when the voltage is or is suspected to have mixed AC and DC.

2) DC is read by many multimeters when the meter is on an AC range, and in some cases as roughly double voltage. Some others see it as about

10-11% high.

Just as an experiment, only for $#!+s and giggles, put one of those zeners in series with the positive DC input to the regulator, "against the flow" to drop that 12V. It should take about half the heat load from the regulator - but could be unable to survive that.

I would say mainly get a bigger heatsink for the 7805.

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

Thanks. Details like that count.

Good.

Use the 104k (100,000 pf, or 10 and 4 additional zeros picofarads, k=10%).

(snip)

That is where it needs to be, with the 470uF cap at the regulator input end of the resistor. But the value is so low that it probably drops only a small fraction of a volt You have more than 10 volts to get rid of. Try something between 47 and 100 ohms.

Keep in mind that there is no clean and simple relationship between the 5 volt supply and the 24 volt winding voltage. I am curious how you are going to connect the 5 volt circuit to the 24 volt AC supply.

You say this is a 24 volt AC relay? You can use a low value resistor, like that 2.7 ohm or a 1 ohm, if you can find one. Put it is series with the current in question and measure the voltage drop across it with your meter. Use Ohm's law to calculate the current from the resistance and voltage drop (I=E/R).

Won't work. They will throw a high current load across the transformer when they break down.

If you find an appropriate value resistor to put in series with the regulator, that drops its input voltage to somewhere between 8 and 12 volts, you might use two of these in series at the regulator input, to limit the peak voltage at the input of the regulator to no more than

24 to 26 volts, if the regulator load gets disconnected. But they won't make the regulator run any cooler under load.

Come to think of it, you might be able to use one of those zeners in place of the dropping resistor, to burn up 12 or 13 volts before the filter capacitor. But then it will get hot. maybe not too hot. But the correct resistor will lower the effective heating load on the transformer more, by extending the charge time for the filter capacitor, each half cycle.

As to your dreams, you might look through some of the hundreds of scopes for sale on eBay, every day. I've bought 3 or 4, cheap, and they mostly work.

I picked up a 22xx Tek ( one of that family - quite basic - for my impecunious client ) for ~ £110 on ebay.

Graham

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It\'s where it should be, but it\'s not dropping enough voltage
because it\'s too small.

If you circuit is drawing 200 mA, that will give you approx 5V drop. That's meaningless when compared to the 25V drop on the regulator.

Try a 10 ohm, it will be hot.

In that case, go for a switching regulator. A typcial 90% switching regulator will be on 10 % to 20%. You will save lots of heat and energy. For example, try the lm3485 circuit at

[...]

The 170deg heatsink says you're taking about 70ma from the +5V. This arrangement should run much cooler john

1N4004 ___ 26Vac o---|___|---->|---o-----o 18V to 7805 in | 56ohm 1 Watt | | [Transformer] | --- --- 2200u | 26Vac o-----------------o-----o 0V

(created by AACircuit v1.28 beta 10/06/04

impecunious

I'm not impecunious, just 'cheap' (I do like that American phrase) .Last week picked up an Advantest 1.5GHz spectrum analyser with built in tracking generator and digital storage. £190 :). Been drooling over an advert for this model since 1988 and watching Ebay intently for the past 3 years.

3 cheers for Ebay!. john

How did you get that number (temperature -> current)?

24V AC would rectify to approx. 35VDC. 70ma on 56ohm would drop only 4V. You still have 31VDC to the 7805.

The transformer isn't going to like the DC passing through it, though.

2.7 ohms at 200 mA will drop .54V and read that much to a voltmeter. (And if this is true will decrease 7805 regulator heating in the previously described application by 2.5%.)

If the resistor is upstream of the positive terminal of the main filter capacitor, it can drop instantaneously and even charge-weighted-average several times that. However, I would agree that more is needed by a factor of at least 10, maybe 20 or somewhat more. (The instantaneous/average current ratio will decrease as this resistor value is increased to decrease current.)

10 ohm may produce noticeable heat, maybe get noticeably quite warm, and probably do so while taking maybe a mere fraction of the heat from the regulator and some heat from the transformer windings. That is, if the resistor is between the rectifier and the filter capacitor. If the resistor is downstream from the filter capacitor, it will experience lower instantaneous current and lower RMS current (by experiencing only the DC load cuirrent and regulator low leg loss current in the form of steady DC) and drop so much less voltage as to heat up less despite passing current more continuously. If you have some need for this added resistor to be downstream of the main filter capacitor, it will need to be of a higher value than if it is upstream of the main filter capacitor. Downstream location will have the resistor take heat only away from the regulator barring malfunction. Upstream from the main filter capacitor will reduce heating of both the regulator and the transformer. Keep in mind that transformer output voltage before any resistive and diode losses times average current, with voltage appropriately weighted for appropriate weighting in such an average (likely to be close to 1.2 times the open-circuit AC voltage), is power consumption in the sum of the load, resistors, rectifier, and transformer secondary winding. If you draw 100 mA DC from a 24 VAC transformer supplying a rectifed-regulated DC power supply with 5 VDC output, plan on close to 2.5 or even maybe closer to the worst-case roughly 3 watts being generated overall - not including heat generated by the load nor the transformer core nor in the transformer's primary winding. Multiply by about 3.42 to get BNTU per hour.

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

The key was that heatsink pic. That particular heatsink style usually runs about 25 degC rise per watt of dissipation. So, (tediously!) convert from the DegF number into DegC and get 76degC. Assume room temperature at 20degC, hence we have a 56 degree heatsink temperature rise. This implies 56/25 or 2Watts of heat being dissipated by the 7805. So ... 34V in, 5v out is a 29V drop. So ... 2W = 29V * ?amps ... =68.96ma

Under -unloaded- conditions, Yes, this would be true. Resistor or no resistor, the cap would charge to peak voltage. But ... when load current is taken, the whole picture changes ... the big cap now needs a significant charge top up every cycle, (remember, only one diode).

This topup current has to flow via that resistor. The resistor drops a voltage which has to be the incoming rising half sine voltage on one end and the (pretty steady)capacitor voltage on the other end. It severely limits the current that can flow into the capacitor, which results in the capacitor being unable to charge to the peak incoming voltage and hence settles out in the 18V area (in this case).

The resistor is essentially dropping 18V of the incoming half sine and the diode input never gets to see an input voltage >18V. These are not clean symetrical waveshapes but relate to raised cosines, smooth pulses etc and normal 'linear' electronic calcs fall over. Component and circuit wise, it's straightforward but an exact mathematical analysis of this apparently trivial circuit could take a chapter. (Spice comes is perfect for these types of circuit :)