why is my full wave bridge overheating?

[Don Klipstein]

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.

The usual 7805 in a TO-220 package can dissipate 20 watts or more, with
an adequate heatsink (quite large). (Not that I consider having a 7805
dissipating so much to be good practice!)

However, I surely agree with comments towards reducing the power
dissipation by means of using lower input voltage.

The Texas Instrument ones have a datasheet saying 35V absolute maximum
input voltage, but recommending to not exceed 25V.

(My newsreader makes me split this into two lines)

http://rocky.digikey.com/WebLib/Texas Instruments/
Web%20data/UA7800%20Series.pdf

I think it's getting towards "anal" in a lower current application to
have input voltage down to 8-10V (although it would be better). However,
it can do a world of good to reduce the input voltage from over 30 volts
to about half that.

And also - have an adequate heatsink, and beware of the pitfalls that I
mentioned before if you want to measure the temperature of the heatsink!

- Don Klipstein ([email protected])

 

[ian field]

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.

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!

 

[Keyser Soze]

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.

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.

 

[Reg Edwards]

Have you connected the diodes the right way round?

======================================

 

[Eeyore]

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.

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

 

[[email protected]]

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:

http://www.ladyada.net/make/x0xb0x/fab/images/to220-heatsink_t.jpg

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.

 

[Eeyore]

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.

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

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.

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

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).

The resistance is too low for it to warm appreciably.

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?

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

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:

http://www.ladyada.net/make/x0xb0x/fab/images/to220-heatsink_t.jpg

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.

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.

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.

That may be your oscillation !

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.

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.

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.

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

 

[amdx]

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:

http://www.ladyada.net/make/x0xb0x/fab/images/to220-heatsink_t.jpg

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.

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

 

[martin griffith]

On 13 Aug 2006 15:55:09 -0700, in sci.electronics.design

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

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

Pokey.

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, tits, BBQ formulae, 42, poisoning
pigeons in the park, but not 7805s

sorry Pokey


martin

 

[Don Klipstein]

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.

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.

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.

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!)

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?

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).

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

Good - same story after a couple hours running?

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:

http://www.ladyada.net/make/x0xb0x/fab/images/to220-heatsink_t.jpg

I do have some white thermal contact paste on it.

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.

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.

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

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.

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.

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.

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 ([email protected])

 

[John Popelish]

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

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

Thanks. Details like that count.

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.
Good.

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.

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

(snip)

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.

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.

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.

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.

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.

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).

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.

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.

 

[Eeyore]

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

 

[John Fields]

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?

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

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:

http://www.ladyada.net/make/x0xb0x/fab/images/to220-heatsink_t.jpg

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.

---
Yes, it does. Measure the resistance of the 2.7 ohm resistor, then
measure the voltage dropped by the resistor with the circuit
operating and the relay closed, then divide that by the resistance
you measured, and that'll be the current through the regulator,
including its operating current.
---

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.

---
I don't understand what the coil power has to do with the contact
ratings. Do you have a URL for the data sheet or can you scan it
and post it?
---

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.

---
So, you're supplying the PIC, the MAX233, and the relay coil with 5V
and you're using the transistor as a low side switch to turn on the
relay?
---

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.

---
It's probably oscillating. You need to connect the smoothing cap
and the cap on the output as close to the 7805 package as you can
and make all your wiring as short as you can. In particular, make
sure that all your ground wiring is _solid_.
---

 

[linnix]

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

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.

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?

Try a 10 ohm, it will be hot.

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.

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 http://linnix.com/smps

 

[John Jardine.]

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

[...]

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 www.tech-chat.de)

 

[John Jardine.]

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

Graham

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

 

[linnix]

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

[...]

The 170deg heatsink says you're taking about 70ma from the +5V.

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.

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 www.tech-chat.de)

 

[John Popelish]

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

[...]

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 www.tech-chat.de)

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

 

[Don Klipstein]

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.

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.)

Try a 10 ohm, it will be hot.

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.

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 http://linnix.com/smps

- Don Klipstein ([email protected])

 

[John Jardine.]

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

[...]

The 170deg heatsink says you're taking about 70ma from the +5V.

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

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

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

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

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 www.tech-chat.de)