Hot resistors

[Barmybaz]

I have built the circuit below, all works fine but I'm a bit alarmed at the resisters as they get very hot,
(you wouldn't want to put your finger on them for too many seconds)
Is this O/K or have I got summat wrong.

Bazza

 

[MikeMl]

Usually, if you spit on your finger, and touch the component, if it sizzles, it is probably too hot. If it doesn't sizzle, then it is likely ok (<70degC)

Otherwise, this is why you can buy 1/2W, 1W, 2W, 5W and 10W resistors...

 

[jjw]

I have built the circuit below, all works fine but I'm a bit alarmed at the resisters as they get very hot,
(you wouldn't want to put your finger on them for too many seconds)
Is this O/K or have I got summat wrong.

Bazza

Do you have a capacitor after the rectifier? If you have, then the DC voltage is about 33V and the power consumed in the resistor is about 0.66W and that will make a 0.25W resistor hot.

 

[Tony Stewart]

The ideal solution for driving LEDs from a voltage source is when the voltage drop on the Rs current limiting resistor is less than the LED drop.

Your results indicate 5.4V thus add one Blue and two Red LED to each string and recompute the R value.

e.g. you had 20mA * 270 Ω= 5.4V and 20mA * 5.4V = 108 mW which is too much for a 1/8W Resistor but should not be for a 1/4W.

Instead 5.4-3.2V = 2.2V becomes 110 Ω @ 20mA and 44 mW which will be much cooler.

added...

However, this design is not a true voltage source and is simply an unfiltered un-regulated rectified sine wave, so the LED's will only turn on with a small duty cycle where the peak sine wave exceeds the String voltage of 19.2V Blue and 18.9V Red.

Meanwhile the transformer has a coupling loss and ESR that will drop the voltage to 24Vac only when loaded at the rated VA rating , which is not the same as the pulsed rating with an average of 450mA, so the conduction losses will rise as duty cycle reduces with rising peak current and higher I^2R losses. Thus you end up with < 60% efficiency for the same transformer temperature rise, as I recall, meaning you cannot use all the 450mA rating as used. Adding a large Cap doesn't improve this either, because now the Caps will only pulse current for low ripple and end up with the same losses.

Conclusion.
This simple design will need a larger power rating resistor with the same number of LEDs.

 

[dr pepper]

You did get summat wrong, the power rating of the resistors, whatever watt rating they are you probably need a couple of sizes bigger.

 

[USER REDACTED]

I have built the circuit below, all works fine but I'm a bit alarmed at the resisters as they get very hot,

In addition to others...

Red LEDs are usually 1.6V, perhaps you got these instead 2.1V?
Assuming there is 20V DC from rectifier, that would be extra 5.6V to handle.

 

[bloki]

At current 20mA resistors disipate 0.1W. Resistors has elevated temperature but it is normally in operating range. If current is higher or not controlled then this is a problem.

 

[audioguru]

The problem is that each strip of LEDs DOES NOT draw 20mA, their current is much higher.
1) The "24V" transformer does not have its rated load of 450mA so its voltage might be 30VAC.
2) If there is a filter capacitor then the DC from the rectifier is the peak minus the rectifier drop which is 40V!
3) LEDs do not have a single voltage rating, they have a range of voltages that might actually be 2.8V for the blue LEDs and 1.6V for the red LEDs.

I betcha the LEDs are almost as hot as the resistors and maybe the LEDs have already burned out.

 

[Tony Stewart]

I suspect the resistor value was chosen to match the measured average current is correct and my assumptions of average power dissipation for a 1/8W resistor would give his observed temperature rise. 24Vac secondary depends exactly on the primary rating and actual values. My calculations for 24Vac rms and ~30Vpeak yield approximately 20 mA with <50%% duty cycle meaning 40mA peak. The Red Vf of 2.1V is accurate for 630nmD LEDs but 660nmD Red will be 1.6. using GaAs which are older technology less efficient types.

THe Vf string of BLue is 3.2V*6 = 19.2V while if the secondary AC is 24V, then the peak voltage out of the rectifier will be 32V and thus the peak voltage drop will be 32V-19.2= 12.8V
The ESR of Blue will be around 15Ω *6= 90Ω,
The ESR of the diode bridge and transformer will be around 2Ω with 1.3V drop on the bridge approximately.
Thus with 270 Rs and the above, the peak current will be 12.8V/(90+2+270) = 35.5mA
With a narrow rectified sine pulse peak of 12.8V above the BLue LED string with a source voltage of 32V or 40% duty cycle, I expect the average BLUE LED current is near 40% of 35.5 mA or 14mA.

With secondary voltage based on full load, I expect the the secondary no load voltage to be 10% higher , and using 160mA peak out of 240mA rating , perhaps < 5% higher than 24Vac * √2

I assume his average current measurements ere done simply with a DMM which will be accurate enough for this.
\Line voltage variation of 10% is provided in NOrth America, while other countries may be worse.

THe temperature of LED junction can be estimated by feeling the Cathode lead and expecting some heat as the reflector of transparent substrate is also the heat conductor to the negative side.
For a more conservative approach 15 mA could be the target, but since LEDs are cheap, you can drive them with 20mA in room temp and derate above 40'C.

I certify my assumptions are reasonable and his results are to be expected with noted tolerances on assumptions.

Adding a large CAP is not recommended as this will only make it more sensitive to AC input variations with loading affecting the Avg DC out with load variations. i.e. wide span of Vdc from no load to full load of 40%

Consider a 270~ 300Ω 1/2W solution and report on measurement details.
If the 270Ω Rs is indeed 1/4W then consider 330Ω 1/2W and maybe 4 or more strings can be added before the transfomer gets hot.

 

[audioguru]

The AC voltage from the transformer should be measured but the non-linear load from the LEDs will distort its sinewave into an oddwave.

 

[Tony Stewart]

4.24V true Avg ( for test only) /270R =16 mA assuming 19.46Vf for string of 6 BLue LEDs
Above shows peak values only.

Simulation

 

[Barmybaz]

Thanks for all your help,
As you may have guessed there are several more strings of LEDs in my circuit,
As I am building the box up, things are cooling down
I realise the power supply is not true DC or regulated so did expect a voltage drop on load
Have taken measurements and at this moment in time.
Voltage at LEDs no load = 26.6
Voltage at LEDs load = 22
Current on red strip = 21.9 mA
Current on blue strip = 20 mA
The transformer does not seem to be getting hot
Am I on the right lines,

Bazza

 

[dr pepper]

Voltage drop on a unreg'd supply sounds about right,
Your resistors are running at about 0.5w, stabdard 0.25w ones are not big enough, I'd suggest 1w ones, or maybe 4 x 0.25w ones in parallel, each one would need to be 4 x the actual resistance you want.

 

[MrAl]

I have built the circuit below, all works fine but I'm a bit alarmed at the resisters as they get very hot,
(you wouldn't want to put your finger on them for too many seconds)
Is this O/K or have I got summat wrong.

Bazza

Hi,

There are more than a couple things to think about here.

First, you can not tell the temperature of a resistor by using your finger. You can use an IR temperature gun however. Fingers are very misleading though so you should avoid that.

Second, the only way to know if the temperature is too high for the resistor to survive is to look up the specs on the data sheet and measure with an IR gun. Some resistors can take very high temperatures like 100 degrees C, while for others that is the top end of their working temperature so you dont want to allow that.

Third, if the resistor can not get free air flow then a 2 watt resistor could overheat and burn out when it only dissipates 1/10 of a watt (one tenth of one watt). That's because heat needs a place to go or else the temperature rises forever until it does get a place to go.

Fourth, sometimes you can not run the resistor at a higher temperature even if the specs say you can. That's because it may be mounted close to another component that could melt or otherwise break down.

So the rules are:
1. Measure the temperature with an IR gun or other probe.
2. Check the temperature specs on the data sheet.
3. Make sure it has free air flow.
4. Make sure it is not mounted close to a heat sensitive component.
5. A rule of thumb is to run the resistor at one half it's power rating or less.

Note there may be times when you just dont want the resistor getting too hot. The temperature rise is closely related to the surface area of the resistor, so the larger the surface area the cooler it stays provided it can get free air flow. That means the bigger the resistor the cooler it stays. I like to keep my resistors cool so i always use overrated parts whenever possible. If you overrate by a factor of 4 they stay nice and cool. This helps a lot when they are enclosed in small packages.

 

[Barmybaz]

Just a thought
but isn't heat a unnecessary wast of energy,

Guess the answer is higher rated resistors.

 

[MrAl]

Hi,

Heat is an inevitable consequence of energizing electronic components.

We dont want heat, but it is always there because we have to have certain components perform certain functions. We try to minimize this loss but we never get it to zero.

Even LED technology is trying to advance by recovering some of the heat energy lost in the process of converting electrical energy to light energy, but it's not perfect either. If we could recover 50 percent of the heat lost we could boost the LED efficacy by maybe 40 percent which would surely make it one of the highest efficiency light sources of all time.

 

[Tony Stewart]

You finally understand. OK.

Increase the power rating of the resistor is preferred and perhaps the increase value to 330R to reduce 25% current.
Then you can add 25% more strings too.
Normally if resistors are intended to run hot, they are elevated above the board to allow convection cooling.

I disagree with others who say Rs is dissipating 1/2W. with your R value , if your average current reading correct.
I^2*R = 110mW for blue at 20mA

With a power supply rated for 24V @450mA and 65mW for each LED wasting 110mW for regulating the current with an unregulated supply is acceptable.
But obviously worse efficiency for example if it was a lower voltage and smaller string and better efficiency if a larger string such as 48V or more.

As I said before, if the voltage is regulated, then the loss should be less than equivalent of 1 LED with correct matching of string to supply with less than 1 LED drop voltage for Rs.

 

[Barmybaz]

Thanks again for all your advice,
I have to be honest some of it was above me, (only a retired gardener with lots of ideas and little knowledge)
The project I'm working on is a light box to give seedlings a good start. I could probably use the heat to my advantage.
As an update,
I would like to thank you all for the advice given to me on my last project, "solar powered snail deterrent" it is up and working well, ( I recognise some of your names)
I think my next project will be a solar powered cat deterrent, PIRs and recordable chips are very cheap, but that's for another day. (watch this space)
Thanks again

Baz

What could possibly go wrong ?

 

[dr pepper]

I wouldnt mind a humane cat deterrent too, I probably have the engineering know but I dont know what would get on a cats wick.

 

[Tony Stewart]

Moth balls for cats.