resistors connected to H11AA1 getting hot
InSilico
Posts: 52
I've implemented a light dimming circuit that uses a H11AA1 for zero cross detection for the SX microcontroller. I have two 15K 1/2W resistors connected to the H11AA1 like this:
The pair of 15K 1/2W resistors get quite hot during operation, but the entire·light dimmer circuit works just fine. By the way, this circuit is designed for 120VAC 60Hz. My calculations indicates that this circuit should be fine. Is this circuit sound, or·should I use 1W resistors instead?·(I can desolder the resistors and drop-in replace it with 1W versons, I still have board space for it)
Much appreciated!
Post Edited (InSilico) : 10/12/2008 9:22:10 AM GMT
15K 1/2W
Hot <----\/\/\-------------
|
Neu <----\/\/\---------- |
15K 1/2W | |
_|__|__|_
| |
| H11AA1 (|
|_________|
| | |
| | 10K
| --------\/\/\----> 5V+
| |
| -----------------> Zero Cross Output
V
GND
The pair of 15K 1/2W resistors get quite hot during operation, but the entire·light dimmer circuit works just fine. By the way, this circuit is designed for 120VAC 60Hz. My calculations indicates that this circuit should be fine. Is this circuit sound, or·should I use 1W resistors instead?·(I can desolder the resistors and drop-in replace it with 1W versons, I still have board space for it)
Much appreciated!
Post Edited (InSilico) : 10/12/2008 9:22:10 AM GMT
Comments
Leon
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Amateur radio callsign: G1HSM
Suzuki SV1000S motorcycle
Based on the datasheet I downloaded, it appears that the H11AA1 is merely an opto-isolator, and not an opto-isolator WITH a zero crossing detector (ZCD). That alone could account for the high amperage passing through those resistors, and the resultant heat you're seeing. With a ZCD, switching only occurs when there is little or no current.
Regards,
Bruce Bates
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When all else fails, try inserting a new battery.
You have 60vac across each resistor, but the same current, so the power is split across the resistors; 0.24W each.
1/4W is a lot of power in a small·space like that, a 1/2W resistor.· It's OK, that's just how it is.
PJ, it seems we are in agreement, since those were the calculations that I've arrived to.
Thanks all for responding... now I can sleep well at night
These posts have helped to me finally figure out how to use the H11AA1 in my own circuit. Thanks everyone.
Regarding the resistor power rating. According to PJ Allen's calculations, using the 15k resistors the wattage calculates out to .24 watts. If you replaced each 15k resistor with a pair of 30k resistors, in parallel, the wattage for each resistor would be halved to .12 watts but the total current through the circuit would remain the same. That should solve the heat problem.
Sandy
Sorry, not correct. True, you have half the current through each resistor, but you have twice the voltage across them so the power dissipation is the same.
You'd get the same benefit by putting four 7.5K resistors in series. This time the current would be the same, but the voltage across each would be half.
-Phil
risk of any single component failure causing an overcurrent situation and reduce the risk of flashover....
Yep, you're right. Misread Sandy's post. My apologies Sandy.
That sounds like a better solution.
Then the customer changed his mind and eliminated the need for zero crossing detection in this circuit haha.
Sandy
BUT I SWEAR I SAW IT MOVE.
I was looking over this thread, and I noticed the 10k resistor on the 5v as pullup.
Isn't that a bit much for the part?
I calculated a current of 0.5ma,... wouldn't the part be better served by a 1k ohm pullup at 5v, allowing 5ma to flow, which is half the transistor base max rate?
So for the propeller, a pullup of 500 ohm at 3.3v? (giving 6.6ma) (or a commonly used 470ohm at 3.3v giving 7.02ma)
I could scope it out and see the rise/fall differences but i figured someone might know from theory/experience.
It certainly was an old thread....
I'm not following the logic here, usually you want the least current in a design, and 5k to 5V for 1mA is more common, with sub-1mA being common now for optos.
At 50Hz/60Hz, there is no demand for speed at all, and higher IC side current just means higher AC drive needed, and that's heat ! (as lamented in #1)
AC input optos look good on paper, but they are quite a lot more expensive than generic optos, and have poor LED:LED matching (often 3:1)
With something like a Prop, you are probably better to use a '817' series opto, with a diode and target 0.1-0.2mA of collector current (20k) and 0.5~1mA of drive level.(56k + 56k)
You then measure the 'lazy AC' result and find the centre of the peak (50% between the _/= and =\_ , that mid-location is aging and CTR tolerant.
Next derive the _/= to _/= as Mains Period, and then delay from the virtual peak P/4, for a derived zero crossing with minimal parts.
Typical voltage drop on h11aa1 is 1.2v.
So the ac line must be...
1.2v or
120v-1.2v = 118.8v ?
Before the led will conduct at all?
From what you wrote (i think) you say it conducts when ac line reaches 118.8v --- up to 120 --- then back down to 118.8v--- then turns off.
mistakenly put it in a new post... ignore this...
Im making crispy bacon, and i gotta get my ac timing right for my element!
Ac Darn phase drifting... not cooking my bacon right.
Ahh, middle of the peak, means my timing is 240hz. Now i get why faster rates were working better. .
When used with the object...
http://obex.parallax.com/object/794
{{ ''******************************************* ''* AC Phase DEMO * ''* Author: Clock Loop @ parallax forums * ''******************************************* }} CON _CLKMODE = XTAL1 + PLL16X _XINFREQ = 5_000_000 ZeroCrossingPin = 0 SolidStateRelayPin = 1 OBJ PWM : "PWM_32_v4.spin" PUB DEMO | PwmRate PwmRate := 50 PWM.Start '1/240 = 0.004166 ' The Period is calculated by taking the inverse of the desired frequency ' 1/240Hz = 4.166ms ; since the value entered is in micro seconds this ' value needs to be multiplied by 1000 so it becomes 4166 Repeat Repeat Until Ina[ZeroCrossingPin] == 0 'Power on ac wire is > 118v. Optoisolator is lit, grounding the pullup resistor, making pin low. Repeat Until Ina[ZeroCrossingPin] == 1 'Power on ac wire is less than optoisolator will light up at, bi-directionally. Pin pulled high. PWM.Duty(SolidStateRelayPin,PwmRate,4166) CON {{ ┌──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐ │ TERMS OF USE: MIT License │ ├──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┤ │Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation │ │files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, │ │modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software│ │is furnished to do so, subject to the following conditions: │ │ │ │The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software.│ │ │ │THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE │ │WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR │ │COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, │ │ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE. │ └──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┘ }}Best to think of optos in current terms & ignore the voltage drop.
Current will start to flow very early, eg with 125k series R, even 2V on the mains will drive a measurable 6.4uA of current.
Of course, CTR down here is very low, so we need to ensure enough LED current flows, to get into the operating region.
Some of the better optos spec at 0.5mA, cheaper ones might need 1mA+
Then, you choose a load current far enough below that, for the worst CTR to deliver it. eg 5:1 allows a feeble 20% min CTR
I am focusing on the AC line voltage, that the transition happens at here, based on the datasheet specs of the H11AA1, and the above attached propeller circuit.
And so, by your comment...
The H11AA1 led won't conduct the needed current to excite the opto and turn on the transistor until 118v? Is that correct?
Heres the exact part datasheet from the manufacturer that I have.
https://www.digikey.com/products/en?keywords=h11aa1sr2vm
https://www.fairchildsemi.com/datasheets/H1/H11AA1M.pdf
If you were running LEDs from logic voltages you would always take the forward voltage drop at the rated current into consideration however.
WARNING: CRISPY BACON AHEAD! Maths and stuff.. TWO transisions per cycle.. (-120 and + 120.) 120hz..
Another two, for the fact that you are at the center of the peak during detection.
(because you MUST also consider both the 0v transistions, no current flows to your device during those moments, effectively turning any device under control, off)
240hz.
Ok, now I am going to yank out the scope damit. Im done.
I think peeps are missing this attachment i made above..
Repeat Repeat while Ina[ZeroCrossingPin] == 0 'Power on ac wire is enough to make Optoisolator lit, grounding the pullup resistor, making pin low. ++cnttmp1 cnt01 := cnttmp1 cnttmp1 := 0 Repeat while Ina[ZeroCrossingPin] == 1 'Power on ac wire is less than optoisolator will light up at. Pin pulled high. ++cnttmp2 cnt02 := cnttmp2 cnttmp2 := 0The pin state Ina[ZeroCrossingPin] == 0 ends up with higher counts of 400,
The pin state Ina[ZeroCrossingPin] == 1 end up with lower counts of 60.
This means the led conducts (thus turning the output transistor ON) for the majority of the time, it turns on and off near the zero transition, not the peak.
240hz won't work, its no longer a neat center timed signal just at the peak voltage, but kinda arbitrary based on part heat, line noise/stray, and program execution...?
EXCELLENT!
So the answer is: The Led turns on when the ac line voltage is ABOUT 1v-1.5v, it stays on all the way up to 120v and then back down to ABOUT 1v-1.5v, where it turns off.
Based on my timing measurement, on counts was 400 and off counts was 60 total loop of 460.
60 / 460 = 0.1304347826086957 = 13.1%
400/ 460 = 0.8695652173913043 = 86.9%
13.1% of 120v = 15.65v
But then i think i need to divide 13.1 in half because half of it is the fall and half is the rise.
6.55% of 120v = 7.86v
If we do some (very rough)maths...
Assuming 1.2v led drop, and a total voltage of 7.86v with a resistor of 30k, i calculate the led starts to light up at 0.2222 mA.
Hmm, so the led conducts more like around 7.86v up to 120v and then back down to 7.86v, and is OFF from 7.86v to 0v back to 7.86v.
My timing code might be an issue? I need to check my circuit to verify my H11AA1 led resistor is in fact a total of 30k.
These are kludge numbers and are not even close due to lack of rms etc...
Not counting any kind of rms or other calcs, program delays... <--thats in bold so no one gets on a soap box and starts yelling.
But if you want to educate on the best method and code to do this stuff.. go ahead...
Repeat If powercnt > powercnttop 'Is the new count higher than max count? powercnttop := powercnt 'Make max count equal new count This ends up as 120 if the counts are right, and you reset the count every second. (120hz) ++powercnt 'Count ac phase change every time it happens. 'The time is right to commit/sync pwm changes to the pins, while the phase is 0v but on the rise. PWM.Duty(SSRPin,PwmRate,7241) '7241 is 86.9% of 1/120 (8333) *my ON time* Repeat While Ina[ZeroCrossingPin] == 0 'Power on ac wire is not 0v. Optoisolator is lit, grounding the pullup resistor, making pin low. Repeat While Ina[ZeroCrossingPin] == 1 'Power on ac wire is less than optoisolator will light up at, bi-directionally. Pin pulled high.BECAUSE we are talking about programming a prop chip and AC MAINS.
A PARALLAX WX device is really recommended here so you can debug and wirelessly program the prop on mains..
I created a post on how the prop is connected, programmed, and debug output viewed wirelessly using the parallax wx device.
https://forums.parallax.com/discussion/comment/1434618/#Comment_1434618
I LOVE WIRELESSLY CRISPIFIED BACON! DON'T U?
The approach is good, but a quick sanity check on your claimed 222uA to the needed > 3mA to register LOW (1.5V on 500 ohms), gives a CTR well above H11AA1 specs to typically 100%, MIN 20%
If you do add in the AC peak (ie sine wave) and correct for bogus halve, I get closer to 700uA LED drive.
That still indicates a high CTR, and suggests the pullup may not be 500ohms, which is far too low anyway.
It is so low, that a worst-case H11AA1 will not work : ((3.3/2)/500/0.2)*30k = 495V
(H11AA4 is better choice, with CTR of MIN 100%, allows you to have better margins, and lower power.)
As already mentioned above, you can/should raise the pullup, and raise the 7.4k x 4, to greatly lower the energy this uses.
Partly right.
It is center timed, (the width was always a variable), but the CTR skews of H11AA1 mean you have two differing center-widths.
Run those captures over a whole mains cycle, (capture +/- separately) and you will get differing half cycle widths due to that MAX CTR skew of 3:1
(Typically it will be better than 3:1, but I think the physical die design means one LED is always better located than the other...)
The center location is stable, for each half-mains peak, however the width is indeed 'kinda arbitrary' and will vary with heat/LED aging/component batch/ and +/- LED side.