# Resistor Temperature

#### Stupdiav

Joined Jul 29, 2011
10
I have a circuit that is detecting the presence of 120VAC. I am trying to figure out why R1 in the attached image is getting hot enough to discolor it. R1 is currently a 1/4w (for bread boarding purposes) which I would think would be overkill, but I know I am missing something and can't think of what it would be. Any help is greatly appreciated!!!

#### dl324

Joined Mar 30, 2015
9,577
The 1/4W resistor is dissipating over a watt.

#### OBW0549

Joined Mar 2, 2015
3,142
I am trying to figure out why R1 in the attached image is getting hot enough to discolor it. R1 is currently a 1/4w (for bread boarding purposes) which I would think would be overkill, but I know I am missing something and can't think of what it would be
What you're missing is that when judging power dissipation and ratings, don't guess-- CALCULATE.

The power dissipated by your resistor is equal the the voltage across it squared divided by its resistance:

Pd = 120V * 120V / 10 kΩ = 14400 / 10000 = 1.44 Watts.

Overkill? Hardly. You're lucky it didn't burst into flames...

The important thing to remember is that these things can be calculated; it's a LOT better than blind guessing.

#### SamR

Joined Mar 19, 2019
1,623
Did you do the math? Where is the step down xfmr?

#### ericgibbs

Joined Jan 29, 2010
9,320
hi stupdiav,
You MUST add an isolation transformer on the 120Vac input, else the Thread will be Deleted, as power supplies without isolation contravene AAC T&C's
Moderation.
E

#### crutschow

Joined Mar 14, 2008
24,077
You MUST add an isolation transformer on the 120Vac input, else the Thread will be Deleted, as power supplies without isolation contravene AAC T&C's
The opto isolator provides isolation and this is not a power supply. (?)

#### SamR

Joined Mar 19, 2019
1,623
You might also note that the DB107 is rated for 1A and the 4N35 less than 100mA.

#### Stupdiav

Joined Jul 29, 2011
10
I thought I might be missing something (I can't remember everything from 30+ years ago). Thank You!

Unfortunately I can not use a stepdown transformer in this project. I am not only limited by space constraints but also have Cellular Communications right above. If you have any other alternatives I am all ears. I was going to use a relays but both worried about life span and the noise I am staying away from those as well. For the time being I will just have to go to a bigger resistor.

#### ericgibbs

Joined Jan 29, 2010
9,320
hi Carl,
I appreciate its not an actual mains supply, but the High side of the opto is at mains potential.
I would consider it prudent to err on the side of safety when dealing with a TS about which we know very little of his technical abilities,

E

#### MrChips

Joined Oct 2, 2009
19,915
Use a small 5VDC wall brick and you're done.

#### Dodgydave

Joined Jun 22, 2012
8,674
I have a circuit that is detecting the presence of 120VAC. I am trying to figure out why R1 in the attached image is getting hot enough to discolor it. R1 is currently a 1/4w (for bread boarding purposes) which I would think would be overkill, but I know I am missing something and can't think of what it would be. Any help is greatly appreciated!!!

View attachment 196333
You can reduce the Heat in your circuit, by using a 330nF capacitor ( 450V AC )in series with the Mains supply to the bridge rectifier, you can then reduce R1 to 2K2 at 1/2 W

At 60Hz a 330nF will be 8K ohms,

#### KeepItSimpleStupid

Joined Mar 4, 2014
3,825
Like @Dodgydave said, you change into a reactive power supply. See http://www.gravitech.us/buacinbo.html Look at the manual link.

There are probably better IC's for detecting AC presence. Those boards are ready made. The values would be slihtly different for 50 Hz.

#### Stupdiav

Joined Jul 29, 2011
10
Thanks SamR, yeah I figured I had better double check that. I will have to refigure the whole thing, since the current is much higher than I originally had thought.

Dodgydave, I will have to look into that.

#### WBahn

Joined Mar 31, 2012
25,068
The opto isolator provides isolation and this is not a power supply. (?)
The intent of that rule is related to circuits that create too high a risk of inexperienced folks from getting tangled with mains voltage, usually because the mains-connected components are too intermingled with low-voltage circuits. The fact that this resistor is in a breadboard almost certainly means that the risk in this circuit is on that too-high side, particularly coupled with the TS's lack of awareness of how to calculate the power dissipated in that resistor, probably reflecting a general lack of knowledge about basic circuits and circuit safety.

#### MrChips

Joined Oct 2, 2009
19,915
I recently had to rewire my kitchen. I needed to confirm that the AC power was off even though I knew that I had opened the circuit at the breaker panel.

Solution: I used an incandescent table lamp and plugged it into the outlet before proceeding with electrical alterations. (Of course I confirmed that the lamp was on and in perfect working condition.)

Edit:
What is the point of this post?
You can do the same thing with a 4W night-light lamp. Then use an LDR (light dependent resistor) to detect that the lamp is shining. There you have your own DIY AC mains detection circuit with complete electrical isolation.

#### ci139

Joined Jul 11, 2016
1,088
You can read your input by something 10µA transferred by opto that would increase your resistor to
$\frac{120·\sqrt{2}-1.4}{10µA}≈16.8MΩ$ well the ~240V not too hot resistor begins somewhere over 240k (240mW)
ups! it's a current driven device (the opto) . . . but the small drain won't make much difference

with 10k you are actually exposing your opto to 10.7 to 16.8 mA ↑↑ can use 100k or a diac (at high) + integrator (at low side)