R1 heating in opto back indication

Hello

I have a problem with my circuit as enclosed. R1 (25 W / 15 Ohms) is heating up too much. I intend to get back indication that
my lamps 1 and 2 have failed. Where it says (LP_Fail To PIC), thats the pin that is pulled HIGH when the optocoupler shuts off in the
event of a lamp failure. I'm using 24VDC. The problem is not programming nor the PIC, but is infact optocoupler back indication
that I want and it seems that there is a flaw in the circuit. R1 over heats too much too quickly. My lamps each draw 1amp current @
24VDC. If I by pass the optocoupler and R1 and connect the 24VDC directly, then i have no issues and the lamps work great with no
over heat issues. I have tried high/low values of R1 and with higher watts as well but the heating up of R1 is far to great and u can smell it burning! Please can someone suggest a fix to the over heating problem or suggest another slicker way of back indication to the
PIC in the event of my lamp failures.

Haseeb

If each lamp draws 1A when 24 VDC is applied, then each lamp has a resistance of:

V = IR -> R = V/I -> R = 24/1 = 24Ω Let's call that RL.

When the lamps are placed in your circuit, they will draw less current because of the series 15Ω /25W resistor. Neglecting the voltage drop across the switching transistors (TR3 and TR4), the current flowing through R1 can be calculated like this:

V = IR -> I = V/R -> I = V/(R1 + RL||RL) -> 24/(R1 + RL²/2RL) = 24/(15 + 24²/48) ≈ .9A

or .9A/2 = .45A per lamp

The entire .9A flows through R1 (neglecting the flow through your optocoupler) so we can calculate the power dissipated by it using:

P = IR² -> P = .9*15² ≈ 202W

That about 8 times R1's power dissipation rating. No wonder it is getting very hot very quickly.

So basically, you need to change the value of R1 so that it dissipates less power but still has a high enough voltage drop across it to turn on your optocoupler.

According to the KB815 data sheet, its input has a forward voltage of 1.2 - 1.4V. That means that the voltage drop across R1 must be 1.2 - 1.4V at a minimum to turn on the optocoupler and pull down Lp_Fail.

If we make R1 1Ω for example then the current flowing through it will be:

V = IR -> I = V/R -> I = V/(R1 + RL||RL) - > I = 24/(1+12) ≈ 1.85A

That means that R1 would have a voltage drop of:

V=IR -> V = 1.85A*1Ω = 1.85 V. That's greater than 1.4V so that means it would turn on the optocoupler.

The power dissipated would be:

P = IR² = 1.85A*(1Ω)² = 1.85W

So, a 1Ω/5W resistor would probably work just fine. However, since you didn't provide and information about TR3 and TR4, I don't know what VDS is for them. 1.85V is pretty close to 1.4V so it may not be OK to neglect VDS. I would try a 1Ω/5W resistor first and if that doesn't work I'd try a 2Ω/10W resistor.

You'll also have the change the value of R2 to compensate for the new value of R1. Once again, from the KB815 data sheet the maximum forward current is listed as 50 mA. We don't want to run our components at their absolute maximum ratings so lets pick a forward current of 20 mA for the KB815 input.

So, for the case of the 1Ω resistor, R2 should be:

V = IR -> R = V/I -> R = (VR1 - Vf)/(If) -> R = (1.85V - 1.4V)/(.02A) ≈ 22Ω

You can calculate the necessary value of R2 if you decide to go with a 2Ω resistor for R1.

Thank you so much!

I never thought of it that way....and you explained very well and it makes perfect sense.

I will try what you say and if any problem will get back to you.

Once again, thank you very much

Haseeb

You're welcome. Good luck.

vne147 said:

P = IR² -> P = .9*15² ≈ 202W

Click to expand...

Close, but no cigar.

The equation is P = I²R so the power would be only 12.15W.

crutschow said:

Close, but no cigar.

The equation is P = I²R so the power would be only 12.15W.

Click to expand...

Doh! Thanks for the correction. It is a 25W resistor. I wonder why it's getting so hot so quick then.

If you want to reduce the value of R1 even further you could use a PNP transistor to sense the voltage drop and drive the optocoupler. The transistor will turn on at about 0.7V, thus the required value for R1 would then be less than an ohm.

Hi



I have just successfully solved the problem. I had to feed in the circuit 18V DC instead of 24VDC and I also changed the resistor to 3R0 / 25W and this solved the heating problem successfully.



Now I have the same R1 over heating problem but this is now for a 24V AC version circuit board. I'm

powering two halogen light bulbs at 24V 50W each. Thats about 2Amps per lamp. The circuit diagram

is enclosed. I have tried feedng in 18V AC and changing R1 to 1R0 25W but still the over heating of R1 is taking place.



Please can someone look at my circuit diagram (enclosed) and suggest what might be wrong and

why is R1 heating up too quickly and too hot??



Thanks

Haseeb

Well again look at the values.
2 amp each means 4 amps through resistor and this means the voltage across it is
4 x 1 ( I x R ) = 4 volts so power it has to dissipate is 4 x 4 ( V x I ) = 16 watts.
What you need to remember is that this has to be lost as heat. Now 16 watts is within the spec for your 25 watt resistor but the physics demands that the temperature of the resistor rises until the heat it gains from the current flowing equals the heat radiated into the air.
This depends on the surrounding air temp and airflow etc. So you can expect it to be quite hot. Now the higher the power rating of the resistor the larger it is and so the more heat it can loose to the surroundings per second and therefore the lower it's final temperature will be. So if you are happy with the circuit otherwise you can do several things:
1/ Higher power resistor. ( 50 watt or more )
2/ get a better airflow around it. ( A fan, a heatsink or a chimney so more air flows past it)
3/ make sure the surrounding air is colder somehow. ( lowest part of your project, colder area )
4/ Do 2 or more of the above.

Yes thanks for that it makes sense now...

cheers!