[SOLVED] Light Regulating Circuit. Opamp keeps dying on me :(

[qbone]

Hey Guys.
So, I have made this light regulating circuit for a small PSU.
But I have a problem. Whenever I adjust my OFFSET potmeter to around 4volts out, my opamp dies.
I am obviously guessing I am killing the opamp by draining too much current, but I dont know how to solve this problem, and I don't exactly know how to calculate the input and output current.
I have added a picture of my schematic with the values of the components I use.
The opamp is a MCP6272E/P - Datasheet: https://elektronik-lavpris.dk/files/sup2/21810d.pdf


I hope you can help me here. Both with the calculations and an indication of why it dies.
P1 is the SPAN and P2 is OFFSET, which adjusts the output when total darkness.
Total darkness = ~5V when maxed.
Direct sunlight = 0V.

I power the circuit with +5V and GND.

Feel free to ask if you need more info.

 

[FvM]

It's absolutely wrong to connect a capacitor to an OP output. You'll need a series resistor (> 100 ohm) to isolate it. What's connected to the circuit output?

 

[qbone]

The circuit output goes to the base of a BC557 transistor when it is connected to the PSU.
But when I am testing and it's dying it isn't connected to anything besides my multimeter.

I didn't know about the rule of not putting a capacitor at the output, it was put there to remove some noise.

 

[KerimF]

Sorry, you missed to show us the most important part of the system. It is the load and how it is connected to the output of your circuit. It seems the load IS the killer

I just noticed just your opamp could be fried without connecting anything at the output port. In this case, the relatively large capacitor at the opamp output is the cause. As it was suggested in a previous post there must be a resistor to isolate this capacitor from the opamp output. How did you calculate its value?

 

[qbone]

It seems the load IS the killer

But how can this be when there is no load except the multimeter?

I will see if I can find the schematic for the PSU and throw it up during the day. But it is a a regulated PSU created from the MCP34063.

 

[KerimF]

If you will post the load too... we can find out together how to rebuild the output part (from the opamp output to the input of the load).
Meanwhile you may like removing the 100uF and things will be fine (if it will be necessary to filter some noise... we will try to calculate its value and add other parts to it like resistors to drive the load properly).

Note: Now the 100u capacitor acts like a short circuit anytime the signal at the output tries to change rather quickly.

 

[qbone]

Ok, so I tried removing the capacitor, and it didnt affect much on the noise, so I think we'll just leave that out, however at the end of this I would love to know how to calculate the proper value and how to isolate the cap from the opamp.

Anyways, my opamp still dies on me

Here is the input, atleast the part I figure is relevant.

Datasheet for the IC is here: MC34063 - http://www.datasheetcatalog.org/datasheet2/c/0hayflawhgpa39l5u6yat3y8157y.pdf

 

[qbone]

I just came to think about the LDR input at the forst opamp and did a little calculations.
When it is affected by direct sunlight it's resistance is 50Ω
And with a supply of 5V and a 1kΩ in series it would make a voltage divider with ~0,24V across the 50Ω resistor and a currect of ~4,7mA.

This would most likely be my problem as the inputs of my opamp have a maximum rating of 2mA :sad:

 

[betwixt]

The output capacitor is what is killing it. It isn't wrong to use a capacitor but that isn't how to connect it.
Two things are happening:
1. although the capacitor doesn't conduct DC (except leakage and initial charging) it has a low reactance to changing (AC) signals.
2. it's value is big enough that the ICs output stage can't charge or discharge it very quickly. This makes the output lag the input and the delayed signal connected through the feedback resistor makes an oscillator. The oscillator is what produces the AC signal that ultimately kills it.

Two solutions:
1. as suggested, isolate the IC by connecting a resistor (around 100 Ohms) directly at the output pin so it is between the pin and the capacitor. It will make almost no difference to the following transistor stage but will remove the reactive load from the IC.

2. Omit the capacitor altogether and fit a much smaller value one at the input of the IC, across R6. It will do exactly the same job and the value will only have to be 100nF or so because the impedance at that point is much higher.

Brian.

Just read your final post - the input rating of the IC is the maximum you can force feed it if you apply too much voltage, it won't draw any significant current if you use it within supply voltage limits.

 

[KerimF]

Here is the input, at least the part I figure is relevant. MC34063

What is the maximum voltage at pin 5? (it is an input pin).

 

[qbone]

Pin 5 is the inverting input of a comparator in the IC. Maximum rating is -0.3 to 40V, but what I give it varies. But that doesnt matter much, it is part of a perfectly working circuit, my problem was the Opamp dying.
I tried replacing the opamp and it still didnt work, but it turned out that when I had killed the opamp earlier I also killed a potmeter, so I couldnt adjust anything - which was what lead me to believe it was still getting killed.

But I can now happily say that my circuit works like a charm, thank you very much, both of you And thanks for the theory behind not putting a capacitor directly at the output op an opamp, makes total sense