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Task: to make a square wave with a frequency of 1 kHz in the voltage range (-12V - +12V), using a PWM signal with voltage (0V-5V).

Circuit: enter image description here

Originally LM258WYDT (ST) was used as an operational amplifier, it did not heat up, but since it was not from the Rail-to-Rail series, there was a 1.5V drop at the output, which did not allow us to give out a voltage close to 12V. So I decided to switch to Rail-to-Rail, for these purposes I found OP284FSZ-REEL7 (AnalogDevices). After replacing with this amplifier, the value of the output voltage became close to 12V (+-11.85V), but the operational amplifier began to heat up (after 5 minutes its temperature was about 45-60 degrees).

I did not find a short circuit, but noticed that at the output of the voltage divider (R1, R2), instead of 2.5V, the voltage became 4.5V. On the LM258WYDT, this voltage was 2.5V. Perhaps there are elements on this input of the operational amplifier that should lift it?

Unfortunately, at the moment I do not have an oscilloscope and I can not see the output signal.

My suspicions:

  1. The amplifier is not designed for such a load or, conversely, it is not enough for it, and it begins to oscillate.
  2. Is this an accurate amplifier, perhaps additional elements are needed for its operation?
  3. The amplifier is damaged.

Parameters of currents from datasheets:

LM258WYDT

  1. Input current: 5mA (DC)
  2. Source output current (Isource): 20-60mA
  3. Output sink current (Isink): 10-20mA

OP284FSZ-REEL7

  1. Supply current: 2.25mA
  2. Output current: 10mA

Question: What could be the reason?

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1 Answer 1

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It sounds like your amplifier might be damaged. Add bypass capacitors (10 or 100nF ceramic) near each supply pin to ground though, and double check the connections before you conclude that. Make sure the unused op-amp is properly dealt with (eg. a voltage follower with input grounded). The typical < 2.2mA total supply current for the two amplifiers should lead to less than 55mW of dissipation which is something like 8°C rise at the chip and 5°C rise at the case.

What you are observing on the inverting input is a result of the type of op-amp you are using - a super-\$\beta\$ bipolar type. To protect the input transistors against reverse Vbe breakdown, they put diodes across the inputs. That's of little consequence in most op-amp circuits, however you are trying to use it as a comparator.

Note that this particular amplifier does not have series resistance on the inputs, so if you were to short R1 or R2 even momentarily, the high current between the inputs could be enough to damage the op-amp. To prevent this you can add another resistor in series with the input signal but it would be better to use a more appropriate part.

enter image description here

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  • \$\begingroup\$ Thank you for the answer! Next to Vs + and Vs- are ceramic capacitors with a capacity of 1μF each. The free channel of the operational amplifier hangs in the air. Unfortunately, this is already displayed on the board, so I will review this with the next version of the board. As I understand it, the cause of the failure can be a large voltage difference between the inverted and non-inverted inputs? Initially, it was assumed that 5.1V would come to the non-inverted input, and 2.4V to the inverted input. The voltage difference is 2.7V, which exceeds 0.6V. \$\endgroup\$
    – Delta
    Feb 1, 2020 at 7:09
  • \$\begingroup\$ Or, perhaps, the operational amplifier is still not damaged, but heating can be caused by a large current of internal diodes, which are trying to reduce this difference to 0.6V? \$\endgroup\$
    – Delta
    Feb 1, 2020 at 7:09
  • \$\begingroup\$ @Delta No, because only a small current will flow with the 100K+100K divider \$\endgroup\$ Feb 1, 2020 at 7:14
  • \$\begingroup\$ You can exceed the +/-0.6V iff you limit the current to no more than 5mA. \$\endgroup\$ Feb 1, 2020 at 7:16
  • \$\begingroup\$ I thought that in the voltage divider the current could be limited, but the non-inverting input is connected directly to the microcontroller, which can give a current of up to 20mA. \$\endgroup\$
    – Delta
    Feb 1, 2020 at 8:10

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