If you are producing an "industrial control" 0-10V signal, then you will only need a couple of milliamps at the most. Industrial controls tend to have 10kΩ to 47kΩ inputs.
This is one of the few applications where an LM358 is good enough!
PWM signal through an RCRC filter into the non-inverting input of the LM358 set to a gain of 2 for 5V pwm or 3 for 3.3V pwm.
Don't forget the "build-out" resistor on the output of the LM358 otherwise it will oscillate if it has to drive any length of cable.

 

You can think of an Op-Amp as having a Resistor built-into it's Output.

The only reason why the Op-Amp has an "effective" low Output "Impedance"
is because of extremely high Gain in combination with Negative-Feedback.

With a heavy Load, and maybe only 1-Volt of Output-Voltage,
the Output-Stage may be approaching one of the Power-Supply-Rails,
this means that all that Voltage-Drop is INSIDE the Chip,
generating tremendous amounts of HEAT.

The "Open-Loop" Output-Impedance,
and the "Closed-Loop" Output-Impedance,
( which may only "appear-to-be" ~0.5-Ohms ),
are two vastly different things.
.
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View attachment 292581.

Hi!
Sorry to come back to reply after a while.
I was looking again at this graph and the schematic of the op777.

1) With 10V and 70 ohm, at the output I would have a current of 142mA which in the curve you attached is not present .. does this mean I should not even consider such an "extreme" case for this type of op-amp? Because I was curious to calculate the power dissipation, but from the graph it's not possible.

2) If, on the other hand, the load were 1k I'd have 10mA (under the 30mA maximum) and therefore 1V (y-axis) ... Power_diss inside the chip = 1V*10mA=10mW. According to your experience .. would 10mW be a lot? Few?

 

Don't forget the "build-out" resistor on the output of the LM358 otherwise it will oscillate if it has to drive any length of cable

Can you explain this concept better to me? Thanks!

 

Can you explain this concept better to me? Thanks!

Op-amps in general don't like capacitive loads. It adds phase shift to the feedback network, and once you get to 180° phase shift you have an oscillator.
The LM358 really doesn't like capacitive loads, and a piece of cable is a capacitor.
So that the load doesn't appear as a capacitor on the output of the LM358, add a resistor in series (100Ω is the usual choice) and it's known as a "build-out resistor".

The resistor and the cable capacitance form a low-pass filter - so don't expect to get high frequencies to the end of a long cable driven by an LM358.

 

.. does this mean I should not even consider such an "extreme" case for this type of op-amp?

The OP777 specifies 'indefinite' output short circuit to ground, which means it is able to limit the current it outputs. From the datasheet, the limits are somewhere around 10ma at 5V and 30mA at 15V. You extreme case is inapplicable to the chip.

10mW is not very much. Page 4 of the datasheet gives dissipation capabilities of different package types.

 

Opamps will not generally drive low impedance loads. I think ±10-15 mA might be typical and an occasional part might go a bit higher.
Here is the section for the LM358

LM158A LM358A (LM158, LM258)
min typ min typ min typ



Note carefully the test conditions, and for this part it can sink more than it can source. This is to be expected with a bipolar part.

 

Hi!
Sorry to come back to reply after a while.
I was looking again at this graph and the schematic of the op777.

1) With 10V and 70 ohm, at the output I would have a current of 142mA which in the curve you attached is not present .. does this mean I should not even consider such an "extreme" case for this type of op-amp? Because I was curious to calculate the power dissipation, but from the graph it's not possible.

2) If, on the other hand, the load were 1k I'd have 10mA (under the 30mA maximum) and therefore 1V (y-axis) ... Power_diss inside the chip = 1V*10mA=10mW. According to your experience .. would 10mW be a lot? Few?

Opamps are generally NOT intended to power things -- they are intended to process signals. If you need to use the signal to power something, you need to insert a suitable power amplifier in the signal chain.

 

There are specialized power opamps that can source/sink many amps or hundreds of volts, but they are very expensive.

Separate from that, this thread would make much more sense if you would post a schematic of the circuit you are trying to discuss. Paraphrasing Rear Admiral Joshua Painter:

"Engineers don't take a dump, son, without a schematic."

ak

 

Op-amps in general don't like capacitive loads. It adds phase shift to the feedback network, and once you get to 180° phase shift you have an oscillator.
The LM358 really doesn't like capacitive loads, and a piece of cable is a capacitor.
So that the load doesn't appear as a capacitor on the output of the LM358, add a resistor in series (100Ω is the usual choice) and it's known as a "build-out resistor".

The resistor and the cable capacitance form a low-pass filter - so don't expect to get high frequencies to the end of a long cable driven by an LM358.

So, the advantage of the build-out resistor is to avoid getting to a 180° phase shift (which creates an oscillator) ... but it has the disadvantage of 'creating' a low pass that cuts high frequencies.
Have I understood correctly?

(because I understand little english and would not like to have misunderstood)

 

So, the advantage of the build-out resistor is to avoid getting to a 180° phase shift (which creates an oscillator) ... but it has the disadvantage of 'creating' a low pass that cuts high frequencies.
Have I understood correctly?

(because I understand little english and would not like to have misunderstood)

You are correct