Hi! I have a theoretical doubt that will probably amuse many of you.

I am thinking of an opamp that amplifies an input voltage, producing, for example, 10V.

If I connect a 70ohm resistor (to ground) to the output of the opamp, I would end up with about 150mA on the resistor.
Without considering 1.5W on the resistor, not the point of the question, I wonder if that 150mA would damage the opamp.

The current would come from the feedback branch, however, if the output impedance of the ideal opamp is low (say 0), why is the current flowing into the resistor and not into the opamp? Since the current always chooses the path of least resistance.

EDIT:
I have the analog PWM output of an Arduino that goes from 0 to 5V (depending on how I decide via software the signal .. there is a whole PID control etc) and I need a stage that amplifies that voltage to, say, 0-10V.
It is important to isolate the load of a few ohms from the microcontroller because of the currents that might flow. I had thought of a buffer, but still I need amplification as well, so I think a single opamp with feedback (feedback resistor etc) is sufficient. A non-inverting opamp also acts as a buffer .. right?
If my reasoning is correct, all that's left is to choose this amplifier, and back to the main question we were discussing.

Thanks to @ericgibbs and @Ian0 for reminding me to update the post with all the available information I had

 

Check your op-amp datasheet. It will specify the maximum output current, which is generally a few tens of milliamperes. Any load requiring more output current will cause the output stage to current-limit.
It is unlikely that the current would come via the feedback resistors. If so, it would have to come from whatever is driving the input.

 

Hi! I have a theoretical doubt that will probably amuse many of you.

I am thinking of an opamp that amplifies an input voltage, producing, for example, 10V.

If I connect a 70ohm resistor (to ground) to the output of the opamp, I would end up with about 150mA on the resistor.
Without considering 1.5W on the resistor, not the point of the question, I wonder if that 150mA would damage the opamp.

The current would come from the feedback branch, however, if the output impedance of the ideal opamp is low (say 0), why is the current flowing into the resistor and not into the opamp? Since the current always chooses the path of least resistance.

A schematic of the circuit you are talking about would sure help make for a more coherent discussion. Without it, there's no way to talk meaningfully about what current is going to or coming from where.

Forget this "current always chooses the path of least resistance" crap. That is a gross oversimplification. Current will flow through all paths if there is a voltage across them and finite resistance along them.

 

Check your op-amp datasheet. It will specify the maximum output current, which is generally a few tens of milliamperes. Any load requiring more output current will cause the output stage to current-limit.
It is unlikely that the current would come via the feedback resistors. If so, it would have to come from whatever is driving the input.

Let's assume the op777 .. which parameter indicates the maximum output current?
I can't find it in the tables

 

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.
.
.
.

 

Let's assume the op777 .. which parameter indicates the maximum output current?
I can't find it in the tables

The line called "Output Circuit". It has different values depending on the supply voltage.

 

Let's assume the op777 .. which parameter indicates the maximum output current?
I can't find it in the tables

There are two sets of specs, one for a supply voltage of 5 V and one for ±15 V (so 30 V total). Since you mentioned 10 V output, let's use the latter:



There's a typo and it says "Output Circuit" where it means "Output Current".

If your total supply voltage is less than 30 V, the maximum output current will be less than 30 mA. At 5 V it is only 10 mA.

This opamp is not suitable for driving your 70 Ω load directly.

 

With a ~70-Ohm Load,
You will not be able to get ~10-Volts of Output,
and the Chip will slowly over-heat until it finally fails.

"" Output Short Circuit
The output of the OP777/OP727/OP747 series amplifier is protected from damage against
accidental shorts to either supply voltage,
provided that the maximum die temperature is not exceeded on a long-term basis
(see Absolute Maximum Rating section).
Current of up to 30 mA does not cause any damage.""
.
.
.

 

Most semiconductor datasheets have little section that looks like this, usually "early" in the datasheet.


It doesn't fully answer your question but tells you how to not kill you op amp.

 

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.
.
.
View attachment 292581.

Thanks for your answer.
It is not clear to me what the y-axis "Δoutput voltage" indicates.

 

LM675 can source/sink 3A. Make sure that it is properly mounted on a heatsink.

 

That Triangle symbol is the Greek-Letter "Delta"
and is commonly used to indicate that the following number
is the "difference" between 2 other numbers.

In this particular case,
the number is the Voltage-Drop between the Power-Supply-Rail, and the Output-Voltage.

The Current at the Output,
times the Voltage-Drop shown in the Graph,
equals the number of Watts of HEAT generated inside the Chip.
.
.
.

 

Thanks for your answer.
It is not clear to me what the y-axis "Δoutput voltage" indicates.

If you divide the Δoutput voltage by the output current then you will get the equivalent output resistance at that point.

 

That Triangle symbol is the Greek-Letter "Delta"
and is commonly used to indicate that the following number
is the "difference" between 2 other numbers.

I this particular case,
the number is the Voltage-Drop between the Power-Supply-Rail, and the Output-Voltage.

The Current at the Output,
times the Voltage-Drop shown in the Graph,
equals the number of Watts of HEAT generated inside the Chip.
.
.
.

Thank you. So if I have 1mA and Vs=+-15V .. the Δoutput voltage=40mV and so 40uW dissipated in the opamp

 

LM675 can source/sink 3A. Make sure that it is properly mounted on a heatsink.

1) Before buying the component, I would like to simulate it. I cannot find the model for LTSpice (or Multisim) of the LM675 ... do you have it?
Maybe I should use one of these two simulators (I don't know what they are) that TI provides?


2) If we there is not the LM675 spice model ... are there alternative opamps that you recommend that I can test via software before buying them?

@crutschow @LowQCab @MrChips @DickCappels @WBahn @Ian0

 

Hi Andrew,
It would help if you told us what is the purpose of this 'OPA' project that requires say 1.5Amps.

There maybe other ways to achieve your goal.
E

 

2) If we there is not the LM675 spice model ... are there alternative opamps that you recommend that I can test via software before buying them?

Happy to recommend some - what is the application?

 

What SPICE program are you using?

 

What SPICE program are you using?

LTSpice and multisim

 

Hi Andrew,
It would help if you told us what is the purpose of this 'OPA' project that requires say 1.5Amps.

There maybe other ways to achieve your goal.
E

I am now editing the main post, I thought I had already done so but forgot.

@Ian0 @ericgibbs