I don't understand the maximum power is at load voltage equal to halv the supply voltage.

So you need to understand the Maximum Power Transfer Theorem:

"Maximum Power Transfer Theorem explains that to generate maximum external power through a finite internal resistance (DC network), the resistance of the given load must be equal to the resistance of the available source."

This means that the maximum power is transferred to the output when the source and load resistances are equal, at which point the load voltage is 1/2 the source voltage and the source dissipation equals the load dissipation.
Thus the maximum power dissipated in the op amp occurs when the output voltage is 1/2 the supply voltage.

That's, of course, for a fixed output load resistance, which is the common condition.

Make sense now?

 

Show us your circuit schematic and then we can give you a more informed answer.

 

A continuous output current at low Vout could lead to the power dissipation of more then 2 Watt. Usually the thermal resistance is around 40 to 60 degC per Watt which when used in calculating the junction temperature that could be above 100 degC.

I don’t think so. Not for a normal opamp.

Let's say supply is 15V and output is 1V.

To get 2W at 14.5V (15-0.5), the current would be 142mA. Normal, non-power, opamps top out at 20-30mA.

In the max power mode (Vout = 7.5V) the current would have to be 267mA.

 

So you need to understand the Maximum Power Transfer Theorem:

"Maximum Power Transfer Theorem explains that to generate maximum external power through a finite internal resistance (DC network), the resistance of the given load must be equal to the resistance of the available source."

This means that the maximum power is transferred to the output when the source and load resistances are equal, at which point the load voltage is 1/2 the source voltage and the source dissipation equals the load dissipation.
Thus the maximum power dissipated in the op amp occurs when the output voltage is 1/2 the supply voltage.

That's, of course, for a fixed output load resistance, which is the common condition.

Make sense now?

Yes got it. Thanks

 

I don’t think so. Not for a normal opamp.

Let's say supply is 15V and output is 1V.

To get 2W at 14.5V (15-0.5), the current would be 142mA. Normal, non-power, opamps top out at 20-30mA.

In the max power mode (Vout = 7.5V) the current would have to be 267mA.

There are operational amplifiers offering more then 100 mA output current. I found AD8397. There are two operational amplifers in the package. Let's say I have a power supply +12V and -12V for each operational amplifier. And I am having inverting configuration which basically invert the polarity with a gain set by the resistors R2/R1. Consider if the load on each operational amplifier draw continuous current 75 mA. The other end of the load is connected to ground.

Calculating the power dissipation for Vout = -3V.

P = V x I = 75/1000 x (12-3) = 675 mW

The same is true for the other operational amplifier.

Would that be a problem ? What consideration do I need to consider in PCB layout.

The total power dissipation would be 1.35 W for two operational amplifiers.

According to datasheet: the thermal resistances are given below for two packages with and with exposed pad.

8-lead SOIC_N: θJA = 157.6°C/W
8-Lead SOIC_N_EP: θJA = 47.2°C/W

Let's say I use the one with exposed pad so the thermal resistance would be 47.2 °C/W. Let's consider the ambient temperature 40 °C.

Junction temperature = ambient_temperature + thermal_resistance x power-dissipation
Junction temperature = (40 + 47.2 x 1.35) °C
Junction temperature = (40 + 63.72) °C

This is clearly above 100 °C.

Question: How can I use AD8397 for 2 x 75 mA continuous current through two operational amplifiers connected to two loads ? Is that impossible or still can be done by some PCB design consideration ?

 

How can I use AD8397 for 2 x 75 mA continuous current

And what is the load that will absorb this 75mA?
You keep taking about theoretical output currents, but you haven't specified what the load is.

 

And what is the load that will absorb this 75mA?
You keep taking about theoretical output currents, but you haven't specified what the load is.

This is still not clear to me about the load and what is the load impedance. Except the load require -3V with contineous maximum current 75 mA.

 

This is still not clear to me about the load and what is the load impedance. Except the load require -3V with contineous maximum current 75 mA.

Simple question: What is the load? In other words, what device is the op amp driving?

 

Simple question: What is the load? In other words, what device is the op amp driving?

This suppling power to the sensor module. I only know the voltage and the maximum current.

 

This suppling power to the sensor module. I only know the voltage and the maximum current.

Why would you use an op amp to supply power to a sensor module?
Usually one would connect the sensor module directly to the power supply.

 

Good practice to list the conditions. Possibly for audio maybe list power output measured at blah blah blah
and up blah blah power at xx.x % distorsion.
However not everything is audio. Looking at page. 20 find references rather than saying according to all about circuits.
Stating a reputable source as reference especially on a published paper looks good. Not as acceptable now with spice simulations.
Handbook of Operational Amplifier Applications (Rev. B)

Lecture with peer review might be tough when they throw tomatoes.
The simulation shown using overhead projector, it is also of interest on this forum. One of the most asked questions is where is your simulation file.
and now asked even for subcircuit of the op amp. No more BS. The change is that the probabilty of improvement based on specific data. See page 8
Microsoft Word - L240-Sim&Measo (gatech.edu)

Personally, It would be difficult to find the right op amp without simulation, It is so easy to pull up a working simulation and see how the op amp performs. last week I saw a circuit that specified a powe op amp, great on impedance and offset but it was terrible on noise,
The power modules sold as diy many would never make it as a finished product. Simulation is showing this dilemma.

 

Why would you use an op amp to supply power to a sensor module?
Usually one would connect the sensor module directly to the power supply.

The other requirement is adjusable voltage and there are several to fit on one PCB.

 

The other requirement is adjusable voltage and there are several to fit on one PCB.

You don't use op amps to send power at adjustable voltages in this manner.
Show us the entire application and we can guide you in a proper manner.

Unfortunately, we are now playing the 101 Q & A game.

 

This suppling power to the sensor module. I only know the voltage and the maximum current.

In that case, you use the opamp as a voltage follower with an external transistor. Or just use a voltage regulator.

A schematic would put an end to most of this tedious guessing game...

 

So you need to understand the Maximum Power Transfer Theorem:

"Maximum Power Transfer Theorem explains that to generate maximum external power through a finite internal resistance (DC network), the resistance of the given load must be equal to the resistance of the available source."

This means that the maximum power is transferred to the output when the source and load resistances are equal, at which point the load voltage is 1/2 the source voltage and the source dissipation equals the load dissipation.
Thus the maximum power dissipated in the op amp occurs when the output voltage is 1/2 the supply voltage.

That's, of course, for a fixed output load resistance, which is the common condition.

Make sense now?

This chain of reasoning relies on a couple of fallacies.

First, using the maximum power transfer theorem to conclude that max power transfer occurs when the source and load resistances are equal, is based on adjusting the load resistance to achieve max power transfer. But you just proclaimed that the load resistance is fixed, thus any adjustment occurs by changing the source resistance. When that is the case, max power transfer occurs by making the source resistance as small as possible.

To illustrate this consider the simple case where a fixed Vcc is across a variable source resistance and a fixed load resistance (in series). The maximum power is delivered to the load when the source resistance is set to zero and the full source voltage appears across the load, while under these conditions there is no power dissipated in the source.

Second, it is relying on the assumption that the Thevenin equivalent of the source is somehow equivalent in terms of power dissipation of the source it is modeling. This is not the case -- the Thevenin equivalent only attempts to match the voltage vs current characteristic of the source at the terminals and it makes no claim whatsoever about the power dissipated by the actual source being modeled.

 

This is beginning to seem like another instance of not providing adequate information, but rather asking how to make work the solution that the TS has picked, while the actual problem is rather different.

 

See

 

This chain of reasoning relies on a couple of fallacies.

They were assumptions, not fallacies.
The assumptions were a fixed load and the op amp output voltage varying from maximum to zero.

But you just proclaimed that the load resistance is fixed, thus any adjustment occurs by changing the source resistance. When that is the case, max power transfer occurs by making the source resistance as small as possible.

I was using the Maximum Power Transfer Theorem to illustrate when the maximum power dissipation would occur in the op amp, since the output looks like a varying resistance, not when the maximum output load power occurs, which is obviously at the maximum op amp output voltage.

it is relying on the assumption that the Thevenin equivalent of the source is somehow equivalent in terms of power dissipation of the source it is modeling. This is not the case

I made no mention of the Thevenin equivalent of the op amp, and did not make that assumption, so why did you bring that into the mix?
The dissipation is for the equivalent resistance of the op amp from power supply voltage to output (which varies), not the Thevenin equivalent.

 

They were assumptions, not fallacies.
The assumptions were a fixed load and the op amp output voltage varying from maximum to zero.
I was using the Maximum Power Transfer Theorem to illustrate when the maximum power dissipation would occur in the op amp, since the output looks like a varying resistance, not when the maximum output load power occurs, which is obviously at the maximum op amp output voltage.

You are using the Maximum Power Transfer Theorem to make say things that it says nothing about. The theorem ONLY talks about maximizing power transfer to a load. It says absolutely nothing, zero, nada, about power dissipation in the source.
[/QUOTE]
I made no mention of the Thevenin equivalent of the op amp, and did not make that assumption, so why did you bring that into the mix?
The dissipation is for the equivalent resistance of the op amp from power supply voltage to output, not the Thevenin equivalent.
[/QUOTE]

And yet you just equated the internal opamp circuity to it's Thevenin equivalent, namely an equivalent resistance in series from a supply voltage to the output.

Your entire premise is that the power dissipated in this equivalent resistance is the same as the actual power dissipated in the actual opamp internal circuitry. That equivalence is the fallacy.

 

For each instant the power dissipated in an op-amp is AT LEAST the instant output current multiplied by the internal voltage drop in the device. There will be a bit more current drawn by the preceding driver stages. That is where the additional power use is.