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If I setup an opamp in a non-inverting configuration as follows:

enter image description here

Vrail+ is 7.5V, Vrail- is GND (0 V)

Vin is 2.5V, Vout = 3.3V (in other words, the gain, aka, 1+R2/R1 is 1.32 [V/V])

Iout = 100mA (connected to some load)

What are the source of my losses? How efficient are these types of configurations? Do I only take the quiescent current as the only loss?

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100mA is a pretty heavy load to put on the output of an op amp. It would be cheaper to just use a 3.3V buck regulator than an op amp that can put out 100mA – Matt Young Feb 26 '14 at 13:58
Can't use a buck because I don't have enough space to put an inductor and diodes and all the jellybeans – hassan789 Feb 26 '14 at 16:25
up vote 2 down vote accepted

The inherent efficiency will be the same as any other linear regulator, give or take, depending on the op-amp quiescent current (could be uA to mA for the op-amp just sitting there with no load).

\$P_D = (V_{in} - V_{out}) \cdot I_{load} + I_q \cdot V_{in} \$

100mA will require an expensive op-amp or a booster stage on the output.

There's another problem- linear regulators are designed and specified for capacitive loads such as the bypass capacitors on whatever you're going to connect to the 3.3V rail. If you connect an op-amp as shown in your schematic to such a load it will probably oscillate and/or overshoot. That could damage your load, and even if the bypass capacitors kill the apparent oscillation it will greatly increase the power consumption of your circuit. I suggest that unless you have very special requirements, you should use voltage regulators to regulate voltage and op-amps to manipulate signals. It's not that you couldn't compensate such a circuit with additional components so it would be stable, it's just that I don't see the point.

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Opamps can't usually supply 100 mA. But, they can still be used to control a power voltage if you add some current gain to their output. If you really want a opamp to control a power voltage that can supply 100s of mA, here is a simple way:

The opamp still does the controlling and still provides voltage gain from Vin to Vout. However, Q1 provides most of the Vout current. R2 and R3 are the voltage feedback and the overall gain is the reciprocal of their attenuation ratio. In this example R2 and R3 are equal, so the voltage gain from Vin to Vout is 2.

You might think all you need is the transistor and R2 and R3, but that could be unstable. To make the opamp stable, you put a small cap directly between its output and inverting input. But, you also need to let the opamp output not be loaded directly with the end load, especially if that load is capacitive. R1 decouples the opamp output from the load such that the opamp should be able to reach whatever output voltage it wants to. Without that, the stability feedback provided by C1 won't work right and the thing can oscillate anyway.

The best value of C1 is hard to predict, so start with something vaguely plausible, like I show, and then experiment. Making it lower will invite instability and result in oscillations. Making it higher will dampen the response to transient loads. Test by connecting the worst case non-resistive load you expect, find the value of C1 just where it starts to be unstable, and then double it in the real circuit.

However, stepping back a couple of layers in this case, you don't need a opamp voltage regulator since the voltage you want is both a common value and known up front. There are plenty of 3-terminal linear regulators specifically designed for this task that can source 100 mA. If it's OK to waste the extra power as heat in the linear regulator, then this is clearly the way to go in your case. In this case you're dropping 7.5V - 3.3V = 4.2V, which times 100 mA is 420 mW. A linear regulator in a TO-220 or similar case will get warm, but won't require a heat sink.

If efficiency or getting rid of heat is a issue, then use a buck regulator to make the 3.3 V supply from the existing 7.5 V supply.

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Opamps are linear devices, and as such, they're going to be as (in)efficient as any other linear regulator.

The load current at the output pin is supplied from the opamp's power supply pins, and any difference in voltage between the two pins appears across the driver transistors in the opamp's output stage. This voltage, multiplied by the load current, represents power that will get dissipated as heat inside the opamp.

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You won't be able to supply 100mA from all standard op-amps - maybe 20mA for most and up to 50mA for more specialist devices. There are such things as power op-amps that can deliver amps but these are specialist devices.

Regarding efficiency - the power delivered to the load might be 3.3V x 20mA = 66mW and the power into the opamp will be 7.5V x (20mA + ~5mA for the op-amp) = 187.5mW.

The quiescent current of the device may be 3mA rising to 5mA to produce the output current. There's a bit of hand-waving but even if you ignored any current taken by the op-amp you still have 7.5V powering it and 20mA being delivered to the load = 150mW.

If you ran the op-amp from a 5V supply it would be better but remember it is a linear device that isn't like a switch mode power supply - if your output is 20mA this 20mA gets taken from the op-amps power rails.

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This circuit gives the output voltage as is set for (maximum output is limited by the rails) but it will not be well regulated anymore at higher load. This also depends on the strength of the op amp of interest. There are lots of high current capability (Amperes) op amps available now out there.

This circuit isn't a voltage regulator by any chance. It'll oscillate with high capacitive load, and can't sustain line or load transients.

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