If I setup an opamp in a non-inverting configuration as follows:
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?
Best Answer
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.