# Tradeoff between low-noise and low-power design

Can anyone help me to understand why this sentence below (which I had find it in a paper) is correct?

Designers must address the tradeoff between low-noise and low-power designs of the amplifier.For an ideal thermal-noise-limited amplifier with a constant bandwidth and supply voltage, the power of the amplifier scales as $1/(vn^2)$ where $vn$ is the input-referred noise of the amplifier.

But I had read in a lot of books that refer that the thermal noise given by this expression $P = K \cdot T \cdot \Delta f= \dfrac{vn^2}{4 \cdot R}$ so power and $vn$ are proportional. So which statement is correct?

• It doesn't make a lot of sense to me without more context. I mean, one can always add gain stages (after a low noise first stage) if you need more power. – George Herold Jun 30 '17 at 18:42

## 2 Answers

Two different parameters. In your equation, P is the power of the signal itself, but in your excerpt, P is power required for the amplifier.

In MOSFET operational amplifiers, as in bipolar, the primary focus is on DC precision: matched input differential devices (FET or bipolar), serviceable matching of high frequency nodes so as to preserve some common-mode rejection, some gain, some output drive strength and stability with capacitive loads.

That first requirement --- matched input differential devices --- leads to NOISY operation of MOSFETS if the IDD is low. Conversely, high current MOSFET opamps can be (but need not be) designed for low noise.

The very lowest noise opamps seem to be bipolar, perhaps SiGe bipolar for fast and lownoise opamps.