Thermal Noise Power
Pn = k × T × BW
Quick answer
This calculator solves Pn = k × T × BW for Noise Power. With Temperature (K) 0.5 K, Bandwidth 10 MHz, the result is 6.9 × 10⁻¹¹ µW. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Johnson-Nyquist thermal noise is the fundamental noise floor generated by the random thermal motion of charge carriers in any resistive element. It is white noise (flat spectral density) and is unavoidable in any circuit at temperatures above absolute zero. Boltzmann's constant k = 1.380649 × 10⁻²³ J/K connects thermal energy to temperature. At room temperature (290K), the noise power spectral density is -174 dBm/Hz, which is the ultimate sensitivity limit for any receiver. Reducing noise below this floor requires cooling the receiver, as done in radio astronomy.
Variables
- Pn — Available noise power (W)
- k — Boltzmann constant (1.38 × 10⁻²³ J/K)
- T — Temperature in Kelvin (K)
- BW — Measurement bandwidth (Hz)
Practical Notes
At 290K: noise floor = -174 dBm/Hz. In 1 Hz BW: -174 dBm. In 1 kHz: -144 dBm. In 1 MHz: -114 dBm. In 20 MHz (WiFi): -101 dBm. Noise voltage across a resistor: Vn = √(4kTRBW). A 10kΩ resistor at room temperature generates about 13 µV RMS noise in a 1 MHz bandwidth.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| Pn | Noise Power | W | µW, mW, W, kW |
| T | Temperature (K) | — | K |
| BW | Bandwidth | MHz | Hz, kHz, MHz, GHz |
Step by step
- 1. What you know: Temperature (K) 0.5 K, Bandwidth 10 MHz
- 2. The relation: Pn = k × T × BW
- 3. Substitute: T = 0.5 K, BW = 10 MHz
- 4. Result: Noise Power = 6.9 × 10⁻¹¹ µW
Worked examples
Minimum range
Given: Temperature (K) 0.025 K, Bandwidth 100 kHz
Result: Noise Power = 3.45 × 10⁻¹⁴ µW
Low range
Given: Temperature (K) 0.25 K, Bandwidth 1 MHz
Result: Noise Power = 3.45 × 10⁻¹² µW
Typical
Given: Temperature (K) 0.5 K, Bandwidth 10 MHz
Result: Noise Power = 6.9 × 10⁻¹¹ µW
High range
Given: Temperature (K) 0.8 K, Bandwidth 100 MHz
Result: Noise Power = 1.1 × 10⁻⁹ µW
Maximum range
Given: Temperature (K) 8 K, Bandwidth 1 GHz
Result: Noise Power = 1.1 × 10⁻⁷ µW
Solving for each variable
Solve for Noise Power
Temperature (K) 0.025 K, Bandwidth 100 kHz → 3.45 × 10⁻¹⁴ µW
Solve for Temperature (K)
Noise Power 10 mW, Bandwidth 100 kHz → 7.24 × 10¹⁵ K
Solve for Bandwidth
Noise Power 10 µW, Temperature (K) 0.025 K → 2.9 × 10¹⁰ GHz
How Noise Power changes with Temperature (K)
| Temperature (K) (input) | Noise Power (result) |
|---|---|
| 0.05 K | 6.9 × 10⁻¹² µW |
| 0.125 K | 1.73 × 10⁻¹¹ µW |
| 0.25 K | 3.45 × 10⁻¹¹ µW |
| 0.375 K | 5.18 × 10⁻¹¹ µW |
| 0.5 K | 6.9 × 10⁻¹¹ µW |
| 0.75 K | 1.04 × 10⁻¹⁰ µW |
| 1 K | 1.38 × 10⁻¹⁰ µW |
| 2 K | 2.76 × 10⁻¹⁰ µW |
| 3.5 K | 4.83 × 10⁻¹⁰ µW |
| 5 K | 6.9 × 10⁻¹⁰ µW |
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<p><a href="https://vbatpower.com/formulas/thermal-noise/">Thermal Noise Power</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Thermal Noise Power?
Use Pn = k × T × BW. With Temperature (K) 0.5 K, Bandwidth 10 MHz, Noise Power works out to 6.9 × 10⁻¹¹ µW.
What is the formula for Thermal Noise Power?
The formula is Pn = k × T × BW. Johnson-Nyquist thermal noise is the fundamental noise floor generated by the random thermal motion of charge carriers in any resistive element.
How do you find Temperature (K) when you know the other values?
Rearrange for Temperature (K): with Noise Power 10 mW, Bandwidth 100 kHz, Temperature (K) is 7.24 × 10¹⁵ K.
How do you find Bandwidth when you know the other values?
Rearrange for Bandwidth: with Noise Power 10 µW, Temperature (K) 0.025 K, Bandwidth is 2.9 × 10¹⁰ GHz.
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