Increasing the heat of one of the two resistors will increase the Johnson noise in that resistor. This will generate a noise voltage across the resistor, will this essentially convert the heat energy into electrical energy? If the heat and resistance was high enough, could the second resistor be replaced by a diode and capacitor so the noise could be rectified?

  • \$\begingroup\$ If there is a voltage on a resistor, will the current flow through it? Herr Georg Simon Ohm claims it will. \$\endgroup\$ – Eugene Sh. Jan 5 '16 at 16:43
  • \$\begingroup\$ Ah! I feel I've worded the question wrong, thank you there! \$\endgroup\$ – Pyrohaz Jan 5 '16 at 16:44
  • \$\begingroup\$ The noise level of a typical resistor would be so tiny that an oscilloscope would be needed to see it at all. Here is a heavy on physics link to an experimental setup. \$\endgroup\$ – rdtsc Jan 5 '16 at 17:00
  • \$\begingroup\$ With an ideal diode, yes. But increasing the resistance also reduces the available current, so don't expect useful power. With any practical diode, the forward voltage required for conduction will exceed thermal noise voltage, so, practically, no. \$\endgroup\$ – Brian Drummond Jan 5 '16 at 17:07

Johnson noise is just the statistical variation of the voltage across a resistor that is due to the random motion of the charges within it.

If you work out the numbers, the amount of power that this represents is very very tiny — even when measured over a 1 GHz bandwidth, we're just talking about 4 picowatts (-84 dBm). If you wanted to produce 3.3V from this, you'd only get about 1 picoamp, assuming you could find a diode whose leakage is significantly less than that.

Note that in the case of two resistors with the same value, the Johnson noise power will divide equally between them, with half dissipated in the source (hot) resistor and half dissipated in the load (cold) resistor. But also note that any real circuit will have nonzero values of parasitic inductance and capacitance, and these will serve to limit the bandwidth of the noise transfer, causing more of the power to stay in the source and less to be transferred to the load.

  • \$\begingroup\$ A very well explained answer with physical examples, thank you. \$\endgroup\$ – Pyrohaz Jan 5 '16 at 17:14

Yes, electrical power will flow from the hot resistor to the cold one. This does not violate the 2nd law of thermodynamics -- it is basically another mechanism for heat conduction from a hot to a cold object. In this case, the heat is conducted by the electrons (which is the normal mechanism for metals and conductors anyway).

As the above answer mentioned, the noise power generated is 4.k.deltaT.BW (independent of resistance), and at maximum power transfer, only 1/4 of this will be deliverable to the load. Note that this power transfer equation includes a temperature difference term -- not only will the 'hot' resistor generate voltage and send power to the cold one, the cold one will generate (less) power and send it the hot one. The difference in temperatures is what matters. At 1 GHz, this is only 4 pW for a 300 K difference. If the temperature difference between transmitter and receiver was only 10 deg, the power would be even smaller.

Rectifying this with a (semiconductor) diode is theoretically possible, but not even close to practical. A diode conducts asymmetrically at all bias values, but for small signals as this has (<< uV), the asymmetry isn't 'sharp' enough, so there is very little difference between the forward current and reverse leakage.

A better approach would be to build a thermopile which generates a DC voltage directly because of differing electron mobilities in different materials.


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