Electronic – Johnson noise potential divider

If you draw 1mA from the resistor divider circuit you mentioned, it will output one volt (the upper resistor will have 1.1mA flowing through it, thus dropping 11 volts; of that 1.1mA, 0.1mA will go through the bottom resistor while the remaining 1mA will go into your load). The 6K resistor would drop 6 volts, thus feeding 6 volts into a 100mA load.

If either the load current or the load resistance is a known constant value, one can calculate a series resistance which will convert a known input voltage into any desired known, lower, load voltage. If the load current or resistance isn't known precisely, however, deviations from the ideal will cause the load voltage to vary from what is intended. The greater the difference between the input voltage and the load voltage, the greater the variation in load voltage.

Adding a load resistor will effectively add a known fixed load in addition to the potentially-variable one. Suppose one had a 12-volt source and the intended load were 10uA +/- 5uA at 6 volts. If one just used a series resistor sized for the 10uA case (600K), it would drop only 3V at 5uA (feeding 9 volts to the load) and 9V at 15uA (feeding 3 volts to the load). Adding a 6.06K resistor in parallel with the load would cause the total current draw to be about 1.000mA+/-0.005mA, requiring the upper resistor be changed to 6K; since changes in the load current would only affect the total current by about 0.5%, they would only affect the voltage drop of the upper resistor by about 0.5%.

If the source voltage is stable, and the output current is small, a voltage divider may be a practical means of generating a stable voltage. Unfortunately, for the voltage divider to generate a stable voltage, the amount of current fed through the lower resistor (and thus wasted) must be large relative to the possible absolute variation in load current. This is usually no problem when the output current is on the order of picoamps, is sometimes acceptable when the output current is on the order of microamps, and generally becomes unacceptable when the output current is on the order of amps.

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.