Nuts and Volts - January 2009

cannot run without a load. It will feed a low voltage element in a water heater. When that thermostat is satisfied, it will feed a low voltage element in a small swimming pool. When that thermostat is satisfied, it will feed low voltage lights equal to the required load. I need a controller to insure that the water heater is always served first, then the pool, and then the lights.

— Bill Edgerton

AIf your thermostat switch is single pole, single throw (the usual case), then some relays will be needed. We can't rely on the voltage from the generator to operate the relays, so I am proposing 24 volt types. In the circuit (Figure 8), when the water heater is calling for power, voltage is removed from the other loads and applied to the water heater. When the water heater is not calling for power, voltage is passed on to the swimming pool. If it is not calling for power, the voltage is applied to the lights. The 24 volt transformer and relays should be available at your local plumbing shop.

HIGH IMPEDANCE PREAMP

QI need an ultra low noise high impedance preamp for an instrument application. The preamp should be reasonably flat over 5 kHz to 500 kHz. and be located on the sensor, fully shielded, battery powered, and driving a 50 ft. shielded cable to the remaining instrumentation. Short battery life is not a problem as the test runs are 30 minutes or less.

— James Ward

■ FIGURE 9

AI am going to assume that your signal source is inductive with low resistance because that will give the best noise performance. I chose the MAX4448 because it has low noise and wide bandwidth; see (Figure 9).

Resistors have noise = 4k TR where k = Boltzman's Constant = 1.38x10^-23T = Kelvin temperature = 273 at room

This equation takes into account all the input noise sources except 1/F noise which is important below 1 kHz. Since your band starts at 5 kHz, we can ignore the 1/F noise.

Et = √(en²+ (Rp + Rn)² In²+ 4k T*(Rp + Rn)) where: Et is the equivalent input noise voltage in volts per root Hz en is the specified input noise voltage density = 4. 5 nV/√(Hz) Rp is the non-inverting input

QUESTIONS & ANSWERS

■ FIGURE 8

resistance (assume five ohms) Rn is the inverting input equivalent resistance = R1*R2/(R1 + R2) = 986 ohms In is the specified input noise current density = 0.5 fA/√(Hz)

Going through the math, I come up with Et = 5. 9 nV/√(Hz)

The gain of this amplifier is 36 dB which leaves about 1 MHz bandwidth. To find the wideband noise, multiply by the square root of the bandwidth: 5. 9 nV*10³= 5. 9 µV. To find the output noise, multiply by the gain. Since the gain doubles every 6 dB, 36 dB is a gain of 64, but this gain is 69.1 (rounding error!); 5. 9 69.1 = 407 µV. If the 1/F noise is a problem, put 470 pF across R2.

The MAX4448 won't have any trouble driving a 75 ohm cable at that level, but I added 75 ohms in series to isolate the capacitance of the line which could cause the preamp to oscillate. You can shut down the preamp between measurements to prolong battery life.