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 = 1.38x10-23
T = 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.
PIC PROGRAMMING
QWhat program do you
use to program the PIC
12F675? I'm new to PIC
programming and have
been following your examples. I'm
January 2009 29