Self-Powered Digital Voltmeter
circuit that would draw an input current of less than 1
milliamp.
Common Mode Range
One of the advantages of a digital meter over an
analog one is that its input can measure positive and
negative voltages without reversing the input leads. To do
so, the meter must have a bipolar power supply. Since a
dual supply is not a feasible approach in battery-powered
equipment, the solution that engineers decided on is to
have the return (negative) input referenced to a mid-supply
level. Since the supply for battery-powered meters is
floating and independent of any external reference, it is of
no concern that the meter’s negative sense input is not tied
to the battery negative.
The positive input resistor voltage dividers are also
referenced to this mid-supply, which usually converts the
measured voltage to the ±200 millivolt measurement
range. Additionally, as long as the return input is
referenced to this mid-voltage within its common mode
range — usually plus or minus 1 volt — the meter will be
able to function.
In a 9 volt powered meter, the battery’s negative lead
sits 4. 5 volts below this mid-supply reference — way too far
from its common mode range for the meter to function
properly if the negative input was tied to it. Some meters
(especially those functioning with 3 to 6 volt batteries), use
a charge pump to artificially generate a negative supply
voltage. Although this is a practical solution, those circuits
consume currents in excess of that of the meter. Since a
very low current consumption was one of the goals, this
option was ruled out.
Opamps to the Rescue
It is a sure bet that almost everyone who is interested
in electronic circuits has experimented with operational
amplifiers. It is common knowledge that these devices are
easily used as voltage amplifiers, but it is not very well
known that they may also work as voltage level shifters.
Thus, a voltage which is referenced to the negative supply
lead may be applied to the meter’s floating input. The trick
is explained in the sidebar.
Having a digital voltmeter that doesn’t draw power
away from your vehicle is a very simple and straightforward
project, as seen in the schematic in Figure 1.
The complete circuit is fed from a low dropout
voltage regulator, U1, which maintains the regulated 8
volts. The current consumption is so low that no
heatsink is required. The tantalum capacitor, C3, is
required for stability.
Although the meter — having been designed
for portable operation — is rated for 9 volt nominal
Teaching an Old Dog a New Trick
If you are reading this, it is likely that you are very familiar with
the classic opamp gain formulas:
Inverting gain Av(-) = -Rf/Ri
Non-inverting gain Av(+) = 1 + Rf/Ri
What this essentially means is that, if you have easy round
numbers — a 100K feedback resistor and a 20K input resistor —
you will obtain a gain of - 5 if your signal is fed to the inverting input
and a gain of + 6 if the signal is fed to the non-inverting input. It does
not get any simpler than that.
Of course, for this to be true, it is assumed that the signal input
is grounded for the output to be also referenced to ground. Of
course, for this to be true, it is assumed that a 0 output voltage implies
a 0 input voltage — since they share the same ground reference.
What if you would like to have a nonzero output voltage for
a zero input voltage? What if you want it to be 4 volts to
accommodate the digital meter’s common mode range? Then you
must also level shift the signal. You can simultaneously level shift and
add gain to an input signal.
In our particular case, we need to convert from a range of
roughly 9 to 15 volts to 90 to 150 millivolts to accommodate the
meter’s 200 mV range. This means a gain of -1/100. (Yes, you can have
gains lower than 1.0.) If we select a 1 MΩ input impedance, then the
first equation will show that the feedback resistor must be 10 KΩ.
Following the second equation, the non-inverting gain with
those resistor values is now 1 + 1/100 or 1.01. The compensating
voltage (shown as a battery in Figure A) will be connected to that
terminal and will see that gain value.
What should the compensating voltage be? Simple: the desired
output voltage (ex., 4 volts) divided by that gain or 4/1.01 = 3.9604.
Where am I going to get such an odd voltage? Again, this is
simple; I’ll use a resistive voltage divider from the 4 volt reference
with the same gain-setting resistor values as the opamp. The end
result is shown in Figure A.
The only problem with such a circuit is that the output is inverted;
an increasing input voltage will result in a decreasing output voltage.
To make it the proper polarity, an additional inverting section must
follow it. To keep all resistors equal then, it is wise to split the gain in
the two sections — in other words, -1/10 and -1/10.As the signal fed
to the second opamp is already level-shifted, its non-inverting input
is simply returned to the reference voltage.
Figure A
JULY 2004
53
