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Experiment #55 Current/Voltage Converters
From August 2007 QST � ARRL
H. Ward Silver, N�AX PO Box 927, Vashon, WA 98070 n0ax@arrl.org
Experiment #55
Current/Voltage Converters
N�AX
HANDS-ON RADIO
Can the handy op-amp spin straw into
gold? Not quite, but as the electronic
equivalent of the Philosophers Stone it can
transmute current and voltage without much
trouble at all!
Terms to Learn
Floating load
Load that has no ground
connection.
High-side/low-side drive
Circuit that
drives a load connected to ground (high-side)
or to a power supply (low-side).
Transimpedance amplifier
Amplifier
that acts an impedance to convert an input
current to an output voltage.
Virtual ground
Point in a circuit that
is maintained at ground potential, but is not
actually connected to ground.
Introduction
Buried deep inside many electronic
gadgets are circuits that output a voltage
(or current) based on a current (or volt-
age) input signal. They are often used in
radio for biasing, regulation, detection, and
other functions. In fact, you use a current-to-
voltage converter every time you change the
TV channel with an infra-red remote control
signal. The receivers photodiode converts
light pulses to a weak current that is then
converted to a voltage signal by an op-amp
circuit for the TVs digital electronics.
Voltage-to-Current Converters
Heres a secret whats the simplest cir-
cuit that converts current to voltage and vice
versa? Its a resistor! Ohms law describes the
process: I = E / R and E = I � R. If thats the
only consideration you had in mind, youre
done! In many circumstances, however, using
a resistor is unsuitable. Inserting resistance
into a current path may upset the operation
of the circuit. The current may be too high
(or too low) for realistic resistor values or
power dissipation. In these cases, the op-amp
converter can be used instead.
Figure 1 shows the simplest voltage-to-
current converter. These are also called cur-
rent regulators
or voltage-controlled current
sources (VCCS)
in different texts and articles.
A reference voltage, V
REF
, is applied to the
op-amps non-inverting (+) input. Until the
voltage at the inverting input () equals that
the conversion sensitivity is 0.1 mA / V, with
V representing the input voltage. Lets build
the circuit and test it. First, calculate the value
of R1 for a conversion sensitivity of 2 mA / V.
(500
) If a standard value of 510 is used,
what is the actual conversion sensitivity?
(1.96 mA/V.)
Build the circuit of Figure 1 using a 10 k

potentiometer (pot for short) to generate
the variable reference voltage and a 100

load resistor (R2). If you have two DVMs,
use one to measure input voltage at the
op-amps + input and the other to measure
current through R2. There are two ways to
measure current in R2 directly and indi-
rectly. To use the direct method, connect the
voltmeter (configured to measure current) in
series with R2. Indirectly, measure the volt-
age across R2 and use Ohms law to calculate
the current.
Set the pot to approximately mid-travel
and apply power. Adjust the pot for 0 V at
the op-amps + input. Confirm that current
through R2 = 0, as well. Now adjust the pot for
+1 V and measure the current through R2. It
should be pretty close to 1 � 1.96 = 1.96 mA,
with the direction from the op-amps output
toward R1. Now set the pot for 1 V and
repeat the measurements. Current flow should
be reversed, but of the same value.
Is the current flow really independent of
the value of R2? Try different values of R2
below 100
, including a short circuit. (Dont
worry you wont hurt the op-amp!) In
each case, you should get the same value of
current, unless R2 increases to the point at
which the op-amp output voltage reaches its
maximum limit. What value is that? If your
power supply delivers
�12 V, the LM741
op-amps output may be able to reach
�10.5
to 11 V. The op-amps output voltage V
OUT

= V
REF
+ I
R2
� R2 = V
REF
+ (V
REF
/ R1)
� R2 = V
REF
(1 + R2/R1). If V
REF
is set to
1 V and R1 = 510
, what value of R2
results in V
OUT
reaching 10.5 V? (4.8 k
.)
Try a 4.7 k
value for R2 and verify that
V
OUT
is close to 10.5 V. Then increase R2 to
6.2 k
without changing anything else. Can
the op-amp supply the required amount of
current? Continue to experiment by changing
the conversion sensitivity (change R1) and
the value of the load resistor (R2).
One drawback of this circuit is that neither
Figure 2 The current-to-voltage
converter is also known as a trans-
impedance ampli er. By requiring the
current in R1 to equal that owing into
or out of the input, the circuits output
voltage is equal to R1 � I
IN
.
Figure 1 The voltage-to-current circuit
works by balancing voltages at its
+ and terminals. The value of R1 sets
the circuits conversion sensitivity from
voltage to current in the load resistor, R2.
at the + input, the op-amps output voltage
increases, causing current to flow through R1
and R2 as shown.
With both op-amp inputs at the same
voltage (V
REF
), the current through R1 must
be I
R1
= V
REF
/ R1. Furthermore, the very
high input impedance of the op-amp means
that approximately no current flows into
the input. Therefore, I
R1
must equal I
R2

otherwise, where would it go? If R2 is
the load resistor, the current through it will
always
equal V
REF
/ R1. It doesnt matter
what value of R2 is, even a short circuit, as
long as the voltages at the output and the
input are within the op-amps power supply
limits. V
REF
can be positive or negative the
current flow follows V
REF
s polarity.
The conversion sensitivity to input voltage
is measured in amperes per volt (A/V) and is
equal to 1 / R1. For example, if R1 = 10 k
, From August 2007 QST � ARRL
HANDS-ON RADIO
terminal of the load resistor can be grounded.
This is called a floating load. There are a
number of more complicated circuits that
allow one end of the load to be grounded
(high-side drive) or connected to the power
supply (low-side drive). If an op-amp cant
supply enough current through its output,
there are also external circuits that can supply
higher currents. These can be found in the
references and in on-line Web sites.
Current-to-Voltage Converters
The complement to the previous circuits
has a current signal for its input and produces
an output voltage instead. Figure 2 shows the
basic form of this circuit. At first glance, this
circuit, known as a transimpedance ampli-
fier
, doesnt look like it should do much of
anything. But it relies on the op-amps high
input impedance and the circuits current
balancing capabilities.
The current into the op-amps terminal
must be balanced by an equal and opposite
current through the feedback resistor, R1.
(Remember to use the current-balance ap-
proach when figuring out op-amp circuits.)
Since the op-amps + terminal is grounded,
the voltage at the terminal will also be at
ground potential, called a virtual ground.
Given these two constraints, its easy to
figure out the output voltage: V
OUT
= I
IN

� R1. This balancing act is only limited by
the power supply voltages and the op-amps
ability to supply or sink enough current.
This circuit also has a conversion sensi-
tivity, measured in volts per ampere (V/A)
and equal to the value of R1. This circuit is
called a transimpedance amplifier, a fancy
name for an active resistor. An ordinary
resistor would also act as a current to voltage
converter, but the input source might not be
able to supply enough voltage or power. The
amplifier circuit of Figure 2 has a very high
input impedance, so whatever is supplying
the input current, such as a sensitive detec-
tor, is very lightly loaded. The high gain of
the op-amp ensures that enough voltage is
applied to R1 such that the currents balance
at its terminal.
Build the circuit of Figure 2, supplying
input current through a 1 k
resistor (R2)
connected to the same pot used in the previ-
ous part of the experiment. Use a 10 k
re-
sistor for R1, giving a conversion sensitivity
of 10,000 V/A or 10 V/mA. Set the pot to
mid-scale a