Two instrumentation amps make accurate voltage-to-current source
Build a 0.01%-accurate voltage-controlled current source.
Frank Ciarlone, Analog Devices, Wilmington, MA; Edited by Martin Rowe and Fran Granville -- EDN, May 14, 2009
Many designs require precise voltage-controlled current sources, especially in the presence of variable loads. Common approaches, which use a few op amps and a handful of passive components, have inherent errors due to nonideal component characteristics, such as finite open-loop gain, common-mode rejection, bias current, and offset voltage. Designs using operational amplifiers may require precision resistors to set gain and additional capacitors for stability. In addition, some circuit designs provide currents that are not directly proportional to the input voltage. The voltage-to-current converter in Figure 1, for example, relies on the fact that the collector current is approximately equal to the emitter current and provides current in only one direction.
With two instrumentation amplifiers and two transistors, you can build a 0.01%-accurate voltage-controlled current source (Figure 2). This current source features a ±10V input-voltage swing that is directly proportional to the output current. It maintains high accuracy, even while delivering as much as 90 mA of output current. The AD620 low-power, low-drift instrumentation amplifiers from Analog Devices provide circuit control and error correction but are not part of the output circuit. Thus, you can substitute higher-power transistors for Q1 and Q2 to achieve higher output currents. You can configure the instrumentation amplifiers for any gain of one to 10,000 to accommodate input signals lower than 1 mV. Simply connect a resistor across the inputs of both IC1 and IC2 to achieve the desired gain.
The first instrumentation amplifier, IC1, controls the base voltage of the push-pull output stage. The resistors and diodes provide bias to Q1 and Q2 to eliminate crossover distortion. IC2 provides error correction and accounts for deltas in the base-to-emitter voltage. The error voltage, which you measure differentially from the D1/D2 junction to the output voltage, feeds into the reference pin of IC1, summing it with the input voltage. The result is an output current that is directly proportional to the input voltage. This circuit achieves a 0.01% typical dc accuracy across a ±10V input span and 1.5% typical ac accuracy at 1 kHz with an output voltage of ±5V p-p.
The equations for calculating the output current are:
where
Therefore,
or
This circuit provides a wide output range, as well as output current that is directly proportional to the input voltage and high linearity and precision (Figure 3).
-
But in the EDN circuit RL is NOT in the feedback circuit. The diode/transistor buffer is in the feedback loop, so IC2 cancels any offsets that occur there. The circuit is a buffer amplifier with output impedance RL.
Nick Allen - 2009-24-9 12:35:00 PDT -
This design seems to be based on those shown in the LT1167 and INA128/129 data sheets and Linear Tech Design Note 182.
It is adding a transistor current booster and substituting a second instrumentation amp IC2 for a voltage follower which was used to buffer the reference current flowing from the instrumentation amp in order to increse the resolution. And these designs are substituting an instrumentation amp for the single op amp of the modified Howland design shown at :
michaelgellis_dot_tripod_dot_com/howland_dot_html
There it is shown that the output current does indeed flow through an output resistor (RL in the EDN circuit) and the output current does depend on the value of that series resistor because it is in the feedback loop of the control. Thus the current can be set by changing that resistor value, by changing the amp gain or by changing the input control voltage.
Steve - 2009-18-9 08:57:00 PDT -
Another instrumentation amp design is shown on the LT1167 data sheet. The Linear Technology Design Note 182 shows this as an 8 decade current source with 3 picoamp resolution. Output current though is limited to 20 mA by the LT1167. The INA128/129 data sheet shows a similar circuit.
Steve - 2009-18-9 08:16:00 PDT -
"V=I/R" can NOT be correct --> the units do not match up properly! V=IR!
Jahoka Fall - 2009-15-6 08:54:00 PDT -
Either this is a test or I'm totally missing the circuit function here! The equations for the circuit are correct and V=I/R. So the current will CHANGE with R. Rl usually means the load resistance. In this circuit the current out goes through the load and the Rl in the circuit. So when the load chnages the current will change - not the function of a constant current supply. This is even reflected in your performance graph where the current is different for diffrent Rl. I would say that this circuit is a current buffer for a high impedance voltage source.
Clark Boyd - 2009-11-6 05:13:00 PDT
