Most inexpensive hand-held multimeters have measurement ranges from several amps to single-digit milliamps. While generally handy, such meters are insufficient for sensitive current measurements. There is a solution though. With the following project, you can extend the current measurement range from milliamps down into the nanoamp and picoamp range with simple, low-cost circuits.
Most inexpensive hand-held multimeters have current measurement capability. The measurement range for these meters extends from several amps down to single-digit milliamps and sometimes into the microamp range. While this measurement capability is sufficient for many applications, there are times when more sensitive current measurement is required. There are meters available which measure much lower levels of current, but these meters are also more expensive and are often dedicated to just this one measurement function.
The goal of this article is to extend the basic hand-held current measurement range from milliamps down into the nanoamp and picoamp range with relatively simple, low-cost circuits. First, I’ll describe several types of current sources with their relevant performance characteristics that affect measurement circuits. Then, I’ll present practical circuits that deliver high performance at low cost. Each of these circuits will be analyzed to determine what level of measurement performance can be expected. General design issues common to all of the circuit techniques will also be discussed and recommended circuit component and layout techniques will be provided. Finally, two of the circuits will be discussed that were built and tested to demonstrate the desired goal of measuring nanoamp and picoamp currents.
Figure 1 shows several types of current sources. Figure 1a is the symbol of what electrical engineers call an ideal current source. It can have any level of current at the output and the output impedance is infinite. The result is that the output current is not affected at all by the characteristics of the measurement circuit. Of course no actual current source is ideal, but this is still a useful concept for approximating actual circuits and is used as a source in circuit simulation programs such as PSPICE.
Figure 1: The current sources: ideal (a), semiconductor (b), and resistive (c)
Figure 1b shows a current source which uses semiconductor transistors (either bipolar or FET). In this circuit, the output impedance is not infinite but can still be quite high (megohms). This means that varying voltages at the collector of the bipolar transistor or drain of the FET transistor have little effect on the output current as long as the voltage is not large enough to affect transistor operation.
Figure 1c is the least ideal of the current sources. The current is generated by the voltage difference across the resistor R2. Any measurement voltage developed by the measurement circuit directly affects the voltage difference across the source resistor. This changes the current that is measured which can result in measurement error. Having said that, Keithley Instruments in Low Level Measurements Handbook (6th edition, page 2-20) uses this current source model in defining what they call the feedback ammeter (also transimpedance amplifier). It may well be that, in real-world circuits, this model is the one that describes the majority of practical applications.
Although the current sources just described are not part of the measurement circuit itself, it is helpful to understand their limitations so that the measurement circuits can be designed to disturb the current source as little as possible. In this way, measurement error is minimized.
CURRENT MEASUREMENT CIRCUITS
In the past, current was measured directly with a moving coil meter. Now using semiconductor technology, voltage is the parameter that is measured directly. The current to be measured is first converted to a voltage by flowing through a load resistor. The resultant voltage is then measured and along with the load resistor is used to calculate the input current.
Figure 2: Current measurement circuits: resistive (a), transimpedance (b), and integrator (c)
Figure 2 and Figure 3 show several circuit techniques which are used to convert current to a voltage. Figure 2 shows the basic techniques, while Figure 3 shows modifications to two of these basic circuits, which give more accurate results and extend the measurement range. The symbol for the input current used in these circuits is the same as the symbol for the ideal current source used in Figure 1; but in this case, it is used to show where the input current connects to the measurement circuit and can represent any of the described current sources.
Figure 3: Modified current measurement circuits: modified resistive (a) and modified transimpedance amplifier (b)
Figure 2a is the least complicated of the measurement circuits. In this circuit, the current source is connected to one end of the load resistor R1 and the other end of the resistor is connected to ground or some other reference point. The voltage developed across this resistor is measured with the voltmeter and used to calculate the input current. This circuit is very simple and is often used on the spot with an available resistor for quick measurements at the workbench or in the field.
The voltage that is developed across the load resistor is called the burden voltage. For a current source that is nearly ideal (such as the transistor source), the burden voltage has relatively little effect on the current being measured unless it is large enough to change the internal working of the current source. For resistive current sources, the burden voltage can directly interact with the current source and give erroneous current readings. This occurs because the load resistor becomes part of the current generating resistance which reduces the current. To minimize this interaction, the load resistor should be much smaller that the output resistance of the current source. The corresponding burden voltage will then also be small.
When a simple hand-held voltmeter is used, the measured voltage cannot be too small because these meters rarely measure below 1 mV. Thus, a compromise is needed between measurement accuracy and a low voltage burden.
Most hand-held meters have a 10-MΩ input impedance on the voltage scale. The load resistor R1 will be in parallel with the meter impedance and needs to be selected appropriately to give the desired equivalent measurement resistance. An example in Table 1 shows the resistor values needed to give a measurement voltage of 50 mV for the given currents. This fairly low value of measurement voltage significantly decreases the burden voltage while at the same time providing enough voltage to give measurement accuracy on the order of 10%.
Table 1: Resistor Values versus Input Current for Resistive Circuit (*Rounded off value. Less than 0.1% error.)
When making a current measurement with the resistive circuit, it is always a good idea to try several resistors of different values to see what voltage results. If changing resistor values by a certain amount changes the measurement voltage by the same amount, then the source current is not being affected by the measurement (burden) voltage. In this case you can use the higher value of resistance to get more output voltage and more measurement accuracy. Conversely, if the corresponding measurement voltage increases less than the amount of the resistor value change, the source current is being affected by the measurement circuit and the smaller resistor value should be used.
A small modification to the circuit as shown in Figure 3a gives improved performance by removing the burden voltage at the expense of adding a variable power supply. This can be particularly useful as a quick measurement tool using an available workbench power supply. The power supply is adjusted until VOUT is 0 V. The value of the current is then obtained by dividing the measured value of the power supply voltage by the resistance R1. In this way the burden voltage is removed and the load resistor can be increased so that the power supply voltage can be greater than 50 mV. This will give more accuracy in the measurements.
Because VOUT is zero, the leakage current going into the hand-held voltmeter is zero and the finite input impedance (10 MΩ) of the meter does not affect the measurement. Even when VOUT is not exactly zero, the leakage current is still small. For example, for VOUT < 5 mV, the leakage current will be less than 500 pA. This gives measurement accuracy of 1% or better for input currents greater than 50 nA. Since the resistor R1 in this modified circuit is not developing a burden voltage, the resistor value is decoupled from the input current and can be any practical value depending only on the maximum voltage of the power supply.
Figure 2b shows the circuit for a transimpedance amplifier. This is perhaps the most versatile of the current measurement circuits in that it can cover a large current measurement range using a simple circuit. In this circuit, the output from the current source is connected to the negative input of the operational amplifier while the positive input of the amplifier is connected to a reference voltage. This reference voltage is typically the circuit ground when there are bipolar power supplies and some intermediate voltage when there is a single power supply.
The inputs of an operational amplifier have very high input impedances (greater than 1 GΩ) so that little current goes into the amplifier. Thus, the input current drives the negative input toward one of the power supply voltages depending on the polarity of the input current. This causes a voltage difference between the amplifier inputs which is then amplified with the large internal open loop gain of the amplifier. As a result, the amplifier output voltage moves in a direction to provide current through resistor R2 which is opposite to the input current. Equilibrium is achieved when the amplifier output voltage is such that the current through R2 is equal in magnitude to the input current. With an ideal amplifier and no offset voltage, this results in 0 V at the negative terminal matching the voltage at the positive terminal. Only the value of resistance R2 and the amplifier output voltage are needed to determine the input current. Since we know the resistance value and can measure the output voltage, we can calculate the current through R2 which will equal the magnitude of the input current.
Since there is no burden voltage, the input current is not affected by the value of the feedback resistor R2 or the magnitude of the output voltage. The output voltage is constrained by the power supply voltages, but in principle there is no constraint on the value of the feedback resistor. Table 2 shows R2 resistor values for several nominal input currents for an output voltage (VOUT) of 1 V.
As seen in Table 2, the resistor values get quite large for small currents. These large value resistors are expensive and are often also physically large. Also, circuit constraints like stray capacitance can have an appreciable effect on the circuit when the resistor value is large.
Table 2: Resistor values for nominal input currents and VOUT = 1 V
There are two ways to reduce the resistor value required for a particular input current. One way is to allow smaller voltages than 1 V to represent the input current. This is acceptable as long as all anomalous voltages in the circuit due to circuit imperfections are calibrated out. This calibration can be either physical using potentiometers to cancel offset voltages. Alternatively, data calibration can be used by measuring the output voltage without the input current and then subtracting that data from the output voltage with the input current. Thus, as the input current (and corresponding output voltage) are reduced, the measurement voltage will still have sufficient accuracy.
The second way to measure lower values of current with a lower resistor value is to use the modified transimpedance amplifier shown in Figure 3b. Here, the output voltage is reduced by the voltage divider consisting of R3 and R4 before driving the feedback resistor R2. If the feedback resistor here is the same as the feedback resistor without the divider, the current flowing to the negative input terminal will be less than before. The internal gain in the amplifier will make the output voltage larger to compensate. For a voltage division of 10, the input current is 10 times lower for the same output voltage as before. One caution: although this circuit does give flexibility in the design, care is needed because there is an amplifier voltage gain equal to the divider ratio. The internal offset voltage and noise voltage of the amplifier are multiplied by this amplifier gain along with the current signal.