The transconductance amplifier (TA) is a widely used, stable, and accurate source of electrical current. TA’s are commonly used for testing and calibration of current sensors such as shunts, current transformers (CT’s), and Rogowski coils. The TA is also used in basic research applications requiring accurate and stable source of current.
The TA takes a voltage input (and in some cases current) and produces an accurate current that is nearly independent of electrical load impedance. Consider a TA that has a 100 A full scale range with a 1 V input. If one places 0.5 V into the same amplifier the expected output will be 50 A. TA’s are an ideal device to use when one cares about having a constant current source for calibration or other testing purposes. However, it must be used with knowledge about its limitations to assure its specifications are met.
The TA is able to maintain a constant current into a varying load within some specified limits. This limit is determined by a parameter known as compliance voltage. The compliance voltage is the maximum voltage at the TA output terminals that is available to produce the desired current into the connected load. Load variations may occur as connection resistances (contact resistances) change with time, heating of conductors as current is applied, load heating, and any other causes or changes in resistance or inductance. (Inductance is discussed further in the following.)
Typical TA maximum compliance voltages range from approximately 2 volts to as high as 7 volts. Specifications for compliance voltage are typically broken into two categories: dc and ac. Further, the voltages may be different on each TA current range with ac compliance voltage typically specified in rms (root-mean-squared) units.
Direct current
Let’s look at the simple case of producing a 100 A dc into a load with a resistance of 0.04 Ω. Using simple Ohm’s law the voltage that is required to accomplish this is:
V=IR=100*0.04 = 4 volts 
For the amplifier to be able to produce the desired current into the 0.04 Ω load it must have a compliance voltage of at least 4 V. If the TA only has a compliance voltage of 3 V on the 100 A range then significant error in the current produced will result. Many TA’s will also detect the over compliance and stop producing current, that is go into Standby condition.
The dc load that the TA sees is the sum of any loads being driven, such as a shunt or resistor, the cable resistances and contact resistances. In other words the sum of all resistances connected to the TA output terminals. We can easily calculate the cable resistances from available conductor resistance tables. Contact resistances must usually be estimated or measured with specialized test equipment.
Let’s consider a typical circuit for testing a 0.01 Ω current shunt at 100 A. Assume the cable resistance is 0.0027 Ω for both cables (+ and – cables combined). We might estimate the contact resistance at 0.002 Ω. Our total load will be: Rtotal= 0.01+0.0027+0.002= 0.0147 Ω. The amplifier will need a compliance voltage of
V=IRtotal =100 A*0.0147 Ω= 1.47 V
which is well within the range of most TA’s.
Alternating current
We now consider a TA generating an alternating current. When we want to produce an alternating current we must consider the ac characteristics of the load and cabling. This involves looking at the inductance of the load as well as the resistance connected to the TA output terminals. (We assume for simplicity that the dc resistance and ac resistance are the same for this analysis.)
In the simplest case the TA sees a load which is the series sum of Rtotal and Lt where Lt results in an inductive reactance at frequency f of XL. Producing the current in the load will require a compliance voltage given by:
Vcomp=I*Z=I*(Rt2+XL2 )1/2
where
XL=2πfLt the inductive reactance,
f = frequency in Hz,
Lt = the total of all inductances connected to the TA,
Z= the total circuit impedance magnitude,
I = the current rms value,
Rt = the sum of all resistances.
In high current circuits the dominant source of inductance is created by either the load or the cabling. The load inductance must be provided by a device manufacturer, estimated or possibly measured. In the case of shunts, Hall-effect current sensors and Rogowski-type sensors the inductive component is negligible compared to the inductance created by the cabling.
The interconnection length, shape, and conductor size are the main contributors to the inductance.
The most common interconnection geometries are either a circular path or a rectangular path, aka structures. We consider the circular structure as it is the simplest to calculate. The circular path inductance is given by [1]:
L=µ0 a[loge (8a/R) - 1.75]
where:
L = the inductance in henries,
a= loop radius in meters, and
R = radius the conductor carrying the current, in meters.
Let’s consider testing a 0.001 Ω shunt, the load, driven by a TA at a current of 100 A @ 50 Hz. We assume the connections are such that a nearly circular path is created with a total cable length of 2 m with a conductor cross section of 25 mm2 yielding a radius of 0.0028 m. Using the above inductance equation and well known circumference and area relationships yield the following:
a=0.32, R=0.0028, µ0= 4πx10-7 H/m resulting in an inductance of L =2.04 uH.
Rt is the sum of the load resistance, cable resistance and an estimate for connection contact resistances.
Rt= 0.001+0.0016+0.002 = 0.0046 Ω
Now use Rt and L to determine the required compliance voltage.
Vcomp=I*Z=I*(Rt2+X_L2 )1/2
Vcomp=I*Z=100*(0.00462+(2π*50*2.04x10-6)2)1/2
=100* (0.00462+(0.00064)2)1/2
=100*0.0046 Ω =0.46 V
Note that there is very little difference at this low frequency between Rt and Z.
At 500 Hz we see that XL increases to 0.0064 from 0.00064 and Rt and XL are nearly equal. Above 500 Hz XL is larger than Rt and dominates the value of Z in the above equation. At 500 Hz a compliance voltage of 0.79 V is required.
Vcomp=I*Z=100*(0.00462+(0.0064)2)1/2=0.79 V
These calculations are repeated for multiple frequencies and currents and plotted. We assume we have a maximum compliance voltage of 6.0 V up to a frequency of 10 kHz and plot this with the multiple calculations. We can now see how current and frequency interacts to stay below the maximum compliance voltage. Our selection of test configuration (current path size and load) is very important to whether we can use the TA to test the shunt at 100 A and 10 kHz. At 10 kHz we cannot reach a full 100 A current for the present configuration and load.
If we reduce the length of the conductors, thereby reducing the radius, a, of our structure, the inductance decreases and the required compliance voltage decreases, especially for higher frequencies. This allows reaching the full 100 A rating at 10 kHz into our load. This is left to the reader to verify.
What have we learned?
Generation of direct currents at the maximum currents for TA’s is easily done without much consideration to the total resistance for the cables but the load resistance should still be considered. When we try to generate alternating currents the size (total length and separation of the positive and negative leads) of the current path is very important, especially for the highest currents and highest frequency. In order to generate the maximum current at the maximum frequencies requires minimizing the inductance – keeping the current leads as short as possible and as close together as a test configuration allows. Finally, another method commonly used to further reduce the inductance is to spiral twist the cables between the TA and the load.
For additional information and analysis of a more typical rectangular cable path, we are working on an application note and perhaps a video and will post links to those materials when available.
[1] Thompson, Marc T., Inductance Calculations Techniques – Part II: Approximations and Handbook Methods, Power Control and Intelligent Motion, Dec 1999.