Can a 4 to 20 mA current loop be measured without using a resistor?

Question

Background

There is an electrical system typically consisting of a 4 to 20 mA transmitter and receiver.

^{simulate this circuit – Schematic created using CircuitLab}

So far, everything is working smoothly.

Now, I have a project in which I must build a device to measure the current that passes through the loop, for which I must place a resistor, something similar to the following schematic

^{simulate this circuit}

Apparently this should work without any problems.

Question

By adding an additional resistor (since the receiver is the only resistor in the loop), the current source, i.e. the transmitter, will have to produce a higher voltage at its terminals to maintain the current levels.

My questions are.

1. Could this higher voltage level, which the current source must generate, be a problem?
2. Could there be some interference by adding an extra resistor to the system.
3. Is there any method to measure the current without being too invasive to the system?

For example, it could be the Hall effect, but this method seems to be practical at current levels in amperes, in milliamps, it may not be practical.

Answer 1

My questions are.

1. Could this higher voltage level, which the current source must generate, be a problem?

Yes. The transmitter will have a maximum output voltage. If the voltage drop across Rs gets too large, it can limit the voltage available at the receiver which may affect its performance.

2. Could there be some interference by adding an extra resistor to the system.

Possibly. Changes in loop current will cause changes in voltage at the transmitter and receiver, which may affect the performance of both.

3. Is there any method to measure the current without being too invasive to the system?

Yes, there are "non-contact" or "non-invasive" ways to measure current, such as hall-effect and flux-gate, however, these are much more expensive than a simple "current-to-voltage" transducer, aka resistor.

The trick is to select Rs such that its voltage is sufficiently low as not to affect the system. Try to keep this voltage below a certain % of the maximum system voltage under worst-case operating conditions. In your application, let's assume the maximum output voltage of the transmitter is 24VDC, I would suggest keeping V on Rs (VRs) below 5% of 24VDC, so below 1.2V.

At 20mA, this means Rs = 1.2V / 20mA = 60-ohm (significantly lower than the 250-ohm shown in your diagram).
Power in the 60-ohm resistor = 1.2V * 20mA = 24mW.

You could even try to keep VRs(max) below 1% of operating voltage (24V /100 = 0.24V), so:
Rs = 0.24V / 20mA = 12 ohm. Power = 0.24V * 20mA = 4.8mW.

The problem we now have is that VRs can get quite small when the sensed current is at its minimum. For the 4-20mA system, we are lucky that the minimum current not too small, it is 4mA, ie: 20% of the maximum current (20mA). If we choose Rb=12ohm, the expected voltage range for VRs is: 48mV minimum (at 4mA) to 240mV maximum (at 20mA).

These voltages seem quite small and may seem difficult to measure; however, there are amplifiers designed specifically to deal with the problems of this type of current measurement. I have designed such systems that measure hundreds of amps, where the signal levels at the minimum current level are below 1mV.

Measuring current in this manner is quite a common application, so it's no surprise that chips have been designed for this specific function. TI makes a series of amplifiers for this purpose, look up "INA1x9" datasheets.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Modified circuit to use the existing receiver shunt resistor.

You already have a shunt resistor, R, so you can measure the voltage dropped across it. The very high input impedance of the op-amp will ensure that there is no significant loading on the shunt and the receiver reading will not be affected.

Answer 3

In a 4-20 mA instrumentation loop, the "current source" is not normally also the source of the voltage that drives the current through the loop. The voltage source is usually associated with the receiver, and it can be as high as it needs to be to deal with the voltage drops associated with the transmitter (which will have a range of compliance voltage that it can deal with), the wiring, and the receiver(s) that might be in the loop.

Also, you can limit the drop across a receiver by using a smaller resistor and a higher gain in the buffer amplifier.

Answer 4

As you've already decided against using the existing \$R\$ to sense current, as proposed by @Transistor, here are my thoughts about your own design.

Could this higher voltage level, which the current source must generate, be a problem?

The source of that 4-20mA current will have some limit to the maximum voltage it can apply across the sense resistance of the receiver, a maximum "compliance voltage". If for example that source is using an internal 12V DC supply, then that maximum could be near +12V.

Whatever sense resistance is employed, it must never develop a voltage exceeding that, so assuming a maximum of 12V, the largest permissible resistance would be calculated according to Ohm's law:

$$ R = \frac{V}{I} = \frac{12V}{20mA} = 600\Omega $$

In practice you would aim to develop significantly under this maximum compliance voltage, to avoid hitting that cap by accident, should the signal source's supply be only 11.5V, for example.

In your case, you have two current sense resistances in series, \$R\$ and \$R_S\$, for a combined value of \$R+R_S\$. This combined resistance must not develop a voltage exceeding that maximum. Staying with a 12V maximum, this would mean:

$$ R + R_S < 600\Omega $$

Therefore, the first thing to do is determine the maximum compliance voltage of the 4-20mA source, which you could do by removing all loads, and replacing them with a large resistance like 10kΩ, while the source is trying producing 20mA. Or you could read the manual. The voltage across that resistance will be the maximum \$V_{MAX}\$ that the source can produce.

Now you have some idea of the largest permissible value for \$R_S\$:

$$ \begin{aligned} R_S + R < \frac{V_{MAX}}{20mA} \\ \\ R_S < \frac{V_{MAX}}{20mA} - R \\ \\ \end{aligned} $$

If you find that \$R_S\$ is prohibitively small, requiring excessive gain \$\frac{R_f}{R_e}\$ from your differential amplifier, then you'll need to find another solution.

Could there be some interference by adding an extra resistor to the system?

If you haven't hit the compliance voltage limit, then there's no "interference" by \$R_S\$. However, you still must consider the your differential amplifier's influence. It might to be the biggest "invasive" element here, because it will divert current away from \$R\$, by an amount related to \$R_e\$. If you could guarantee that current entering one of its inputs was equal to current returning via the other, then this wouldn't be a problem, but sadly that's not going to be the case. You'd have to make \$R_e\$ (and consequently \$R_f\$) extremely large in comparison to \$R\$ to minimise the error introduced by your amplifier, which comes with its own issues.

Is there any method to measure the current without being too invasive to the system?

While not addressing methods here, I will propose that you consider using an instrumentation amplifier (IA), like the INA128, to reduce input currents to near zero, removing the influence I just described. IAs tend to have great precision even at high gain, so you'll probably find you can employ a very small \$R_S\$ without worry.

The precision of your own differential amplifier depends very much on good matching of peripheral resistors \$R_e\$ and \$R_f\$. An IA will not require any such matching.

Answer 5

Any 2-wire transmitter will have a minimum voltage drop and you have to subtract that from the supply voltage used and the maximum current multiplied by the total resistance in the circuit (including wiring). You should avoid getting too close to the maximum to maintain noise immunity.

For example, if the transmitter drops 8V, there is a 24V supply and you wish to accommodate 25mA (to allow for broken sensor detection and to allow some headroom and supply tolerance) then the maximum total loop resistance would be 640 ohms. With a 2-wire transmitter you may have the option of increasing the supply voltage, but there will be a maximum for the transmitter.

With a 4-wire transmitter, it's simpler- there's just a maximum voltage specification which will correspond to a maximum loop resistance.

There are several ways to avoid adding resistance- a current sensor that drops no significant voltage, for example, but that might not be accurate enough and will certainly be more expensive. You could use a transimpedance amplifier (which would have virtually no voltage drop in normal operation) or retransmit the 4~20mA so you'd have another loop. But it's easier to just use a lower value precision resistor and amplify the voltage. Even 100-200mV is plenty of signal for modern precision op-amps in most situations (assuming 0.1%-ish accuracy requirements) and it represents a negligible fraction of the compliance of a typical industrial loop. You would use Kelvin connections to the sense resistor at that 5-10 ohm level, and preferably a resistor with such connections provided at the component level. You would use a differential amplifier circuit (you could use an instrumentation amplifier but that would be a waste of money and likely have poorer noise performance and require dual power supplies).

Answer 6

In a current loop, a voltage signal is converted to a 4 mA to 20 mA current in the transmitter, this current can then be connected through long wires to the receiver where it is typically converted back to a voltage by passing it through a 250 Ω resistor. The voltage across this resistor can then be measured by an ADC or voltmeter. As long as the impedance of the ADC or voltmeter is several orders of magnitude higher than 250 Ω the reading should be accurate. The voltage across the resistor will range from 1 V for 4 mA to 5 V for 20 mA.

Since the current through series connected devices is the same, the current won't be dependent on the resistance of the connecting wires or any other resistance in series. As long as the transmitter can adjust the voltage high enough to generate the correct current through the loop it should work.

So let's say the wires were really short so their resistance was negligible, the transmitter will output 1 V to 5 V. If you then inserted a second 250 Ω resistor into the loop in order to measure the current (as in your diagram) the transmitter would have to adjust the voltage to 2 V to 10 V. The most common supply voltage used in these systems is 24 V, but 12 V is also common. So assuming the it can use the full supply voltage the maximum loop resistance for 24 V would be around 1200 Ω, and 600 Ω for 12 V. By adding the second resistor you're reducing the leeway the system has for wiring resistance, and in a 12 V system you would be cutting it close.

What you could do instead is measure the voltage across the receiver resistor, as long as you use a high impedance measurement circuit it shouldn't be very invasive. Alternatively you could use a smaller second resistor, which would give you a lower voltage reading but that can be accounted for. Using 10 Ω for instance would give you a voltage range of 40 mV to 200 mV and not affect the loop very much.

Answer 7

The method you show of how to measure the current is exactly the approach 4-20mA circuits are designed for.

The power source to the current loop, say of 12V or 24V (typical values), sets how much headroom the transmitter has to work with. Some of this range is used to power the transmitter itself. You also have the existing receiver which drop a certain voltage as they are again typically just a sense circuit like you show themselves.

A 250R sense resistor for example requires at least 5V (20mA x 250R) of the loop supply.

Given you have your existing receivee device in the loop (R), depending on it's size and if a 12V supply is used, your 250R might be too large to have enough headroom. Say for example the transmitter and existing receiver need 5V each, then on a 12V supply you couldn't handle another 250R because it would starve the transmitter of power. This wouldn't be an issue for a 24V circuit.

Dropping it to 100R would limit the max voltage for your measurement circuit to only 2V which might be more reasonable. Going lower still if you can will further limit the impact of your sensor on the circuit.