Monitor the current


Monitor the current with an operational amplifier, a BJT and three resistors

This article explains the functionality of an intelligent circuit that accurately measure the supply current.

First of all, I must admit that the title is slightly misleading. The circuit presented in this Article requires only an op-amp, a transistor and three resistors. It is not, however, an independent power monitor, perhaps “measuring current” it would be more accurate “current monitor”, but even this definition is accurate. Can I assume that in the end the circuit is just over a “current-voltage converter”, but keep in mind that converts the current into voltage in a manner compatible with the supply current monitoring applications. So maybe we should call it a “current to voltage converter for the output current monitoring applications”


There are various situations in which you want to measure current consumed by your project.

Among the many comes to mind to me dynamically adjust the functionality of a subsystem based on the current consumption of another subsystem, estimate the battery life, or to establish the smallest possible IC regulator capable of providing adequate output current. It is also possible to use the measurements of current consumption recorded as a minimally invasive way of tracking the transitions between states of a microcontroller of the upper and lower power.


As discussed in the opening paragraphs, this circuit converts the current into voltage. If you need a circuit that is more autonomous in its ability to record and / or respond to the current consumption behavior, you probably want to digitize the measurements using a microcontroller. If you are required only basic functionality and do not need any other processor, you can use a comparator or an analog detector windows.

The circuit

The circuit presented in this article is based on a circuit found in an application note entitled “Op Amp Circuit Collection”, published (in 2002) da National Semiconductor. My version looks like this:

And here is my implementation LTspice:

This may seem a little confusing at first sight, but the system is really quite simple:

  • The current flows from the supply, through R1, To the load. R1 works like a typical resistor current sensor, as such has a very low resistance so as to reduce the power dissipation and minimize its effect on the measurements and on the load circuit.
  • The voltage applied to the input terminal of the operational amplifier non-inverting input is equal to the supply voltage minus (supply current × R1).
  • Do not let the PNP transistor will detract from the fact that the operational amplifier has a negative feedback actually.
  • Since the upper end of both the resistors R1 and R2 is connected to the supply voltage, to bring the operational balance, there must be an equal tension on both these resistors, consequently the current through R2 is equal to the current through R1 divided by the ratio between R2 and R1. In the circuit shown above, R2 is 1000 times larger than R1, which means that the current through R2 will be 1000 times smaller than the current through R1.
  • The base current of the BJT is very small, then we can say that the current through R3 is more or less equal to the current through R2. Then, R3 we use to create a voltage that is directly proportional to the current flowing in R2, which in turn is directly proportional to the current flowing through R1.

This is the scheme that should help to clarify and reinforce this explanation:

The final equation for VOUT is

What exactly he is doing the PNP?

You can think of the transistor in two ways: as an adjustable valve that allows the operational amplifier to increase or decrease the current flowing through R2 and R3, or as a device with a variable voltage drop that the operational amplifier can be used to establish the correct voltage VOUT on node. In both cases the end result is the same: the transistor is the means by which the operational amplifier can force the voltage on the inverting input terminal to match the voltage on the noninverting input terminal.

The transistor is really the most interesting part of this circuit. We often use in applications BJT “on/off” and it is important to recognize that the situation is completely different in this circuit. The operational amplifier (with the help of negative feedback, obviously) He is actually making small and precise adjustments to the emitter-base voltage of the PNP (WEBSITE). The following graph shows VEB for a range of load currents (corresponding to the load resistors 50 Ω a 300 Oh).

Note how all these voltages are close to the typical threshold of ignition (~ 0,6 V) for a pn junction silicon. This shows that the op-amp is very carefully operand in the region of the BJT threshold to produce the changes required to produce a large voltage drop between emitter and collector. The entire range of values ​​is about VEB 50 mV

Compare this variation around 50 mV with the change of approximately 4 V in the emitter-collector voltage:


If use is made of resistors low tolerance and a good operational amplifier, this circuit can be considered quite accurate. I treated an interesting and effective circuit that accurately converts the supply current into a voltage that can be measured, digitized or used as input to a comparator.

Amilcare Greetings

9 replies
  1. schottky
    schottky says:

    The article is very interesting and the discussion that arose even more, Of course these circuits, in the present moment, are limited to solve a temporary problem sensa have the chance of a good provider because now all semiconductor companies produce special CI, some of which follow the same topology that allow precise revelations, with simple and inexpensive circuits and especially in need of infinitesimal voltage drops. I have not yet figured out how to fit the figures in the comments (I'm stupid), but a good part of the schemes to which I refer can be found here for what concerns the Linear IC, sifting through the datasheet of the other houses T / Infineon etc are the other solutions

    • schottky
      schottky says:

      E’ hardly necessary to say that with these IC will solve all the problems related to the integrated circuit supply voltages and circuit to the fact that you want to measure high side or low side etc.

      • theremino
        theremino says:

        Very interesting AN105 which you linked, I did not know her.
        Why do not you do a little article with a list of links pointing to the Application Notes interesting you know?
        Eventually we may update the list even later, with the collaboration of all.

        Anyway, although there are special integrated, it is always good mage with the basic operational. You could for example have a free cross-section of a quad LM324 and use it to measure the current. Or you may want to end up in the day, without waiting for the delivery of a special integrated, which by the way, only to order one from Mouser and Farnell, It would cost an exaggeration of postage.

      • theremino
        theremino says:

        Since you have the eagle eye for these integrated, Would you help me find the most suitable to our system?
        So far the best I have found is that the LT1999 is powered at 5 volt, It is a common way to -5 a +80 volts and comes in three versions, 10v / v, 20v / v e 50v / v
        Too bad it's only in MSOP and SOIC and costs 3 to 8 euro (to buy a few) so for many hobbyists is a bit’ uncomfortable.
        Do you know that they cost less and there are even DIL8?

      • theremino
        theremino says:

        I found MAX4173 which is also better than the LT1999
        – Less external components
        – Comodo case SOT23, small but pretty easy to solder
        – Food 3 a 28 volt
        – Common mode by 0 a 28 volt (independent of the power supply)
        – Gain 10x, 50x or even 100x (possible to use a piece of track as a resistor)
        – Low cost 1 euro (circa) instead of 2 the LT1999

  2. theremino
    theremino says:

    Here is another small change that could be useful in some cases.
    With two resistors in most are unable to measure the current even if the voltage of the high branch (High Side) It is greater than the supply voltage of the operational amplifier.

    Hi-Side Hi-Voltage Current Monitor

    With this variant you can also be used not operational Rail-To-Rail, but only if they have the output drops to zero volts or nearly. For example, the common LM324 are fine.

  3. Amilcare
    Amilcare says:

    To be brief, perhaps I exaggerated!!
    Try the simulator or even in a real test to power the operational amplifier with a floating power supply, hooked to the positive but floating with respect to the mass.
    Maybe even with a Zener diode in series with the limiting resistance to ground.
    You will see that whatever the value of the main power supply the circuit operates.
    The transistor serves precisely to translate the voltage levels and to give voltage references to ground even if the test is done on the positive supply.
    Going to be provided only for R3 a divider that face to see the circuit downstream of it a maximum voltage compatible with the downstream circuits

    • theremino
      theremino says:

      All non-operational Rail to Rail, for example, the LM324 and the like, They have a maximum common-mode voltage at the inputs does not exceed their supply voltage less 1.5 volt circa. But the input voltage in this circuit + (and so over -) They are equal to the supply voltage. Under these conditions the first transistor, that they are at the entrance of the operational non-Rail to Rail, they depolarize and no longer function.

      In the original circuit of National this problem does not highlighted because in those years it was customary to feed the operational with high voltages, often duals, eg +/- 15 volt, even if the signals to be treated were small, for example a few volts.

  4. theremino
    theremino says:

    Just recently they asked me how to measure the current of a servo motor, why this article came at the right time.

    Thinking about it a bit’ I found a variant that could be useful to simplify assembly.

    Hi-Side Current Monitor

    The operation is similar to the original circuit, but replacing the transistor with a resistor, that is easier both to find that to be welded.

    The operational need not necessarily be a LMC6482, but it must be a Rail-To-Rail (note that it had to be also in the original circuit, otherwise the voltage on the operational amplifier inputs exceed the maximum common-mode voltage permitted).

    Both the circuit Amilcare that this, They may not work if the power supply voltage is greater than that of the operational amplifier. But by varying the values ​​of resistors perhaps you may succeed. If I can publish the scheme in the near post.


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