You can change a current mirror into a multiplier/divider by adding transistors on either side. Then the number of transistors defines the current ratio. The transistors have to be integrated on the same chip to have a good precision (good matching), and besides, nobody wants to place 100 discrete transistors on his PCB for this.

Are there ICs for this, like a \$\times\$10 or \$\times\$100 multiplier (or \$\div\$10/\$\div\$100 divider, you should be able to use them for both)?

  • 1
    \$\begingroup\$ If there exist x100 multipliers they will do this in stages, so that you won't need that many transistors. Like x10 (10 transistors) can be done with x5 (5 transistors) + x2 (2transistors). \$\endgroup\$
    – stevenvh
    May 10, 2012 at 14:38
  • \$\begingroup\$ You just want to create a current that is 10 or 100 times some other current? There are lots of ways to do that. The number of transistors somehow being the gain factor doesn't make any sense. \$\endgroup\$ May 10, 2012 at 14:42
  • \$\begingroup\$ @OlinLathrop the idea comes from integrated analog circuits, where multiple transistors are used to create the equivalent of one with a larger channel. This doesn't make it suited for PCB design, though \$\endgroup\$
    – clabacchio
    May 10, 2012 at 14:57
  • \$\begingroup\$ @clabacchio: "doesn't make it suited for PCB design". What do you mean by that? \$\endgroup\$ May 10, 2012 at 15:14
  • \$\begingroup\$ I meant that while in an IC is fine to use several transistors for that function, with discrete components isn't either accurate and convenient...you can do the same job in different ways. \$\endgroup\$
    – clabacchio
    May 10, 2012 at 15:39

3 Answers 3


Zetex (now Diodes, stupid name) has a nice little IC which has been mentioned on this site a few times. The ZXCT1009 is a high side current sensor which works as a current divider, though to some extent (low currents) you'll also be able to use it as a multiplier.

enter image description here

The transfer function is

\$ I_{OUT} = \dfrac{R_{SENSE}}{100 \Omega} \cdot I_{SENSE} \$

so with a 1\$\Omega\$ sense resistor you get a \$\div\$100 divider; a 1k\$\Omega\$ resistor will multiply by 10. The output current is limited, but you can use the schematic to roll your own multiplier for higher currents:

enter image description here

Don't make the sense resistor too small; below 10mV voltage drop accuracy decreases fast.

edit (re your comment)
For high side current measurement I think this is an excellent IC. Choose a 100m\$\Omega\$ resistor for \$R_{SENSE}\$ and a 1k\$\Omega\$ for \$R_{OUT}\$ and you'll have a nice 1V/A output.

  • \$\begingroup\$ Looks great! I need a divider. \$\endgroup\$ May 11, 2012 at 6:29
  • \$\begingroup\$ @FedericoRusso what do you need it for??? \$\endgroup\$
    – clabacchio
    May 11, 2012 at 8:15
  • \$\begingroup\$ @clabacchio: current measurement for power supply. What the IC does, but I wanted to do it with a current mirror/divider. I guess the IC is a better choice, because the voltage drop across the resistor will probably be lower than that of the current mirror. \$\endgroup\$ May 11, 2012 at 8:23
  • \$\begingroup\$ @FedericoRusso so you're addressing the wrong problem: a current mirror isn't the best way to do the job, and Steven's IC is exactly what you need. Forget mirrors and dividers, and use a current sense amplifier. \$\endgroup\$
    – clabacchio
    May 11, 2012 at 8:25
  • \$\begingroup\$ @clabacchio: I guess you're right. Thanks for your answer, though. Sorry for the trouble. \$\endgroup\$ May 11, 2012 at 8:28

There seems to be some confusion about current mirrors. Here is a basic current mirror:

Some part of the circuit produces current, which is dumped onto IN. OUT then becomes a sink for the same amount of current, which is where the term mirror comes from.

Look carefully at how this works. Assume both transistors are identical. The base voltage of Q2 will be whatever it takes so that its collector draws the current being dumped onto it. Since Q1 is identical to Q2, this is also the same base voltage so that Q1 tries to sink the same current. The matching between the transistors must be pretty close for this not to be wildly off, since the base of Q1 is being voltage driven. The B-E junction is a diode, so the current can vary a lot for small changes of voltage.

On a IC this matching will be close enough to be useful, since the whole IC is subjected to the same process variations. This is not a good circuit if you are building it yourself from discrete parts. In that case you add two resistors to make things much more accurate and less dependent on the individual characteristics of the two transistors:

Due to the resistors, the function of base voltage to collector current becomes more predictable. The downside is that the resistors eat up a little voltage, but a little helps a lot. Even if the resistors drop 500 mV at the maximum current, the error from input current to output current will be much better than with no resistors.

Now note that the two resistors need not be equal. If we consider the B-E voltage of the two transistors fixed, then the ratio of output current to input current will be R1/R2. If R2 is half of R1, for example, then the mirror will want to sink twice the current at OUT that you put in to IN. At high ratios you need to let the resistors drop a bit more voltage to get reasonably linear operation since there will be more variation of the B-E drop between the two resistors. It is a tradeoff between voltage lost in the resistors to accuracy, but this circuit will work well enough for many purposes when built from discrete parts.

Note that if the objective is to multiply a current signal by a fixed gain, there are other ways to do this than with a current mirror.

  • \$\begingroup\$ All I would add is that the primary advantages of current mirrors over op-amp-based circuits (which are generally going to be more accurate) is that they've got a higher bandwidth, higher voltage compliance range, and follow the simpler-is-better rule (no power supplies needed). \$\endgroup\$
    – Jason S
    May 10, 2012 at 17:16
  • \$\begingroup\$ Nice explanation. \$\endgroup\$
    – Russell McMahon
    May 10, 2012 at 17:28
  • \$\begingroup\$ Well explained, but Federico's suggestion should also work, and IIRC I've seen it for a series of current doublers in a DAC. Resistors are avoided in IC integration. \$\endgroup\$
    – stevenvh
    May 11, 2012 at 6:32
  • \$\begingroup\$ Cool post, but it doesn't actually answer the question: are there any ICs doing that? \$\endgroup\$
    – clabacchio
    May 11, 2012 at 8:13
  • \$\begingroup\$ @clabacchio: You're right, I was mostly trying to clear up what looked like confusion about current mirrors. I hope the information is useful, but it doesn't actually answer the question. I don't know what ICs are out there for this purpose, so I can't provide a direct answer. Hopefully I've shown how to use discrete parts as a substitute if a suitable IC can't be found. \$\endgroup\$ May 11, 2012 at 12:14

About current mirrors with ratioed branches:

This is a 5:1 current mirror IC, probably you can find others.

But is there a special need for such a component? It's not a common one, and its function can be performed in a different way.

For measuring currents:

Steven's suggestion of a current sense amplifier is good, you can measure the current with a minimum perturbation on the circuit. You can also perform the same task with a differential amplifier (or an In-Amp) and a small resistor, but then you have to carefully check the input and output range of the amplifier, and in any case the precision of the resistor (it's better if you measure it) determines the accuracy of the measurement.


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