Multiple current sources from a single reference

Discussion in 'General Electronics Chat' started by hrs, Feb 10, 2017.

  1. hrs

    Thread Starter Member

    Jun 13, 2014
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    Hi,

    At the bottom of section 5.0 here it says:

    It is not uncommon to see a single reference transistor (Q3 in Figure 7b) controlling two or more current sources (which may be supplying different currents). The reference should always control the current source transistor that has the most stable current. For example, if there were a current source for the input stage and the Class-A amplifier, the reference should always be taken from the source used for the input stage. This minimises any possible cross-coupling between the two current sources.


    So if you need 2 current sources, instead of using 2 transistors for each, you might save 1 transistor by using the same reference. But I'm not what that circuit would look like. Is the attached circuit what the article is referring to? Q1 and Q2 form the normal current source and Q3 would be the additional current source.

    It looks like it works but current source Q3 is not nearly as good as Q2. Searching the net I found multiple current mirrors from the same reference but they all produce the same current (no?) so that can't be it. I would expect the performance of Q3 to be similar if you would swap Q1 for 2 diodes, so while this scheme maybe good enough for some applications, it will be inferior to separate current sources(?).
     
  2. GopherT

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    Nov 23, 2012
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    First, in R3, the current source is not worse, it is just ohms law that prevents your current source from staying in compliance (Max voltage = 10.0v - 0.6V = 9.4V). Trying to pump 2 mA through a 3k ohm resistor works because the voltage drop a roast the resistor is 3k X 0.002 amps = 6 volts. Once that resistor is above about 5.1k, the voltage across it was 5100 X 0.002 mA more than 10 volts and the constant current cannot be kept.
     
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  3. hrs

    Thread Starter Member

    Jun 13, 2014
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    Yes, I did notice that when I increase the supply voltage to 20V, the current sources are flat almost up to 10k for R3, somewhat less if I make R5 variable. And I understand how a higher resistance load will end the linear range of the current source. Looking at the graph, current source Q3 is worse up to 5k because it's not as flat as Q2.

    My main question is if I got the general circuit right. It looks a bit like what Rod Elliott calls "a buffered current mirror" (figure 12a) but it has the buffer transistor on the other side so I think it's a different circuit altogether.
     
  4. MrAl

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    Jun 17, 2014
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    Hello there,

    I think you are right to test it in theory but in practice there are some other things to test.
    If Q3 is located far from Q2 then the thermal coupling will be very bad. If Q3 is close to Q2 better, but if Q3 is in the same package as Q2 then even better.

    If you really want good current sources though go to an op amp based circuit if you can.
     
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  5. GopherT

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    A transistor with higher gain will stay in compliance better (see Mr Al comment immediately above to use op amp).
     
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  6. hp1729

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    Nov 23, 2015
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    Multiple current sources from the same reference? As in the LM346?
     
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  7. hrs

    Thread Starter Member

    Jun 13, 2014
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    Thanks for your comments.

    I'm trying to figure out the parts that make an opamp and after that I'll probably make a discrete one. I'm sure it will be terrible compared to matched transistors on the same substrate but maybe I'll learn something. In any case using an opamp IC inside a discrete opamp would be cheating ;)
     
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  8. WBahn

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    Mar 31, 2012
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    The practice of using a single mirror transistor as the reference for multiple reflectors is very common, particularly in IC design. The output currents are scaled by the relative sizes of the mirror and reflector transistors. In good designs, the size is controlled by the number of identically laid out and oriented transistors.

    One advantage of this approach is the superior ratiometric matching of the different current sources. So if one is 7% high then all are typically about 7% high. If your circuit's performance then depends on current ratios and not absolute current magnitudes, you get very high performance.

    In CMOS imager designs it is not at all uncommon to have a single transistor mirror provide the bias voltage for literally millions of reflected current sources.
     
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  9. GopherT

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    @WBahn, what type of circuit (end application) needs so many constant current circuits?
     
  10. hrs

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    Are talking about something like the attached circuit? I was thinking that IQ1, IQ2 and IQ3 would all be equal to Iref whereas the current sources from post #1 can be controlled by R1 and R4.
     
  11. WBahn

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    Uh... an imager.

    If you have a 1280 x 1024 imager and each pixel has just a single current source in it you have over 1.3 million current sources slaved to the same mirror. In SPRITE arrays (Signal Processing in the Element) there can be many current sources in each pixel and, due to routing constraints as well as matching concerns, you generally try to slave as many of them to the same bias signal as you can.
     
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  12. WBahn

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    That is the basic idea, but with BJT mirrors there is an error due to the finite base current and the more reflectors there are the worse the mismatch. With a beta of 100, you are limited to less then about 10 reflectors before the effect becomes too much for most applications. For constant current sources, CMOS largely avoids that problem because the gate current is so negligible that you can have millions of outputs without needing to be concerned about it. In fact, having that many gates hanging on the bias line stabilized the bias voltage. But it does mean that you have significantly impaired high frequency response (which is often just what you want).

    Ignoring the base-starvation issue, let's say that the transistor providing IQ2 was not the same as Qref? Or image typing the collectors of the Q2 and Q3 together and calling that IQ23? Now you have an output that is twice what Iref is. Let's make Qref be 10 transistors in parallel and let's make Q1 be 5 of the same transistor in parallel and the transistor Q2 be 13 of them in parallel. If Iref is 10 mA, then IQ1 will be 5 mA and IQ2 will be 13 mA.
     
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  13. MrAl

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    Hi,

    Why would using an op amp be cheating, is this for a chip die design or are you just accepting a challenge to do it without an op amp?
     
  14. WBahn

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    It would be cheating in the same way as someone that wants to make a discrete NAND gate to learn something about how they work using a quad NOR gate.
     
  15. hrs

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    I'm accepting the challenge to do without an opamp, trying to get a better understanding of how opamps work. And then I'll put some through-hole bjt's on a perfboard and see if reality agrees with me :) (probably not).
     
  16. WBahn

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    It won't unless you play some games. The design techniques used in discrete circuits are different than those used in integrated circuits, so a circuit designed for implementation in one but that is implemented in the other often performs very poorly.
     
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