Current Mirror Question

Discussion in 'General Electronics Chat' started by blah2222, Jul 18, 2010.

  1. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Hello all! My primary goal this summer is to get a better grasp with transistors and analogue circuits. So far I've gone through the P-N diode and that all makes sense but the transistor is taking a little bit longer. I have gone through the AAC resources and they are great, but some things are still unclear.

    For instance, say for a common-base transistor (npn) set-up with the emitter-base forward bias and the collector-base reverse-bias.

    1. How does the base-current control the collector current? Isn't the emitter current more important because without it there would be zero initial electron flow?

    2. Besides the simple relation of Ic = B*Ib, why is Ic zero, if Ib is zero?

    3. How is does a shorted transistor act as a feedback diode?

    4. Is a current mirror just a way of producing current sinks/sources or are there other uses, and where do you attach other transistors to make an array of current sinks/sources, at the collector or the base of the master transistor?

    Thank you very much!

    J
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    1. There is an obscure model of transistors that says they can be treat as voltage controlled devices. It works, but not very well, partly because controlling the extremely fine voltages is next to impossible. In a current mirror two identical transistors are used, one sets up the voltage that a specific current will cause. This voltage is feed into the other BE junction, and if the transistors are identical the same current is created. However, current mirrors must be the same temperature, as the characteristics for a BE junction vary with temperature. In an IC package this is not a major problem, but in the larger world much more so.

    2. The equation works. Given it does work, why wouldn't it be zero?

    3. This one I have no clue what you're asking. Care to give a schematic example?

    4. Other people may disagree, but my impression is a current mirror is perfect for any circuit that requires a constant current as part of an IC. With only 3 components (2 transistors and a resistor) it is the simplest way to do this job. Since differential amplifiers (the heart of op amps) require this, and since op amps are a mainstay, current mirrors are everywhere.

    It can not be understated how important temperature stability is for a current mirror is. It is just about the only reason they aren't even more common.
     
  3. Ghar

    Active Member

    Mar 8, 2010
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    In a normal configuration having Ib means you have Ie.
    All 3 currents and all voltages are related to each other and this is why it's pretty unimportant whether it's current or voltage or whatever else controlling the transistor. All the models are mathematically interchangeable, you simply choose one that you're comfortable with and it's a good idea to use different ideas in different situations.
    In a current mirror the simplest explanation is that two identical transistors with identical Vbe must have identical Ic. If one transistor is larger than another the current will scale with its area according to the voltage and current relationships with the condition of identical Vbe.

    Ic isn't really zero with Ib zero because of things like leakage.
    If you look at something like ICBO you'll see you don't need emitter current for a collector current to flow.
    http://premiumorange.com/daniel.robert9/anglais/Semi_conducteurs5_Suite2.html

    These are non-idealities of the transistor and aren't part of the basic functionality.

    Ic = B Ib only works in the active region of the transistor and only across small variations. B changes with current level, temperature, and several other factors. It's not really a linear relationship but it's nice and easy to work with.

    Some people also like to extend B into the saturated (on like a switch) mode but the currents are set by the circuit and not by the transistor gain so it isn't really a gain anymore.

    To have multiple current mirrors you share the base and emitter connections. The goal is to have the same Vbe on each transistor because it's related to the collector current.
     
  4. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Thank you again!
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Transistors are odd devices, the voltage on the Collector Emitter has a very slight connection to the Base Emitter. So the Collector Emitter can be 0 Volts (some text books use 0.2V for saturation, but I've seen much less in real circuits) while the BE is 0.6 volts. Basically it is a quantum function of transistors, they really aren't 2 diodes back to back.

    As for #3, you have the current going through the collector emitter as well as the current (and voltage) on the base emitter. This means all the requirements for the second transistor (which interact) are present on the first transistor.
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    I would say it's more than a slight connection.
    Ignoring the internals of the transistor entirely you are still stuck with the relationships Vce = Vbe - Vbc and Ie = Ib + Ic
    A transistor isn't two diodes you're right but they are very similar to diodes giving you exponential relationships between current and voltage. This glues those relationships together pretty strongly.

    The 0.2V in saturation comes from assuming that the difference between Vbe and Vbc when both are fully forward biased is 0.2V. We know that these voltages are related to current and the design of the transistor which can make the true value diverge pretty far from that.
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    Perhaps, but I'm trying to help and staying on topic.

    And you are?
     
  8. Ghar

    Active Member

    Mar 8, 2010
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    I didn't realize that discussing potentially misleading information is off topic.
     
  9. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Okay, so basically my question about the transistor acting as a diode refers to the example experiment they give on this site: http://www.allaboutcircuits.com/vol_6/chpt_5/14.html

    This is the schematic they provide for a simple current mirror:

    [​IMG]

    My question is, how come collector current of Q1 doesn't just flow straight through as the base current for Q2 as there is a short-wire on Q1? What I'm seeing from this schematic is that Ic1 will flow into Q2 as Ib2 and Ic2 = B*Ib2 which would be greater than Ic1, which makes no sense. I get how both transistors are supposed to be the same and have the same Vbe but why is Q1 shorted?
     
  10. Wendy

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    Mar 24, 2008
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    Q1 isn't shorted, the Collector Emitter has the bulk of the current going through it, while the BE has a little. They balance according to the transistors characteristics (the numbers work themselves out), which match Q2's characteristics. Since the BE is identical for both transistors, it sets up the current through Q2 Base Collector.

    Thinking about it Q1 and Q2 have exactly the same current going thought the BE (they split), which means the collector currents match. Just another way of looking at the same thing.
     
  11. bc108

    New Member

    Jul 18, 2010
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    My guess is you are not designing bipolar ICs, so can not arrange identicle transistors next to each other on the chip.

    You could glue two packages together with devices from the same batch to get reasonable matching, or better still just stick as big ballast resistors as you can get away with between the emitters and ground, the vagueries of vbe/ibe vs temperature become much less significant then.

    The circuit without ballast resistors works fine in simulation, but plug in different temperatures for the two transistors and see what happens - with different Vce ands the 'same' Ic the devices dissipate different amounts of power and have different temperatures.
     
  12. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Ok, but I still find it unclear why the collector and base of Q1 have a wire connecting them, why does it need to be there?
     
  13. Norfindel

    Active Member

    Mar 6, 2008
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    Probably because it's the best, less painfull way to do it. If you just use a separate resistor to supply current to the base, Ie will be dependant on that current in addition to Ic. If you calculate the base resistor to saturate the transistor, it would saturate also the 2nd transistor, because they share the same base connection, so it wouldn't be a current mirror.
     
  14. bc108

    New Member

    Jul 18, 2010
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    Ok, but I still find it unclear why the collector and base of Q1 have a wire connecting them, why does it need to be there?

    The transistor with the base-collector link sets the base emitter voltage for its companion. Because the devices are identical the same base emitter voltage draws the same base current. Collector current is just hfe times the base current (independent of collector voltage, away from breakdown and saturation).

    In the transistor with the c-b short, most of the current to be mirrored flows into the collector (hfe times more than goes into the base...).

    The c-b linked transistor isn't saturated, Vbe will be around 0.65V, and saturation wont set in till Vce=0.2V (for old bc108s), less for a modern device.
     
  15. Wendy

    Moderator

    Mar 24, 2008
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    The currents through Q1 are self balancing, there will be the amount of current through the BE to have the rated current through the CE, and the CE matches the BE current (not the same current, but it matches the equation ICE = ß IBE). The equation describes it pretty closely.

    Q2 pick up the same about of current through Q1 and matches the CE current automatically.
     
  16. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Ok so if Q1 is not shorted between Collector and Base then what is that wire called? Everywhere I have looked on-line says that it is acting as a negative feedback diode. I have constructed the experimental circuit given on this site many times now and to no avail.

    I am using two 2N3704 (npn) transistors, a 12V power supply, a 10K resistor as Rlimit connected between high and Q1's collector, not using a pot unlike in the diagram, as well as three 1.8K resistors and one 10K resistor in series, connected between the ammeter and Q2's collector.

    When connected to all resistors (15,400 ohm) Ic2 = 0.7 mA;
    Three 1.8K, Ic2 = 2.2 mA
    Two 1.8 K, Ic2 = 3.3 mA
    One 1.8K, Ic2 = 6.7 mA
    No load on Q2, Ic2 = 207 mA

    Isn't the whole point of this circuit to have a constant current sink/source? Might my transistors be fried or something else wrong?
     
  17. Ghar

    Active Member

    Mar 8, 2010
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    A current mirror only works as long as both transistors are well inside active mode.
    This means the collector voltage of the output transistor must be higher than about 0.3V. It will become nonlinear as you approach this.

    This gives you a maximum resistance that you can drive while still maintaining relatively the same current.

    With your numbers you should have approximately 1.1 mA which gives you a maximum resistance of (12 - 0.3)/1.1m = 10.6 k
    Any resistance higher than that and the current will be dramatically reduced. You need to stay well below that value to get a decently constant current.

    Have you double checked your resistor values, especially the one with Q1? Those numbers are implying your output transistor is saturated (or shorted...)

    And about the 'feedback diode'... a transistor with its collector shorted to its base is called 'diode connected' because it acts almost exactly like a diode. You can actually make a current mirror with a diode and a transistor but because the diode is slightly different than the transistor the current won't match perfectly. You can scale the current between your branches by having more or less diodes/transistors in parallel.
     
    Last edited: Jul 21, 2010
  18. Potato Pudding

    Well-Known Member

    Jun 11, 2010
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    See if you can find an IC package of multiple identical transistors. They are a much better way to try and set this circuit up.

    2 PNP in current Mirror Config

    4x NPN

    Surface mount is cheap at least

    And when you want to set up a (Low Power) bridge try to find them with a PNP and NPN in one package. Another situation where it makes things easier.
     
    blah2222 likes this.
  19. blah2222

    Thread Starter Well-Known Member

    May 3, 2010
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    Okay, figured out that all my signal transistors are pretty fried and causing problems. Switched to an NPN array and everything seems to be making sense.

    By the way, can alpha (Ie/Ic) be greater than 1?
     
  20. Potato Pudding

    Well-Known Member

    Jun 11, 2010
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    Ie=Ic+Ib

    But you have Alpha defined incorrectly.
     
    Last edited: Jul 22, 2010
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