Is base current really required in a Current mirror...

Discussion in 'General Electronics Chat' started by Himanshoo, May 6, 2015.

  1. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    In this circuit a small amount of the input current Iin is diverted from the diode (transistor acting as diode) to feed the transistors base..hence causing further slight mismatch(because now Vbe_Q2 will change and it wont match with Vbe_Q1 for producing required collector current)…since we cannot totally cancel out this base current as this is necessary to ON the transistor Q1 and Q2 (though it bring a slight mismatch)..so what would be the approximate value of this Ib ..should it be a compromise value that makes Iin participate up to its maximum extent in generating needed Vbe across the diode and on the other hand turn ON the transistor Q1 so that Iout could trace Iin equally..???

    courtesy :WBahn
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    The whole point of this is to use the voltage BE control model (yes, it exists). By using two "identical" transistors you create the same current in the second transistor as the first.

    The concept works, is is used in many IC op amp designs. You can build one using discrete transistors. Where the drift comes in is temperature differences, which is much less of a problem on an IC die.
     
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  3. dl324

    Distinguished Member

    Mar 30, 2015
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    Agree with Bill. This is a technique that works well on an IC, or if you use two transistors in the same package. Using individual transistors won't give the same results. The idea is that you're using two transistors with very similar parameters that will experience the same conditions. You don't get that with transistors in different packages. It will work, but not as well. If you're trying to simulate, don't bother unless you're using a dual transistor.
     
    Last edited: May 6, 2015
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  4. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Look up 'Ebers-Molls Equation'. You will discover that the emitter current is dependent upon base voltage -- and temperature and reverse saturation current -- and is independent of base current.
     
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  5. crutschow

    Expert

    Mar 14, 2008
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    A current-mirror works well with matched transistors on the same substrate but one source of error is the finite collector output impedance which will cause a slight change in the mirror current with collector voltage change. This can be minimized by using a three or four transistor Wilson mirror.
     
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  6. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Just for an instance assume perfect ideal condition with no early effect or so and the transistor to be well matched…

    as we know a required Vbe is required to to maintain a particular collector current..but when transistor acts as diode the collector current of Q2 divides and flow towards the bases…hence a reduced Vbe at Q2 is experienced…Now as both the transistors are matched the Vbe of Q2 will track Vbe of Q1..which will reduce Iout since Iout will track a reduced Iin…


    Now my query is as we know that nullifying base current is impossible because it is necessary to make the transistors turn ON….

    and now on the other hand if the magnitude of this base current increases it will decrease Iin as discussed earlier..so now its just a kind of a paradox that at one time base current is necessary and at other time its undesirable…so what could be the approximate value of base current..whether it should be a compromise..??
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Not sure I understand your question. :confused:
    The base current will be whatever is needed to get the desired current. You do not have independent control of the base current versus collector current.
    You put a desired current into the left (diode connected) transistor and the base current will be whatever that transistor requires for that collector current, based upon its current gain. The "current division" between collector and base is determined by this gain.
    The right transistor will then draw the same base current (since the Vbe's are matched) and thus the same collector current (since the current gains are also matched).
    There will be a slight error in the two collector currents (as determined by the current gain) due to both base currents being provided by the left input current.
     
    Last edited: May 6, 2015
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In this circuit Iin = Ic1 + Ib1 + Ib2. So if we assume β1 = β2 = β then we have
    Iin = Iout + 2Ib
    Iout = Iin - 2Ib = Iin*β/(β + 2)
     
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  9. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Yet, Ebers-Moll does not require β1 = β2, as per your assumption. In fact, it says Ic is solely dependent upon Vbe -- given the same temperature and reverse saturation current for each transistor.
     
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  10. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    The larger the base current gets the more it gets difficult for Iout to match Iin....so in order to make Iin trace Iout base current should be as small as possible...as the value of Ib could not be too small so that it could not turn on the transistors..hence the value of Ib should lie somewhere in middle so that it could make Iin trace Iout to maximum extent and on the other hand turn on the transistor success fully..
    am i right?
     
  11. crutschow

    Expert

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    Basically. That's why the transistors are sized so the current density in the junction is optimum (the Goldilock's principle) for the best current-mirror operation.
     
  12. joeyd999

    AAC Fanatic!

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    No. Ebers-Moll works over many decades of collector current. This is why linear log amps work so well.

    Temperature differences between the two transistors (assuming matched) is the largest source of error. For large Ic, differences in Vce will cause one transistor to dissipate far more power than the other -- creating a temperature imbalance. This is why matched pairs are usually fabricated on the same silicon very close to each other. Situated on the same silicon, they are also effected by ambient temperature changes similarly.
     
  13. cabraham

    Member

    Oct 29, 2011
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    The current mirror operates with emitter current as the control variable. For a conventional current mirror, un-weighted, the objective is to match the 2 emitter currents, so that the 2 collector currents will then match. Ic = alpha*Ie. But how should this be done? Some have suggested that the 2 b-e junctions be placed in parallel, so that the 2 values of Vbe match, resulting in matched Ie & Ic. But the relation between Vbe & Ie is: Ie = Ies*exp((Vbe/Vt) - 1). For 2 discrete devices, Ies does not likely match, there are differences in specimen, and with temperature.

    When fabricating 2 devices on 1 substrate, such as with an IC, the Ies values can be well matched, and hopefully the 2 devices are at reasonably matched temperature. Even if Ies values are not perfectly matched, the emitters possess some inherent resistance and that tends to help the 2 devices share current. With discrete current mirrors, it is recommended that ballast resistors be used in each emitter. If Vbe values do not match, the emitter resistors will compensate. If 1 device hogs the current, it gets hotter, its Ies value increases, and hogs more current. The emitter resistors compensate because should 1 emitter current increase, the resistive voltage drop increases and that Vbe decreases.

    With emitter ballast resistors, emitter currents match despite mismatches in Ies & Vbe. Discrete current mirrors need to have emitter resistors.

    Lastly, there is the weighted current mirror. By using differing emitter resistor values, we can force 1 emitter current to be a multiple of the other, and the collector currents are scaled accordingly. If 1 emitter resistor is 10 ohm, the other 100 ohm, the emitter currents scale by a factor under 10, as do the collector currents.

    Current mirrors exploit the bjt property that Ic = alpha*Ie. If we can get the 2 Ie values to match, the Ic values will match as well. For weighted mirrors, the currents scale inversely with emitter resistors. Did this help?

    Claude
     
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