How does a the circuit below help reduce variation due to load resistance

Discussion in 'Analog & Mixed-Signal Design' started by SlitheringSnake, Jul 15, 2017.

  1. SlitheringSnake

    Thread Starter New Member

    Jul 15, 2017
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    Hi,

    I have the following current mirror circuit. Lets assumeI change the load resistance from a small value to a higher value. I dont see how this changes the current Ip. However the Vce of the transistor Q2 changes which may cause the current to change.



    Now , by adding a resistor to the collector , it appears that it improves. I dont understand how this is changing and helping to change load varaition vs current stability.
     
  2. ebeowulf17

    Distinguished Member

    Aug 12, 2014
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    I was hoping someone with the relevant experience would answer this one. I'm curious to know myself.
     
  3. OBW0549

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    Mar 2, 2015
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    The two 1.0 kΩ resistors are added in series with the emitters of the two transistors, not one of the collectors.

    Without the resistors, the current mirror's performance depends on the matching of Vbe and Hfe of the two transistors: identical transistors with equal base-emitter voltages will have roughly equal collector currents, subject only to the Early effect from unequal collector-emitter voltages.

    Adding the emitter resistors drastically reduces the current mirror's sensitivity to mismatched Vbe in the two transistors by inserting an additional resistive voltage drop (≈ Ip * 1.0 kΩ) in series with the emitters.
     
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  4. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
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    It doesn't. Both collector currents are identical, and the right-side collector current is a constant, hence the name constant current source. With theoretically perfect transistors it remains constant up to the point the the right-side transistor saturates.

    The idea here is called dangle-biasing. A transistor has a current gain, the ratio of the base current to emitter current in a non-saturated condition. Let's assume the gain Hfe is 99. When power is applied, current runs from the emitter to the base to the collector resistor to GND. That base current causes collector current into the same resistor. So if there is 1 mA of base current, there is 99 mA of collector current for a total of 100 mA emitter current. But, as the collector current increases, the voltage across the resistor increases, decreasing the base current. At some point (based on the resistor value), the base current decreases to the point where the collector current stops increasing, and the system sits there balanced. With the right power supply voltage and resistance value, the base and collector currents are perfectly balanced at 99:1 and the transistor is not saturated. Now, if you attach a 2nd transistor, it's base current goes through the same resistor. This changes the balance point for the first transistor, but again with the right components, balance is achieved with one collector current and two base currents through the same resistor. Now, since the base currents of the two transistors are identical and *they are theoretically perfectly matched transistors, the second transistor's collector current must be identical to the first.

    The circuit works inside ICs, where the component parameters and temperatures are almost perfectly matched. With discrete parts (2N3906, 2N4403, etc.) it sucks. The text says to expect a 25% variation. I see that as a starting point, not a max limit.

    ak
     
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  5. SlitheringSnake

    Thread Starter New Member

    Jul 15, 2017
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    Hi,

    @OBW0549 How does the added drop reduce sensitivity? Could you explain a little bitmore?
     
  6. SlitheringSnake

    Thread Starter New Member

    Jul 15, 2017
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    @AnalogKid
    Isnt Q1 always in saturation since Vcb is 0?
    The second question I have is 2Ib+Ic flows trhough the collector resistor. However, what make the 2Ib's the same? In other words, how id adding the emitter resistor compensating for mismatc?
    How would it be different if there was no emitter resistor
     
  7. ebeowulf17

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    Thanks so much to both of you for those excellent explanations!
     
  8. AnalogKid

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  9. SlitheringSnake

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    Jul 15, 2017
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    @AnalogKid , What is the effect of adding the emitter resistors? How is it helpin?
     
  10. OBW0549

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    Mar 2, 2015
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    There's a good explanation here.
     
  11. davideather

    Member

    Dec 12, 2016
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    Talking only about the second circuit. The matching of current in the current mirror depends totally on the matching of the b-e junction voltage and to a lesser extent the Hfe of the transistors. In an integrated circuit they can match fairly closely without much effort on the part of the manufacturer. Using discrete components it is much harder and they will often have to be matched by physically measuring the b-e junction voltage of a bunch of transistors and selecting the closest pair. The exact value isn't important, just how close it is to the other transistor. In some cases The tracking of the current match might need to be improved by putting the 2 transistors in physical contact with each other by using a clamp and heat-sink compound.

    In the first circuit the effect of b-e junction voltage mismatch is minimised by using the resistors in the collector - 1k in this case. At 1ma these will drop 1v and make any mismatch virtually irrelevant. (
     
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