Constant Current Source - NPN vs MOSFET

Discussion in 'General Electronics Chat' started by Robin Mitchell, Dec 21, 2015.

  1. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
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    Hi all,

    Quick question and more of a curiosity:
    Is it better to use either an NPN or MOSFET for a constant current source as shown in the image below:
    [​IMG]
    In my practice I found that NPNs give a better max current rating but I may be completly wrong.
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    There at least thee things that i can think of right now that will limit the available current.
    With an NPN, you are limited to roughly the output current of the opamp times 10 or somewhere around that number. However the npn has only 0.7V between base and emitter, so you can get almost all of the current that you would if you took the transistor out.
    With a mosfet you are not limited by the output current of the opamp, but the voltage between gate and source will be at least a few times larger than that of the bjt, which limits the available current with an N channel mosfet in this source follower configuration. With a P type common emmiter (source) the situation would be a bit different.
    Thirdly you are limited by the power dissipation in the transistor, but I don´t see here one type being inherently better than the other.
     
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  3. dl324

    Distinguished Member

    Mar 30, 2015
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    I think it's more a matter of personal preference and/or what parts you have on hand. I prefer to use transistors because I have a dozen different part numbers each for NPN and PNP transistors and 1 each for N/P MOSFETs. If I could find MOSFETs for a nickel, I might be inclined to use them more often.

    For the opamps and supply voltages I typically use, BJTs give me more flexibility.

    These days, MOSFETs have higher current ratings than BJTs.
     
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  4. #12

    Expert

    Nov 30, 2010
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    I think you are wrong. In terms of amp per amp gain, the mosfet has a bjt beat (at low speeds) every day. The limit which I have to consider is available voltage maximum. The bjt can operate will less voltage than a mosfet...in most of my circuits.
     
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  5. crutschow

    Expert

    Mar 14, 2008
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    The BJT maximum current is limited by the op amps output current but it's higher than kubeek stated.
    It equals the maximum op amp output times the BJT current gain which is typically about a 100 or more.
    The gain of 10 is when you use the BJT as a switch and here it's used in the linear region.
     
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  6. Robin Mitchell

    Thread Starter Well-Known Member

    Oct 25, 2009
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    Sounds like a MOSFET may better suite my needs! I am looking to source up to 15~2oA so a MOSFET may do better.

    Quick question, why is the limit 10 x gain for an NPN anyway?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    That would probably only be if you are using a power transistor -- in which case you are probably well past the capability of the opamp to feed it anyway.

    For a normal transistor, since you want to keep it well out of saturation so that it can control the current, you are probably looking at a practical limit that is at least 100x the opamp output limit.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    A common practical definition of "saturation" is when the current gain of a BJT has been reduced to 10. That's for small signal transistors. For power transistors the limit is often even lower. It's a somewhat arbitrary choice, but it works in many/most situations.
     
  9. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    If you wanted to squeeze the aboslute max current possible then you would want to drive the bjt into saturation.
     
  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    A bipolar power darlington is a good compromise. It doesn't require a high drive voltage like a power MOSFET, or a high drive current like a standard power BJT. Depending on the part type and circuit parameters, it can have a current gain of over 1000.

    ak
     
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  11. WBahn

    Moderator

    Mar 31, 2012
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    But if you do that, then how are you going maintain the goal, which is for a constant current source?
     
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  12. ian field

    Distinguished Member

    Oct 27, 2012
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    Its pretty much a matter of personal preference if you have plenty of voltage headroom.

    Unless you use a logic level MOSFET, the voltage at a MOSFET source will never be more than a few volts less than the v-out high of the op-amp. For a typical power MOSFET; the VGSthr is around 6 - 8V. Occasionally the MOSFETs on PC motherboard buck regulators have VGSthr as low as 1.6V - only a fraction more than the Vbe of a Darlington transistor.

    With a bipolar transistor; the op-amp must supply the base current, which is the regulated current divided by the transistor gain.
    The MOSFET of course is voltage driven, so the op-amp pretty much doesn't have to drive any current.
     
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  13. sootydog

    New Member

    Jul 23, 2011
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    I've built quite a few constant current circuits for device testing. My preferred configuration is to use a mosfet but with the load on the drain side to minimize the headroom issue. You could use a BJT in that configuration but there would be an error due to the base current being added to the load current passing through the sense resistor.
     
  14. Sensacell

    Well-Known Member

    Jun 19, 2012
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    The cool thing about using the MOSFET is that there is no base current error, you can use the source or drain as the output and the current is identical- because the gate basically draws no current.
     
  15. Bordodynov

    Active Member

    May 20, 2015
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    When the supply voltage is high enough, the advantage for the FET. When the supply voltage is low (3-5 volts) for the benefit of a bipolar transistor. The supply voltage of the operational amplifier may be greater than that of the transistor. This will reduce the power dissipation of the transistor. I did one of my projects.
    The field effect transistor can provide higher current, which is independent of the power operational amplifier.
     
  16. ian field

    Distinguished Member

    Oct 27, 2012
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    In the circuit posted by the TS; the resistor in series with the load samples *ALL* the current through the load and the op-amp adjusts bias to the power device accordingly - or to put it another way, the sampling resistor includes the BJT base current.
     
  17. sootydog

    New Member

    Jul 23, 2011
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    You're right, I guess that's unavoidable with a BJT - probably why I don't use them for this application.
     
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