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playing around with a current sink.

Discussion in 'General Electronics Discussion' started by (*steve*), Aug 20, 2016.

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  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,501
    2,841
    Jan 21, 2010
    I'm building a circuit with an active load which happens to be a constant current sink.

    One desired factor is a high impedance, and since it's for an audio application I really want it to be flat over audio frequencies.

    The results first (please excuse the mobile phone images.

    At 130Hz:
    WP_20160820_14_26_57_Pro.jpg

    The input signal is 2V/div and the smaller output signal is 2mV/div.

    The input signal represents the change in voltage seen by the current sink. 8V peak to peak represents the largest voltage swing expected. The supply rail is 18V.

    The output signal is the voltage across a 100 ohm resistor, and is a measure of the change in current.

    Before the math. Let's look at higher frequencies:

    1.3kHz
    WP_20160820_14_27_38_Pro.jpg

    13kHz:
    WP_20160820_14_28_00_Pro.jpg

    And finally at 130kHz:
    WP_20160820_14_28_21_Pro.jpg

    As you can see, things are starting to degrade at 130kHz, but I can't hear frequencies that high :D.

    I'm going to call that output voltage 1.5mV. Across 100 ohms that is 15uA.

    If a change of voltage of 8V produces a change in current of 15uA, then the impedance is in excess of 500k.

    Given that it's a very simple current sink, I'm happy with that performance.

    I'll draw the circuit in a moment :)
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Here's the circuit:

    currentsink.png

    My calculations indicate that the 10M resistor should not be able to supply sufficient base current to Q1, but it does.

    I will probably reduce the value of this resistor to allow for devices that have lower performance. Reducing the value of this resistor has the effect of also increasing the impedance of the current sink, so it helps in all sorts of ways.

    C1 helps increase the impedance at higher frequencies by making the voltage at the base of Q1 a little more stiff (and yes, I checked it by removing it and watching the impedance drop at higher frequencies).
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Why 130Hz, 1.3kHz, and 13kHz?

    The signal generator was set for 1300 and I didn't bother changing it :)
     
  4. Alec_t

    Alec_t

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    Jul 7, 2015
    Can you show where/how you are applying the 8V and where the 100 Ohm load is connected?
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,501
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    Jan 21, 2010
    Sure, I can draw a schematic tomorrow, but basically the 100ohm resistor is connected in series with the output of the constant current and then between this and ground is an emitter follower biased so it does about half the available voltage across the transistor. The signal is AC coupled to the base of this transistor.
     
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