2 thoughts on “Jim Williams’ Analog IQ #8

  1. I also posted a version of this question on your YouTube channel.
    You mentioned the 1N4148 diode early in the video as providing a small equivalent resistance of little concern. My other question asked about the purpose of this diode. I have slept since then and now think that it is part of the “programming” of the current source. Specifically, the current through the left leg with the diode is determined by the 4.7K and 15K resistors along with the diode. The right leg has a 4.7K resistor and a p-n junction. I think the diode is placed as a match for the p-n junction. The current through the diode results in a diode voltage drop which is then replicated by the p-n junction setting the current through the transistor and into the capacitor.
    Please confirm if that is correct and if not the purpose for the diode. Thanks.

    • Yes, the diode is intended to match the PN junction of the transistor’s base-emitter. Specifically, this is done for temperature compensation.

      The voltage drop across the transistor’s base-emitter junction (Vbe) varies with temperature, causing the emitter voltage to change, which means that the collector/emitter current would also change over temperature (see my previous video discussing the effects of temperature on Vbe). This is undesirable for a constant current source!

      Ideally, the temperature characteristics of the diode’s PN junction would exactly match that of the base-emitter PN junction, so that the voltage on the transistor’s base would change proportionally with temperature to compensate for the change in Vbe, thus keeping the emitter voltage constant. In practice they will not be matched, so the diode will only partially compensate for the change in Vbe.

      Additional reading: https://coefs.uncc.edu/dlsharer/files/2012/04/D10.pdf

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