In a constant current sink from the two NPN transistors, load and sensing resistor, application of increasing voltage on the base of the transistor produces practically linear increase in output current. How to describe this relation (slope) between applied voltage and output current?

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    \$\begingroup\$ Did you try doing a small signal analysis? \$\endgroup\$ Commented Dec 16, 2016 at 23:53
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    \$\begingroup\$ Your statement does not match the reality of the circuit so either ask a question without your thoughts or change the circuit. \$\endgroup\$
    – Andy aka
    Commented Dec 16, 2016 at 23:59
  • \$\begingroup\$ Why do you think this is a constant current source? \$\endgroup\$ Commented Dec 16, 2016 at 23:59
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    \$\begingroup\$ The circuit comes from talkingelectronics.com/projects/TheTransistorAmplifier/… , I think. See the middle schematic there. It's a current sink. Not a current source. \$\endgroup\$
    – jonk
    Commented Dec 17, 2016 at 0:43
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    \$\begingroup\$ This is a "base current compensated current mirror with current shunting resistor". The output current is approximately the current through R: Vbe / R ~ 0.7V / R. To make a current mirror, attach another transistor with its base connected to R. Ref: "Bipolar and MOS Analog Integrated Circuit Design", p175. I tested this out in LTspice and the output current is pretty much independent of the supply voltage and the voltage on the 1K resistor (once it's over a couple volts). If the question is the initial ramp-up to 2V or the slight increase after, that's harder to compute. \$\endgroup\$ Commented Dec 17, 2016 at 1:03

1 Answer 1


It's a non-linear transfer function. Where impedance ratios affect gain Rin = (hFE * Re) + R1 (10k)

  1. For Vin via series R1 to base < 0.6V, Ic=0.
  2. For Vin >0.6V Ic=(hFE * Re) / (hFE * Re+R1) * (Vin-0.7)/Re
  3. when Ve/Re=Ic reaches 0.65V/Re then Ic remains constant current sink.

If Re=30~33 Ohms or about the same ESR as two 5mm LEDs then current is limited to ~ 20mA.

Then check the ratio of Rb/Re.

  • 10k/33=303 which may exceed the hFE of the transistor such than it never goes into current limiting with 5V . (thats why it says (6V-15V)
    • Thus choose a smaller value such as 2K depending on the Voltage control gain of current you wish. Consider that the voltage gain of a Common emitter with same value impedance on emitter and collector is near unity but voltage control current on LEDs may be 0.3V for full current range so voltage attenuation is need if you want linear current control over most the input voltage range. So a high R1 is useful for a linear controlled brightness with a current sink limit.
  • \$\begingroup\$ It looks like a PWM-able constant current drive. An input signal can turn on/off the current source. \$\endgroup\$ Commented Nov 30, 2018 at 4:46
  • \$\begingroup\$ It's almost constant but not quite. I believe he asked about a linear equation. An ideal CC has infinite source impedance, whereas this does not since the loop gain is not infinite. The Current gain is linear until the feedback voltage gain attenuates the forward hFE to attempt to maintain constant forward base current and thus constant LED current. But the Vin must be >> 1.2V and R1/Re must be less than hFE for tight regulation but it is a tradeoff of shunting input current to output current error. > @EdgarBrown \$\endgroup\$ Commented Nov 30, 2018 at 6:28
  • \$\begingroup\$ @TonyEErocketscientist Without writing down the equations, it seems to me that the loop gain is basically hFE^2. That’s pretty high for such a simple configuration. \$\endgroup\$ Commented Nov 30, 2018 at 6:36
  • \$\begingroup\$ But the Resistor ratio and bias voltage affect that gain.as I said in my answer , \$\endgroup\$ Commented Nov 30, 2018 at 7:00

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