I simulated this circuit from a reference design but I'm not entirely sure how it's working or how you would go about designing such a thing. In simulation it looks like it's designed to hold the current through D1 constant at around 5mA despite having an input voltage range of up to 25V.

I see the gate voltage for M1 is held at about 1.6V, and the base voltage for the BJT rises as the input voltage rises. So as the voltage rises the current through the BJT increases so it's acting like an adjustable impedance there I guess to hold the gate voltage constant. Is that right?

Is this the kind of thing you just do in spice or is some kind of current mirror circuit that's well defined somewhere and I just don't recognize it?

enter image description here


2 Answers 2


This circuit is designed to provide a constant current to the LED independent of the supply voltage.

The MOSFET is turned on by the voltage at the collector of Q1. As soon as the current through R1 (which is the same as through the LED) results in a drop of about 0.6V, Q1 will start to turn on and divert current through R2.

This will then reduce the voltage at M1 gate to control the current through M1 and the LED.

The negative feedback will stabilize the current through D1, M1 and R1 at about 5mA as that will result in 0.6V at Q1 base.

The current will vary slightly as the supply voltage varies but much less than just using a resistor.

Also vary with temperature as the Vbe of the transistor will have ~2.2mV/deg temperature coefficient.

The same circuit can be used where M1 is a BJT (such as 2n2222) rather than a MOSFET. The value of R2 will be more critical as the transistor will require some base current from the R2.

  • 3
    \$\begingroup\$ ah ok so R1 is the control here combined with the 0.6Vbe of Q1 That sets the current to 5mA. \$\endgroup\$
    – confused
    Feb 9, 2017 at 16:47
  • \$\begingroup\$ Yes that's correct - you can change the value of R1 to modify the LED current. \$\endgroup\$ Feb 9, 2017 at 21:15
  • \$\begingroup\$ To be more precise, M1 is a transistor, since the "T" in "MOSFET" stands for "Transistor". It would be better to write "where M1 is a BJT". \$\endgroup\$ Feb 9, 2017 at 21:42
  • \$\begingroup\$ Good point - edited. \$\endgroup\$ Feb 9, 2017 at 21:44

It should be noted that this is not the simplest circuit for a current source. Driving a LED with 5mA current can be done with a single transistor:

enter image description here

Besides being simpler, this schematic has the advantage of current value being dependent on zener voltage (with 2-5% tolerance commonly available) instead of Vbe which can vary as much as 20% from one transistor to the other. There's also an extra diode D for temperature compensation, but it can be left out for devices which don't have high precision requirements or are intended to be used indoors.

The schematic you've found is better suited for high current applications. Because the current trough the load is decided by the Vbe of Q1 and R1, and the current through Q1 is small, you can achieve high load currents without any significant heat (and related parameter drift) in Q1.

For 5mA application however, it's a waste of perfectly good N-MOS.

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    \$\begingroup\$ This circuit will not work well with a 5V supply though as the voltage drop across R2 will be many volts. Zener diodes below 5.6V have very soft knee so are not usually a good choice. To get good temperature compensation the zener should be 5.6V. A 2.5V or 1.2v reference device such as a TL431 could be used instead of the zener. M1 in the original circuit could be replaced with a bipolar transistor with little change to operation. \$\endgroup\$ Feb 9, 2017 at 21:13
  • \$\begingroup\$ @KevinWhite Sure, a reference device is a good idea if the OP can afford one. I didn't consider a 5.6V zener to be a problem since the input voltage range is up to 25V. \$\endgroup\$ Feb 10, 2017 at 13:06

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