Discrete Current Sources

Discrete current sources are constant-current circuits built from just a transistor and a handful of passive components (e.g. no op-amp or other more “advanced” circuitry). They set a fixed current and are commonly used for biasing and simple LED drives. This page covers two popular discrete designs: the Zener diode + BJT source and the JFET constant-current diode.

For the general concept of current sources and sinks (compliance, schematic symbols, independent vs. controlled sources) see the parent Current Sources And Sinks page. Current-mirror-based sources are covered separately on the Current Mirrors page.

Zener Diode And BJT Current Source

Below shows one of the popular methods for constructing a current source from a Zener diode and BJT.

Schematic of a Zener diode + BJT current source.

How It Works

1. The Zener diode $D1$, biased into its voltage regulation range with resistor $R1$, maintains a steady voltage $V_B$ at the base of the BJT transistor.
2. Using the rule that there is a $0.7\unit{V}$ drop across the base-emitter junction of a BJT, this puts a fixed voltage at the emitter also.
3. Since this means there is a fixed voltage across $R_E$, this means by Ohm’s law there is a fixed current through $R_E$.
4. Because most of the current supplied to the emitter of a BJT comes from the collector (only about $1/100$th to $1/800$th of the current comes from the base, depending on the gain $h_{FE}$ of the BJT), this means there is a fixed current through $R_{load}$ also!

The current through the load is given by:

$$\begin{align*} I_{load} = \frac{V_Z - 0.7\unit{V}}{R_E} \end{align*}$$

where:
$V_Z$ is the voltage of the Zener diode in $\mathrm{V}$
$R_E$ is the resistance of resistor $R_E$ in $\mathrm{\Omega}$
$I_{load}$ is the resulting current through the load resistor, in $\mathrm{A}$

Double-Diode And BJT Current Source

This design is similar to the Zener diode and BJT current source above, but uses two diodes in series instead of a Zener diode. This time, rather than sinking current from a load down to ground, we’ll build a high-side current source that pushes current into the load — so we’ll use a PNP transistor above the load. The following schematic shows a current source that will drive approx. $4.7\unit{mA}$ into a load:

Schematic of a double-diode + BJT current source.

This works because there is $1.4\unit{V}$ across $D1$ and $D2$. There is a drop of $0.7\unit{V}$ across the base-emitter junction of the PNP transistor, so there is $0.7\unit{V}$ of drop across $R_E$. The current through $R_E$ is then $I_{R_E} = 0.7\unit{V}/150\unit{\Omega} \approx 4.7\unit{mA}$. Because the base current of the PNP transistor is small compared to the collector current, this means the current through the load will also be approx. $4.7\unit{mA}$.

The exact value of $R1$ does not matter, as long as there is enough current to properly forward bias the diodes but not too much to overheat them.

Two-BJT Current Sink

The two-transistor current sink (also called the two-transistor current limiter) is another popular discrete current source. Unlike the Zener and double-diode designs above, this circuit does not use Zeners or diodes as a voltage reference. Instead, it uses a second BJT to limit the current through the first BJT. It has the advantage of being able to work down to very low voltage and uses two of the same component (the BJTs).

Schematic of a two-BJT current sink.

How It Works

$Q1$ is the pass transistor that carries the load current, and $R1$ feeds base current into it from the top rail. The sense resistor $R_{sense}$ sits in $Q1$‘s emitter, so the entire load current flows through it. The clever part is $Q2$: its base-emitter junction is connected directly across $R_{sense}$, and its collector is tied to $Q1$‘s base.

The loop regulates itself against $Q2$‘s base-emitter voltage:

1. As the load current rises, the voltage across $R_{sense}$ rises with it.
2. When that voltage reaches $Q2$‘s turn-on threshold (≈$0.7\unit{V}$), $Q2$ begins to conduct.
3. $Q2$ pulls $Q1$‘s base voltage down, stealing base drive from $Q1$ and throttling it back.
4. The loop settles at the point where $R_{sense}$ drops almost exactly ≈$0.7\unit{V}$.

The load current is therefore set by the sense resistor and a single base-emitter voltage:

$$\begin{align*} I_{load} = \frac{V_{BE(Q2)}}{R_{sense}} \approx \frac{0.7\unit{V}}{R_{sense}} \end{align*}$$

where:
$V_{BE(Q2)}$ is the base-emitter voltage of $Q2$, in $\mathrm{V}$
$R_{sense}$ is the resistance of the sense resistor, in $\mathrm{\Omega}$
$I_{load}$ is the resulting current through the load, in $\mathrm{A}$

Because the reference is a single, well-defined $V_{BE}$ rather than a loosely-toleranced Zener voltage, the absolute accuracy is reasonable. However, $V_{BE}$ drifts by about $-2\unit{mV/^\circ C}$, so the current has a positive temperature coefficient (roughly $+0.3\unit{\%/^\circ C}$ for a $0.7\unit{V}$ sense drop).

What makes the two-BJT topology notable is that it is likely the most commonly used circuit purely as a current limiter: it normally passes current freely and only intervenes on overload, e.g. a short-circuit protection front-end on power supplies.

The µA723 voltage regulator IC uses a similar circuit to limit the current. The two transistors are highlighted in red. The BJT on the left “steals” base drive from the series pass BJT once a certain voltage is reached, set by a externally provided sense resistor (so it is adjustable).¹

Constant-Current Diode (JFET Current Source)

The constant-current diode (a.k.a JFET current source, current-limiting diode (CLD), or regulating diode²) is a very simple current source made from a JFET and optional resistor.

Below shows the schematic for a constant-current diode:

Schematic for a constant-current diode, a.k.a JFET current source. Current source is very simple, and consists of just an N-channel JFET and (optional) resistor. This two-terminal current source is not referenced to any rail and thus the load can be high-side, low-side, or anywhere in-between!

How It Works

The constant-current diode works by using the resistor $R_S$ to negatively bias the N-channel JFET at a specific operating point. At this operating point the current from drain to source (and consequentially the load) will be relatively constant.³

Tip

Because of the typically large $R_S > 100\unit{\Omega}$ this circuit is only suitable for making low-power current sources up to the $10\unit{mA}-100\unit{mA}$ range. Also, because of the large uncertainty in $V_{GS(off)}$ (e.g. for the 2N5457, the datasheet says it is somewhere between $-0.5\unit{V}$ and $-6.0\unit{V}$!⁴), these don’t make very accurate current sources (without trimming). But they are stable!

How To Calculate The Value Of The Resistor $R_S$

1. Decide on what N-channel JFET you are going to use and your desired constant current, $I_D$.

2. You then calculate the required gate-source voltage⁵:

   $$\begin{align*} V_{GS} = V_{GS(off)} \left[ 1 - \left(\frac{I_D}{I_{DSS}}\right)^{1/k} \right] \end{align*}$$

   where:
   $V_{GS}$ is the gate-source voltage required to bias the JFET at the correct current, in $\mathrm{V}$
   $V_{GS(off)}$ is the gate-source cutoff voltage, a parameter you can get from the JFETs datasheet, in $\mathrm{V}$
   $I_D$ is the current you want the current source to drive at, in $\mathrm{A}$
   $I_{DSS}$ is the zero gate-source drain current, a parameter you can get from the JFETs datasheet, in $\mathrm{A}$
   $k$ is the conduction parameter for the JFET, and depends on the device geometry. This is not normally mentioned in the datasheet, however for this equation you can generally assume it to be $2$.
   

3. You can then find the value of $R_S$ with:

   $$\begin{align*} R_S = \frac{|V_{GS}|}{I_D} \end{align*}$$

Example: 1.0mA Constant-Current Diode

Let’s design a constant-current diode using the ubiquitous 2N5457 JFET (we’ll choose the one from On Semi) for a constant current $I_D$ of $1.0\unit{mA}$.

From the OnSemi 2N5457 datasheet, we get the following values:

$$\begin{align*} I_{DSS} &= 3.0\unit{mA}\ \text{(typ)} \\ V_{GS(off)} &= -6.0\unit{V}\ \text{(max)} \\ \end{align*}$$

Plugging these values into the JFET Vgs equation we can work out the required gate-source voltage (with the common assumption that $k = 2$):

$$\begin{align*} V_{GS} &= V_{GS(off)} \left[ 1 - \left(\frac{I_D}{I_{DSS}}\right)^{1/k} \right] \\ &= -6.0\unit{V} \left[ 1 - \left(\frac{1.0\unit{mA}}{3.0\unit{mA}}\right)^{1/2} \right] \\ &= -2.54\unit{V} \end{align*}$$

Knowing the current $I_D$ through the resistor and the voltage $V_{GS}$ across it, we can find the resistance using $R_S = \frac{|V_{GS}|}{I_D}$:

$$\begin{align*} R_S &= \frac{|V_{GS}|}{I_D} \\ &= \frac{|-2.54\unit{V}|}{1.0\unit{mA}} \\ &= 2.54\unit{k\Omega} \\ \end{align*}$$

Done!

The resistor $R_S$ can be replaced with a potentiometer to create a variable current source, or for tuning/calibration of the current source to a precise value (typically a trimmer-style potentiometer would be used for this).

An important parameter determining the accuracy/stability of a constant-current diode is the output conductance $g_{oss}$ of the JFET. This is the ratio of change in $I_D$ to change in $V_{DS}$. This is a measure of how stable the constant-current source will be to changes in the voltage across it, which occur if either the supply voltage or load resistance changes. The lower the $g_{oss}$ the better (an ideal current source would have $g_{oss} = 0$, which is the equivalent to a resistance of $\infty$).

For more great reading on the constant-current diode, see the Vishay/Siliconix application note AN103.⁵

Depletion-Mode MOSFET Current Source

A depletion-mode MOSFET can be turned into a two-terminal constant-current source in exactly the same way as the JFET constant-current diode above. Unlike the far more common enhancement-mode MOSFET (which needs a gate-source voltage above a threshold before it conducts), a depletion-mode MOSFET is normally on — it conducts at $V_{GS} = 0\unit{V}$. This means it can be self-biased with a single source resistor, just like the JFET.

The attraction over a JFET is voltage and power headroom. Where JFET constant-current diodes top out around $100\unit{V}$ and $100\unit{mA}$, depletion-mode MOSFETs built for this job go much higher. The classic part is the Supertex (now Microchip) DN2540, a $400\unit{V}$, $500\unit{mA}$ device⁶; the LND150 pushes the voltage rating higher still. These are widely used for high-voltage LED strings, vacuum-tube plate loads, and high-voltage startup/bleeder supplies.

Schematic of a depletion-mode MOSFET constant-current source.

How It Works

Self-biasing works identically to the constant-current diode: the source resistor $R_S$ develops a voltage as the drain current flows through it, and because the gate is tied to the bottom of $R_S$ this appears as a negative gate-source voltage. As the current rises, $V_{GS}$ becomes more negative and pinches the channel back, settling at a constant operating current.

How To Calculate The Value Of The Resistor $R_S$

The depletion MOSFET follows the same square-law relationship as the JFET, so the design steps are identical:

1. Choose your depletion-mode MOSFET and desired constant current, $I_D$.

2. Calculate the required gate-source voltage:

   $$\begin{align*} V_{GS} = V_{GS(off)} \left[ 1 - \left(\frac{I_D}{I_{DSS}}\right)^{1/k} \right] \end{align*}$$

   where the parameters mean the same as in the JFET section above ($V_{GS(off)}$ is the gate-source cutoff voltage, $I_{DSS}$ the zero-gate-voltage drain current, and $k \approx 2$).

3. Find $R_S$ with:

   $$\begin{align*} R_S = \frac{|V_{GS}|}{I_D} \end{align*}$$

Example: 10mA Current Source With The DN2540

Let’s set a constant current of $I_D = 10\unit{mA}$ using the DN2540. From the datasheet⁶:

$$\begin{align*} I_{DSS} &= 150\unit{mA}\ \text{(min, at } V_{GS} = 0\unit{V}) \\ V_{GS(off)} &= -1.5\unit{V} \ \text{to} \ -3.5\unit{V} \\ \end{align*}$$

Note that $V_{GS(off)}$ is only specified as a range, not a single value — this is typical for depletion MOSFETs and has a big effect on the result. Solving the square-law for $V_{GS}$ (with $k = 2$):

$$\begin{align*} V_{GS} &= V_{GS(off)} \left[ 1 - \left(\frac{I_D}{I_{DSS}}\right)^{1/2} \right] \\ &= V_{GS(off)} \left[ 1 - \left(\frac{10\unit{mA}}{150\unit{mA}}\right)^{1/2} \right] \\ &= 0.742 \cdot V_{GS(off)} \end{align*}$$

Evaluating $R_S = \frac{|V_{GS}|}{I_D}$ at both ends of the $V_{GS(off)}$ range:

$$\begin{align*} R_S\big|_{V_{GS(off)} = -1.5\unit{V}} &= \frac{0.742 \cdot 1.5\unit{V}}{10\unit{mA}} = 111\unit{\Omega} \\ R_S\big|_{V_{GS(off)} = -3.5\unit{V}} &= \frac{0.742 \cdot 3.5\unit{V}}{10\unit{mA}} = 260\unit{\Omega} \\ \end{align*}$$

So depending on the individual device, $R_S$ needs to be anywhere from about $110\unit{\Omega}$ to $260\unit{\Omega}$ — a greater than 2:1 spread! This is why, for anything needing accuracy, you fix a value near the middle and put a trimmer potentiometer in series, then set the current by measurement. The cutoff voltage also drifts with temperature (about $4.5\unit{mV/^\circ C}$ for the DN2540⁶), so the current is not rock-steady over temperature either.

Comparison Of The Discrete Topologies

The table below summarizes the three discrete current sources covered on this page to help you pick between them.

Property	Zener + BJT	JFET Constant-Current Diode	Depletion-Mode MOSFET
Terminals	3 (referenced to a rail)	2 (floating, not rail-referenced)	2 (floating, not rail-referenced)
Typical current range	mA up to amps (limited by BJT and $R_E$ power)	$10\unit{mA}-100\unit{mA}$	$10\unit{mA}-500\unit{mA}$ (e.g. DN2540)
Max voltage	Set by BJT $V_{CEO}$ (tens of volts typical)	$\sim100\unit{V}$	$\sim400\unit{V}$+ (e.g. DN2540)
Accuracy (untrimmed)	Moderate — set by $V_Z$ and the $0.7\unit{V}$ $V_{BE}$ approximation	Poor — large $V_{GS(off)}$ spread	Poor — large $V_{GS(off)}$ spread ($>$2:1)
Temperature stability	Good if $V_Z$ tempco offsets $V_{BE}$ (≈$-2\unit{mV/^\circ C}$)	Good	Moderate (≈$4.5\unit{mV/^\circ C}$ for DN2540)
Parts count	Higher (Zener, bias resistor, BJT, $R_E$)	Lowest (JFET + optional resistor)	Low (MOSFET + resistor)
Best for	Adjustable higher-current biasing off a rail	Simple low-power floating current limiting	High-voltage / higher-power floating sources

For anything requiring accuracy, all three benefit from a trimmer to set the current by measurement, as discussed in the sections above.

Source Resistor Calculator

The calculator below sizes $R_S$ for either device type. Enter the datasheet $I_{DSS}$, the $V_{GS(off)}$ range, and your target current $I_D$ — it returns the resulting $R_S$ range and a suggested fixed-resistor-plus-trimmer combination. The presets load the datasheet values for the parts discussed above.

Sizes the source resistor R_S for a JFET or depletion-mode MOSFET wired as a two-terminal constant-current source. Because the cutoff voltage V_GS(off) is specified as a datasheet range, the result is a range of R_S. One solution is to pick a fixed value near the low end and trim up with a series potentiometer.

Presets:DN2540 (depletion MOSFET)LND150 (depletion MOSFET)2N5457 (JFET)

I_DSS

A

Zero-gate-voltage drain current (drain current at V_GS = 0), from the datasheet.

V_GS(off) min

V

Gate-source cutoff (pinch-off) voltage, minimum-magnitude end of the datasheet range. Negative for n-channel.

V_GS(off) max

V

Maximum-magnitude end of the cutoff-voltage range. Set both min and max equal for a single-value calculation.

I_D

A

The constant current you want the source to drive.

k

Conduction (square-law) exponent. Almost always 2 — leave as-is unless the datasheet says otherwise.

V_GS

-1.113 V to -2.596 V

Required gate-source voltage at the operating point (V_GS = V_GS(off) [1 − (I_D/I_DSS)^1/k]).

R_S

111.3 Ω to 259.6 Ω

Source resistor: R_S = |V_GS| / I_D.

The spread comes from the V_GS(off) range. To get better accuracy, use a fixed resistor near 111.3 Ω in series with a trimmer of at least 148.4 Ω, then adjust it to hit the target current with an ammeter.

Footnotes

1. Texas Instruments (1999, Jul). µA723 Precision Voltage Regulators [datasheet]. Retrieved 2026-07-18, from https://www.ti.com/lit/ds/symlink/ua723.pdf. ↩

2. Central Semiconductor Corp. What is a Current Limiting Diode?. Retrieved 2022-01-11, from https://www.centralsemi.com/pdfs/selection/leaded/what_is_a_cld.pdf. ↩

3. Electronics Tutorials. FET Current Source. Retrieved 2022-01-11, from https://www.electronics-tutorials.ws/transistor/fet-current-source.html. ↩

4. On Semiconductor (now On Semi) (2010, Feb). 2N5457, 2N5458 JFETs: General Purpose N−Channel − Depletion [datasheet]. Retrieved 2022-01-11, from https://www.onsemi.com/pdf/datasheet/2n5457-d.pdf. ↩

5. Siliconix (now Vishay) (1997, Mar 10). AN103: The FET Constant-Current Source/Limiter [application note]. Retrieved 2021-04-29, from https://www.vishay.com/docs/70596/70596.pdf. ↩ ↩²

6. Supertex (now Microchip). DN2535/DN2540 N-Channel Depletion-Mode Vertical DMOS FETs [datasheet]. Retrieved 2026-07-15, from https://cdn-reichelt.de/documents/datenblatt/A200/DN240N5-G-STEX.pdf. ↩ ↩² ↩³